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Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.1 Questions and Answers.

11th Maths Part 2 Functions Exercise 6.1 Questions And Answers Maharashtra Board

Question 1.
Check if the following relations are functions.
(a)

Solution:
Yes.
Reason: Every element of set A has been assigned a unique element in set B.

(b)

Solution:
No.
Reason: An element of set A has been assigned more than one element from set B.

(c)

Solution:
No.
Reason:
Not every element of set A has been assigned an image from set B.

Question 2.
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}
(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}
(iii) {(1, 3), (4, 1), (2, 2)}
(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}
Solution:
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.
Reason: (2, -1), (2, 2), show that element 2 ∈ A has been assigned two images -1 and 2 from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has been assigned a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason:
3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.

Question 3.
Check if the relation given by the equation represents y as function of x.
(i) 2x + 3y = 12
(ii) x + y² = 9
(iii) x² – y = 25
(iv) 2y + 10 = 0
(v) 3x – 6 = 21
Solution:
(i) 2x + 3y = 12
∴ y = \(\frac{12-2 x}{3}\)
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(ii) x + y² = 9
∴ y² = 9 – x
∴ y = ±\(\sqrt{9-x}\)
∴ For one value of x, there are two values of y.
∴ y is not a function of x.

(iii) x² – y = 25
∴ y = x² – 25
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(iv) 2y + 10 = 0
∴ y = -5
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(v) 3x – 6 = 21
∴ x = 9
∴ x = 9 represents a point on the X-axis.
There is no y involved in the equation.
So the given equation does not represent a function.

Question 4.
If f(m) = m² – 3m + 1, find
(i) f(0)
(ii) f(-3)
(iii) f(\(\frac{1}{2}\))
(iv) f(x + 1)
(v) f(-x)
(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\), h ≠ 0.
Solution:
f(m) = m² – 3m + 1
(i) f(0) = 0² – 3(0) + 1 = 1

(ii) f (-3) = (-3)² – 3(-3) + 1
= 9 + 9 + 1
= 19

(iii) f(\(\frac{1}{2}\)) = \(\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{4}-\frac{3}{2}+1\)
= \(\frac{1-6+4}{4}\)
= \(-\frac{1}{4}\)

(iv) f(x + 1) = (x + 1)² – 3(x + 1) + 1
= x² + 2x + 1 – 3x – 3 + 1
= x² – x – 1

(v) f(-x) = (-x)2 – 3(-x) + 1 = x2 + 3x + 1

(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\)
= \(\frac{(2+h)^{2}-3(2+h)+1-\left(2^{2}-3(2)+1\right)}{h}\)
= \(\frac{\mathrm{h}^{2}+\mathrm{h}}{\mathrm{h}}\)
= h + 1

Question 5.
Find x, if g(x) = 0 where
(i) g(x) = \(\frac{5 x-6}{7}\)
(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
(iii) g(x) = 6x² + x – 2
(iv) g(x) = x³ – 2x² – 5x + 6
Solution:
(i) g(x) = \(\frac{5 x-6}{7}\)
g(x) = 0
∴ \(\frac{5 x-6}{7}\) = 0
∴ x = \(\frac{6}{5}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
g(x) = 0
\(\frac{18-2 x^{2}}{7}\) = 0
∴ 18 – 2x² = 0
∴ x² = 9
∴ x = ±3

(iii) g(x) = 6x² + x – 2
g(x) = 0
∴ 6x² + x – 2 = 0
∴ (2x – 1) (3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

(iv) g(x) = x³ – 2x² – 5x + 6
= ( x- 1) (x² – x – 6)
= (x – 1) (x + 2) (x – 3)
g(x) = 0
∴ (x – 1) (x + 2) (x – 3) = 0
∴ x – 1 = 0 or x + 2 = 0 or x – 3 = 0
∴ x = 1, -2, 3

Question 6.
Find x, if f(x) = g(x) where
(i) f(x) = x⁴ + 2x², g(x) = 11x²
(ii) f(x) = √x – 3, g(x) = 5 – x
Solution:
(i) f(x) = x⁴ + 2x², g(x) = 11x²
f(x) = g(x)
∴ x⁴ + 2x² = 11x²
∴ x⁴ – 9x² = 0
∴ x² (x² – 9) = 0
∴ x² = 0 or x² – 9 = 0
∴ x = 0 or x² = 9
∴ x = 0, ±3

(ii) f(x) = √x – 3, g(x) = 5 – x
f(x) = g(x)
∴ √x – 3 = 5 – x
∴ √x = 5 – x + 3
∴ √x = 8 – x
on squaring, we get
x = 64 + x² – 16x
∴ x² – 17x + 64 = 0
∴ x = \(\frac{17 \pm \sqrt{(-17)^{2}-4(64)}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{289-256}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{33}}{2}\)

Question 7.
If f(x) = \(\frac{a-x}{b-x}\), f(2) is undefined, and f(3) = 5, find a and b.
Solution:
f(x) = \(\frac{a-x}{b-x}\)
Given that,
f(2) is undefined
b – 2 = 0
∴ b = 2 …..(i)
f(3) = 5
∴ \(\frac{a-3}{b-3}\) = 5
∴ \(\frac{a-3}{2-3}\) = 5 ….. [From (i)]
∴ a – 3 = -5
∴ a = -2
∴ a = -2, b = 2

Question 8.
Find the domain and range of the following functions.
(i) f(x) = 7x² + 4x – 1
Solution:
f(x) = 7x² + 4x – 1
f is defined for all x.
∴ Domain of f = R (i.e., the set of real numbers)

∴ Range of f = [\(-\frac{11}{7}\), ∞)

(ii) g(x) = \(\frac{x+4}{x-2}\)
Solution:
g(x) = \(\frac{x+4}{x-2}\)
Function g is defined everywhere except at x = 2.
∴ Domain of g = R – {2}
Let y = g(x) = \(\frac{x+4}{x-2}\)
∴ (x – 2) y = x + 4
∴ x(y – 1) = 4 + 2y
∴ For every y, we can find x, except for y = 1.
∴ y = 1 ∉ range of function g
∴ Range of g = R – {1}

(iii) h(x) = \(\frac{\sqrt{x+5}}{5+x}\)
Solution:
h(x) = \(\frac{\sqrt{x+5}}{5+x}=\frac{1}{\sqrt{x+5}}\), x ≠ -5
For x = -5, function h is not defined.
∴ x + 5 > 0 for function h to be well defined.
∴ x > -5
∴ Domain of h = (-5, ∞)
Let y = \(\frac{1}{\sqrt{x+5}}\)
∴ y > 0
Range of h = (0, ∞) or R+

(iv) f(x) = \(\sqrt[3]{x+1}\)
Solution:
f(x) = \(\sqrt[3]{x+1}\)
f is defined for all real x and the values of f(x) ∈ R
∴ Domain of f = R, Range of f = R

(v) f(x) = \(\sqrt{(x-2)(5-x)}\)
Solution:
f(x) = \(\sqrt{(x-2)(5-x)}\)
For f to be defined,
(x – 2)(5 – x) ≥ 0
∴ (x – 2)(x – 5) ≤ 0
∴ 2 ≤ x ≤ 5 ……[∵ The solution of (x – a) (x – b) ≤ 0 is a ≤ x ≤ b, for a < b]
∴ Domain of f = [2, 5]
(x – 2) (5 – x) = -x² + 7x – 10
= \(-\left(x-\frac{7}{2}\right)^{2}+\frac{49}{4}-10\)
= \(\frac{9}{4}-\left(x-\frac{7}{2}\right)^{2} \leq \frac{9}{4}\)
∴ \(\sqrt{(x-2)(5-x)} \leq \sqrt{\frac{9}{4}} \leq \frac{3}{2}\)
Range of f = [0, \(\frac{3}{2}\)]

(vi) f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
Solution:
f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
For f to be defined,
\(\sqrt{\frac{x-3}{7-x}}\) ≥ 0, 7 – x ≠ 0
∴ \(\sqrt{\frac{x-3}{7-x}}\) ≤ 0 and x ≠ 7
∴ 3 ≤ x < 7
Let a < b, \(\frac{x-a}{x-b}\) ≤ 0 ⇒ a ≤ x < b
∴ Domain of f = [3, 7)
f(x) ≥ 0 … [∵ The value of square root function is non-negative]
∴ Range of f = [0, ∞)

(vii) f(x) = \(\sqrt{16-x^{2}}\)
Solution:
f(x) = \(\sqrt{16-x^{2}}\)
For f to be defined,
16 – x² ≥ 0
∴ x² ≤ 16
∴ -4 ≤ x ≤ 4
∴ Domain of f = [-4, 4]
Clearly, f(x) ≥ 0 and the value of f(x) would be maximum when the quantity subtracted from 16 is minimum i.e. x = 0
∴ Maximum value of f(x) = √16 = 4
∴ Range of f = [0, 4]

Question 9.
Express the area A of a square as a function of its
(a) side s
(b) perimeter P
Solution:
(a) area (A) = s²
(b) perimeter (P) = 4s
∴ s = \(\frac{\mathrm{P}}{4}\)
Area (A) = s² = \(\left(\frac{\mathrm{P}}{4}\right)^{2}\)
∴ A = \(\frac{\mathrm{P}^{2}}{16}\)

Question 10.
Express the area A of a circle as a function of its
(i) radius r
(ii) diameter d
(iii) circumference C
Solution:
(i) Area (A) = πr²

(ii) Diameter (d) = 2r
∴ r = \(\frac{\mathrm{d}}{2}\)
∴ Area (A) = πr² = \(\frac{\pi \mathrm{d}^{2}}{4}\)

(iii) Circumference (C) = 2πr
∴ r = \(\frac{C}{2 \pi}\)
Area (A) = πr² = \(\pi\left(\frac{\mathrm{C}}{2 \pi}\right)^{2}\)
∴ A = \(\frac{C^{2}}{4 \pi}\)

Question 11.
An open box is made from a square of cardboard of 30 cms side, by cutting squares of length x centimeters from each corner and folding the sides up. Express the volume of the box as a function of x. Also, find its domain.
Solution:

Length of the box = 30 – 2x
Breadth of the box = 30 – 2x
Height of the box = x
Volume = (30 – 2x)² x, x < 15, x ≠ 15, x > 0
= 4x(15 – x)², x ≠ 15, x > 0
Domain (0, 15)

Question 12.
Let f be a subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify?
Solution:
f = {(ab, a + b): a, b ∈ Z}
Let a = 1, b = 1. Then, ab = 1, a + b = 2
∴ (1, 2) ∈ f
Let a = -1, b = -1. Then, ab = 1, a + b = -2
∴ (1, -2) ∈ f
Since (1, 2) ∈ f and (1, -2) ∈ f,
f is not a function as element 1 does not have a unique image.

Question 13.
Check the injectivity and surjectivity of the following functions.
(i) f : N → N given by f(x) = x²
Solution:
f: N → N given by f(x) = x²

∴ f is injective.
For every y = x² ∈ N, there does not exist x ∈ N.
Example: 7 ∈ N (codomain) for which there is no x in domain N such that x² = 7
∴ f is not surjective.

(ii) f : Z → Z given by f(x) = x²
Solution:
f: Z → Z given by f(x) = x2

∴ f is not injective.
(Example: f(-2) = 4 = f(2). So, -2, 2 have the same image. So, f is not injective.)
Since x² ≥ 0,
f(x) ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iii) f : R → R given by f(x) = x²
Solution:
f : R → R given by f(x) = x²

∴ f is not injective.
f(x) = x² ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iv) f : N → N given by f(x) = x³
Solution:
f: N → N given by f(x) = x³

∴ f is injective.
Numbers from codomain which are not cubes of natural numbers are not images under f.
∴ f is not surjective.

(v) f : R → R given by f(x) = x³
Solution:
f: R → R given by f(x) = x³

∴ For every y ∈ R, there is some x ∈ R.
∴ f is surjective.

Question 14.
Show that if f : A → B and g : B → C are one-one, then gof is also one-one.
Solution:
f is a one-one function.
Let f(x₁) = f(x₂)
Then, x₁ = x₂ for all x₁, x₂ …..(i)
g is a one-one function.
Let g(y₁) = g(y₂)
Then, y₁ = y₂ for all y₁, y₂ …..(ii)
Let (gof) (x₁) = (gof) (x₂)
∴ g(f(x₁)) = g(f(x₂))
∴ g(y₁) = g(y₂),
where y₁ = f(x₁), y₂ = f(x₂) ∈ B
∴ y₁ = y₂ …..[From (ii)]
i.e., f(x₁) = f(x₂)
∴ x₁ = x₂ ….[From (i)]
∴ gof is one-one.

Question 15.
Show that if f : A → B and g : B → C are onto, then gof is also onto.
Solution:
Since g is surjective (onto),
there exists y ∈ B for every z ∈ C such that
g(y) = z …….(i)
Since f is surjective,
there exists x ∈ A for every y ∈ B such that
f(x) = y …….(ii)
(gof) x = g(f(x))
= g(y) ……[From (ii)]
= z …..[From(i)]
i.e., for every z ∈ C, there is x in A such that (gof) x = z
∴ gof is surjective (onto).

Question 16.
If f(x) = 3(4^x+1), find f(-3).
Solution:
f(x) = 3(4^x+1)
∴ f(-3) = 3(4^-3+1)
= 3(4^-2)
= \(\frac{3}{16}\)

Question 17.
Express the following exponential equations in logarithmic form:
(i) 2⁵ = 32
(ii) 54⁰ = 1
(iii) 23¹ = 23
(iv) \(9^{\frac{3}{2}}\) = 27
(v) 3^-4 = \(\frac{1}{81}\)
(vi) 10^-2 = 0.01
(vii) e² = 7.3890
(viii) \(e^{\frac{1}{2}}\) = 1.6487
(ix) e^-x = 6
Solution:

Question 18.
Express the following logarithmic equations in exponential form:
(i) log₂ 64 = 6
(ii) \(\log _{5} \frac{1}{25}\) = -2
(iii) log₁₀ 0.001 = -3
(iv) \(\log _{\frac{1}{2}}\)(8) = -3
(v) ln 1 = 0
(vi) ln e = 1
(vii) ln \(\frac{1}{2}\) = -0.693
Solution:
(i) log₂ 64 = 6
∴ 64 = 2⁶, i.e., 2⁶ = 64

Question 19.
Find the domain of
(i) f(x) = ln (x – 5)
(ii) f(x) = log10 (x² – 5x + 6)
Solution:
(i) f(x) = ln (x – 5)
f is defined, when x – 5 > 0
∴ x > 5
∴ Domain of f = (5, ∞)

(ii) f(x) = log₁₀ (x² – 5x + 6)
x² – 5x + 6 = (x – 2) (x – 3)
f is defined, when (x – 2) (x – 3) > 0
∴ x < 2 or x > 3
Solution of (x – a) (x – b) > 0 is x < a or x > b where a < b
∴ Domain of f = (-∞, 2) ∪ (3, ∞)

Question 20.
Write the following expressions as sum or difference of logarithms:
(a) \(\log \left(\frac{p q}{r s}\right)\)
Solution:

(b) \(\log (\sqrt{x} \sqrt[3]{y})\)
Solution:

(c) \(\ln \left(\frac{a^{3}(a-2)^{2}}{\sqrt{b^{2}+5}}\right)\)
Solution:

(d) \(\ln \left[\frac{\sqrt[3]{x-2}(2 x+1)^{4}}{(x+4) \sqrt{2 x+4}}\right]^{2}\)
Solution:

Question 21.
Write the following expressions as a single logarithm.
(i) 5 log x + 7 log y – log z
Solution:

(ii) \(\frac{1}{3}\) log(x – 1) + \(\frac{1}{2}\) log(x)
Solution:

(iii) ln (x + 2) + ln (x – 2) – 3 ln (x + 5)
Solution:

Question 22.
Given that log 2 = a and log 3 = b, write log √96 terms of a and b.
Solution:
log 2 = a and log 3 = b
log √96 = \(\frac{1}{2}\) log (96)
= \(\frac{1}{2}\) log (2⁵ x 3)
= \(\frac{1}{2}\) (log 2⁵ + log 3) …..[∵ log mn = log m + log n]
= \(\frac{1}{2}\) (5 log 2 + log 3) ……[∵ log mⁿ = n log m]
= \(\frac{5 a+b}{2}\)

Question 23.
Prove that:
(a) \(b^{\log _{b} a}=a\)
Solution:
We have to prove that \(b^{\log _{b} a}=a\)
i.e., to prove that (logb a) (log_b b) = log_b a
(Taking log on both sides with base b)
L.H.S. = (log_b a) (log_b b)
= log_b a …..[∵ log_b b = 1]
= R.H.S.

(b) \(\log _{b^{m}} a=\frac{1}{m} \log _{b} a\)
Solution:

(c) \(a^{\log _{c} b}=b^{\log _{c} a}\)
Solution:

Question 24.
If f(x) = ax² – bx + 6 and f(2) = 3 and f(4) = 30, find a and b.
Solulion:
f(x) = ax² – bx + 6
f(2) = 3
∴ a(2)² – b(2) + 6 = 3
∴ 4a – 2b + 6 = 3
∴ 4a – 2b + 3 = 0 …..(i)
f(4) = 30
∴ a(4)² – b(4) + 6 = 30
∴ 16a – 4b + 6 = 30
∴ 16a – 4b – 24 = 0 …..(ii)
By (ii) – 2 × (i), we get
8a – 30 = 0
∴ a = \(\frac{30}{8}=\frac{15}{4}\)
Substiting a = \(\frac{15}{4}\) in (i), we get
4(\(\frac{15}{4}\)) – 2b + 3 = 0
∴ 2b = 18
∴ b = 9
∴ a = \(\frac{15}{4}\), b = 9

Question 25.
Solve for x:
(i) log 2 + log (x + 3) – log (3x – 5) = log 3
Solution:
log 2 + log (x + 3) – log (3x – 5) = log 3
∴ log 2(x + 3) – log(3x – 5) = log 3 …..[∵ log m + log n = log mn]
∴ log \(\frac{2(x+3)}{3 x-5}\) = log 3 …..[∵ log m – log n = log \(\frac{m}{n}\)]
∴ \(\frac{2(x+3)}{3 x-5}\) = 3
∴ 2x + 6 = 9x – 15
∴ 7x = 21
∴ x = 3

Check:
If x = 3 satisfies the given condition, then our answer is correct.
L.H.S. = log 2 + log (x + 3) – log (3x – 5)
= log 2 + log (3 + 3) – log (9 – 5)
= log 2 + log 6 – log 4
= log (2 × 6) – log 4
= log \(\frac{12}{4}\)
= log 3
= R.H.S.
Thus, our answer is correct.

(ii) 2log₁₀ x = 1 + \(\log _{10}\left(x+\frac{11}{10}\right)\)
Solution:


∴ x² = 10x + 11
∴ x² – 10x – 11 = 0
∴ (x – 11)(x + 1) = 0
∴ x = 11 or x = -1
But log of a negative numbers does not exist
∴ x ≠ -1
∴ x = 11

(iii) log₂ x + log₄ x + log₁₆ x = \(\frac{21}{4}\)
Solution:

(iv) x + log₁₀ (1 + 2^x) = x log₁₀ 5 + log₁₀ 6
Solution:

∴ a + a² = 6
∴ a² + a – 6 = 0
∴ (a + 3)(a – 2) = 0
∴ a + 3 = 0 or a – 2 = 0
∴ a = -3 or a = 2
Since 2x = -3 is not possible,
2x = 2 = 21
∴ x = 1

Question 26.
If log \(\left(\frac{x+y}{3}\right)\) = \(\frac{1}{2}\) log x + \(\frac{1}{2}\) log y, show that \(\frac{x}{y}+\frac{y}{x}\) = 7.
Solution:

Question 27.
If log\(\left(\frac{x-y}{4}\right)\) = log√x + log√y, show that (x + y)² = 20xy.
Solution:

Question 28.
If x = log_abc, y = log_b ca, z = log_c ab, then prove that \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) = 1.
Solution:

11th Maths Practical Handbook Pdf 

• Functions Ex 6.1 Class 11 Maths Solutions
• Functions Ex 6.2 Class 11 Maths Solutions
• Functions Miscellaneous Exercise 6 Class 11 Maths Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Miscellaneous Exercise 5 Questions and Answers.

11th Maths Part 2 Sets and Relations Miscellaneous Exercise 5 Questions And Answers Maharashtra Board

(I) Select the correct answer from the given alternative.

Question 1.
For the set A = {a, b, c, d, e} the correct statement is
(A) {a, b} ∈ A
(B) {a} ∈ A
(C) a ∈ A
(D) a ∉ A
Answer:
(C) a ∈ A

Question 2.
If aN = {ax : x ∈ N}, then set 6N ∩ 8N =
(A) 8N
(B) 48N
(C) 12N
(D) 24N
Answer:
(D) 24N
Hint:
6N = {6x : x ∈ N} = {6, 12, 18, 24, 30, ……}
8N = {8x : x ∈ N} = {8, 16, 24, 32, ……}
∴ 6N ∩ 8N = {24, 48, 72, …..}
= {24x : x ∈ N}
= 24N

Question 3.
If set A is empty set then n[P[P[P(A)]]] is
(A) 6
(B) 16
(C) 2
(D) 4
Answer:
(D) 4
Hint:
A = Φ
∴ n(A) = 0
∴ n[P(A)] = 2^n(A) = 2⁰ = 1
∴ n[P[P(A)]] = 2^n[P(A)] = 2¹ = 2
∴ n[P[P[P(A)]]] = 2^n[P[P(A)]] = 2² = 4

Question 4.
In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus are
(A) 80%
(B) 40%
(C) 60%
(D) 70%
Answer:
(C) 60%
Hint:
Let C = Population travels by car
B = Population travels by bus
n(C) = 20%, n(B) = 50%, n(C ∩ B) = 10%
n(C ∪ B) = n(C) + n(B) – n(C ∩ B)
= 20% + 50% – 10%
= 60%

Question 5.
If the two sets A and B are having 43 elements in common, then the number of elements common to each of the sets A × B and B × A is
(A) 43²
(B) 2⁴³
(C) 43⁴³
(D) 2⁸⁶
Answer:
(A) 43²

Question 6.
Let R be a relation on the set N be defied by {(x, y) / x, y ∈ N, 2x + y = 41} Then R is
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) None of these
Answer:
(D) None of these

Question 7.
The relation “>” in the set of N (Natural number) is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalent relation
Answer:
(C) Transitive
Hint:
For any a ∈ N, a ≯ a
∴ (a, a) ∉ R
∴ > is not reflexive.
For any a, b ∈ N, if a > b, then b ≯ a.
∴ > is not symmetric.
For any a, b, c ∈ N,
if a > b and b > c, then a > c
∴ > is transitive.

Question 8.
A relation between A and B is
(A) only A × B
(B) An Universal set of A × B
(C) An equivalent set of A × B
(D) A subset of A × B
Answer:
(D) A subset of A × B

Question 9.
If (x, y) ∈ N × N, then xy = x² is a relation that is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalence
Answer:
(D) Equivalence
Hint:
Let x ∈ R, then xx = x²
∴ x is related to x.
∴ Given relation is reflexive.
Letx = 0 and y = 2,
then xy = 0 × 2 = 0 = x²
∴ x is related to y.
Consider, yx = 2 × 0 = 0 ≠ y²
∴ y is not related to x.
∴ Given relation is not symmetric.
Let x be related to y and y be related to z.
∴ xy = x² and yz = y²
∴ x = \(\frac{x^{2}}{y}\) and z = \(\frac{y^{2}}{y}\) = y …..[if y ≠ 0]
Consider, xz = \(\frac{x^{2}}{y}\) × y = x²
∴ x is related to z.
∴ Given relation is transitive.

Question 10.
If A = {a, b, c}, The total no. of distinct relations in A × A is
(A) 3
(B) 9
(C) 8
(D) 29
Answer:
(D) 29

(II) Answer the following.

Question 1.
Write down the following sets in set builder form:
(i) {10, 20, 30, 40, 50}
(ii) {a, e, i, o, u}
(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Solution:
(i) Let A = {10, 20, 30, 40, 50}
∴ A = {x/x = 10n, n ∈ N and n ≤ 5}

(ii) Let B = {a, e, i, o, u}
∴ B = {x/x is a vowel of English alphabets}

(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
∴ C = {x/x is a day of a week}

Question 2.
If U = {x/x ∈ N, 1 ≤ x ≤ 12}, A = {1,4, 7,10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}. Write down the sets.
(i) A ∪ B
(ii) B ∩ C
(iii) A – B
(iv) B ∩ C’
(v) A ∪ B ∪ C
(vi) A ∩ (B ∪ C)
Solution:
U = {x/x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}
(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

(ii) B ∩ C = {}

(iii) A – B = {1, 10}

(iv) C’ = {1, 2, 4, 6, 7, 10, 11}
∴ B ∩ C’ = {2, 4, 6, 7, 11}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(vi) B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}
∴ A ∩ (B ∪ C) = {4, 7}

Question 3.
In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?
Solution:
Let A = set of students who drink apple juice
B = set of students who drink orange juice
X = set of all students
∴ n(X) = 425, n(A) = 115, n(B) = 160, n(A ∩ B) = 80

No. of students who neither drink apple juice nor orange juice = n(A’ ∩ B’) = n(A ∪ B)’
= n(X) – n(A ∪ B)
= 425 – [n(A) + n(B) – n(A ∩ B)]
= 425 – (115 + 160 – 80)
= 230

Question 4.
In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?
Solution:
Let A = set of teachers who teach Mathematics
B = set of teachers who teach Physics
∴ n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4

Since n(A ∪ B) = n(A) + n(B) – n(A ∩ B),
20 = 12 + n(B) – 4
∴ n(B) = 12
∴ Number of teachers who teach physics = 12

Question 5.
(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A × B) ∩ (B × A).
(ii) If A = {-1, 1}, find A × A × A.
Solution:
(i) A = {1, 2, 3} and B = {2, 4}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
∴ (A × B) ∩ (B × A) = {(2, 2)}

(ii) A = {-1, 1}
∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

Question 6.
If A = {1, 2, 3}, B = {4, 5, 6}, check if the following are relations from A to B. Also, write its domain and range.
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Solution:
A = {1, 2, 3}, B = {4, 5, 6}
∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
Since R₁ ⊆ A × B,
R₁ is a relation from A to B.
Domain (R₁) = Set of first components of R₁ = {1}
Range (R₁) = Set of second components of R₁ = {4, 5, 6}

(ii) R₂ = {(1, 5),(2, 4),(3, 6)}
Since R₂ ⊆ A × B,
R₂ is a relation from A to B.
Domain (R₂) = Set of first components of R₂ = {1, 2, 3}
Range (R₂) = Set of second components of R₂ = {4, 5, 6}

(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since R₃ ⊆ A × B,
R₃ is a relation from A to B.
Domain (R₃) = Set of first components of R₃ = {1, 2, 3}
Range (R₃) = Set of second components of R₃ = {4, 5, 6}

(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Since (4, 2) ∈ R₄, but (4, 2) ∉ A × B,
R₄ ⊄ A × B
∴ R₄ is not a relation from A to B.

Question 7.
Determine the domain and range of the following relations.
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Solution:
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}
Range (R) = {b / b = 4} = {4}

(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Since a ∈ Z and |a| < 3,
a < 3 and a > -3
∴ -3 < a < 3
∴ a = -2, -1, 0, 1, 2
b = |a – 1|
When a = -2, b = 3
When a = -1, b = 2
When a = 0, b = 1
When a = 1, b = 0
When a = 2, b = 1
Domain (R) = {-2, -1, 0, 1, 2}
Range (R) = {0, 1, 2, 3}

Question 8.
Find R : A → A when A = {1, 2, 3, 4} such that
(i) R = {(a, b) / a – b = 10}
(ii) R = {(a, b) / |a – b| ≥ 0}
Solution:
R : A → A, A = {1, 2, 3,4}
(i) R = {(a, b)/a – b = 10} = { }

(ii) R = {(a, b) / |a – b| ≥ 0}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ R = A × A

Question 9.
R : {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Check if R is
(i) reflexive
(ii) symmetric
(iii) transitive
Solution:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
(i) Here, (x, x) ∈ R, for x ∈ {1, 2, 3}
∴ R is reflexive.

(ii) Here, (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.

(iii) Here, (1, 2), (2, 3) ∈ R,
But (1, 3) ∉ R.
∴ R is not transitive.

Question 10.
Check if R : Z → Z, R = {(a, b) | 2 divides a – b} is an equivalence relation.
Solution:
(i) Since 2 divides a – a,
(a, a) ∈ R
∴ R is reflexive. .

(ii) Let (a, b) ∈ R
Then 2 divides a – b
∴ 2 divides b – a
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b) ∈ R, (b, c) ∈ R
Then a – b = 2m, b – c = 2n,
∴ a – c = 2(m + n), where m, n are integers.
∴ 2 divides a – c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Question 11.
Show that the relation R in the set A = {1, 2, 3, 4, 5} Given by R = {(a, b) / |a – b| is even} is an equivalence relation.
Solution:
(i) Since |a – a| is even,
∴ (a, a) ∈ R
∴ R is reflexive.

(ii) Let (a, b) ∈ R
Then |a – b| is even
∴ |b – a| is even
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b), (b, c) ∈ R
Then a – b = ±2m, b – c = ±2n
∴ a – c = ±2(m + n), where m, n are integers.
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

Question 12.
Show that the following are equivalence relations:
(i) R in A is set of all books given by R = {(x, y) / x and y have same number of pages}
(ii) R in A = {x ∈ Z | 0 ≤ x ≤ 12} given by R = {(a, b) / |a – b| is a multiple of 4}
(iii) R in A = (x ∈ N/x ≤ 10} given by R = {(a, b) | a = b}
Solution:
(i) a. Clearly (x, x) ∈ R
∴ R is reflexive.

b. If (x, y) ∈ R then (y, x) ∈ R.
∴ R is symmetric.

c. Let (x, y) ∈ R, (y, x) ∈ R.
Then x, y, and z are 3 books having the same number of pages.
∴ (x, z) ∈ R as x, z has the same number of pages.
∴ R is transitive.
Thus, R is an equivalence relation.

(ii) a. Since |a – a| is a multiple of 4,
(a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R
Then a – b = ±4m,
∴ b – a = ±4m, where m is an integer
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
a – b = ± 4m, b – c = ± 4n,
∴ a – c = ±4(m + n), where m, n are integers
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

(iii) a. Since a = a
∴ (a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R Then a = b
∴ b = a
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
Then, a = b, b = c
∴ a = c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

State Board 11th Maths Book Answers 

• Sets and Relations Ex 5.1 Class 11 Maths Solutions
• Sets and Relations Ex 5.2 Class 11 Maths Solutions
• Sets and Relations Miscellaneous Exercise 5 Class 11 Maths Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.2 Questions and Answers.

11th Maths Part 2 Sets and Relations Exercise 5.2 Questions And Answers Maharashtra Board

Question 1.
If (x – 1, y + 4) = (1, 2), find the values of x and y.
Solution:
(x – 1, y + 4) = (1, 2)
By the definition of equality of ordered pairs, we have
x – 1 = 1 and y + 4 = 2
∴ x = 2 and y = -2

Question 2.
If \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\), find x and y.
Solution:
\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\)
By the definition of equality of ordered pairs, we have
x + \(\frac{1}{3}\) = \(\frac{1}{2}\) and \(\frac{y}{3}\) – 1 = \(\frac{3}{2}\)
∴ x = \(\frac{1}{2}\) – \(\frac{1}{3}\) and \(\frac{y}{3}\) = \(\frac{3}{2}\) + 1
∴ x = \(\frac{1}{6}\) and y = \(\frac{15}{2}\)

Question 3.
If A = {a, b, c}, B = {x, y}, find A × B, B × A, A × A, B × B.
Solution:
A = (a, b, c}, B = {x, y}
A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}
B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
B × B = {(x, x), (x, y), (y, x), (y, y)}

Question 4.
If P = {1, 2, 3} and Q = {1, 4}, find sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {1, 4}
∴ P × Q = {(1, 1), (1, 4), (2, 1), (2, 4), (3, 1), (3, 4)}
and Q × P = {(1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3)}

Question 5.
Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}. Verify,
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution:
A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
(i) B ∩ C = {5, 6}
A × (B ∩ C) = = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) B ∪ C = {4, 5, 6}
A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
∴ A × (B ∪ C) = (A × B) ∪ (A × C)

Question 6.
Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.
Solution:
{(x, y) / x² + y² = 100, where x, y ∈ W}
We have, x² + y² = 100
When x = 0 and y = 10,
x² + y² = 0² + 10² = 100
When x = 6 and y = 8,
x² + y² = 6² + 8² = 100
When x = 8 and y = 6,
x² + y² = 8² + 6² = 100
When x = 10 and y = 0,
x² + y² = 10² + 0² = 100
∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

Question 7.
Let A = {6, 8} and B = {1, 3, 5}. Show that R₁ = {(a, b) / a ∈ A, b ∈ B, a – b is an even number} is a null relation, R₂ = {(a, b) / a ∈ A, b ∈ B, a + b is an odd number} is a universal relation.
Solution:
A = {6, 8}, B = {1, 3, 5}
R₁ = {(a, b)/ a ∈ A, b ∈ B, a – b is an even number}
a ∈ A
∴ a = 6, 8
b ∈ B
∴ b = 1, 3, 5
When a = 6 and b = 1, a – b = 5, which is odd
When a = 6 and b = 3, a – b = 3, which is odd
When a = 6 and b = 5, a – b = 1, which is odd
When a = 8 and b = 1, a – b = 7, which is odd
When a = 8 and b = 3, a – b = 5, which is odd
When a = 8 and b = 5, a – b = 3, which is odd
Thus, no set of values of a and b gives a – b as even.
∴ R₁ has a null relation from A to B.
A × B = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
When a = 6 and b = 1, a + b = 7, which is odd
When a = 6 and b = 3, a + b = 9, which is odd
When a = 6 and b = 5, a + b = 11, which is odd
When a = 8 and b = 1, a + b = 9, which is odd
When a = 8 and b = 3, a + b = 11, which is odd
When a = 8 and b = 5, a + b = 13, which is odd
∴ R₂ = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
Here, R₂ = A × B
∴ R₂ has a universal relation from A to B.

Question 8.
Write the relation in the Roster form. State its domain and range.
(i) R₁ = {(a, a²) / a is a prime number less than 15}
(ii) R₂ = {(a, \(\frac{1}{a}\)) / 0 < a ≤ 5, a ∈ N}
(iii) R₃ = {(x, y / y = 3x, y ∈ {3, 6, 9, 12}, x ∈ {1, 2, 3}}
(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
(v) R₅ = {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
(vi) R₆ = {(a, b) / a ∈ N, a < 6 and b = 4}
(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
(viii) R₈ = {(a, b)/ b = a + 2, a ∈ Z, 0 < a < 5}
Solution:
(i) R₁ = {(a, a²) / a is a prime number less than 15}
∴ a = 2, 3, 5, 7, 11, 13
∴ a² = 4, 9, 25, 49, 121, 169
∴ R₁ = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}
∴ Domain (R₁) = {a/a is a prime number less than 15}
= {2, 3, 5, 7, 11, 13}
Range (R₁) = {a²/a is a prime number less than 15}
= {4, 9, 25, 49, 121, 169}

(iii) R₃ = {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
Here y = 3x
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴ R₃ = {(1, 3), (2, 6), (3, 9)}
∴ Domain (R₃) ={1, 2, 3}
∴ Range (R₃) = {3, 6, 9}

(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
Here, y > x + 1
When x = 1 and y = 2, 2 ≯  1 + 1
When x = 1 and y = 4, 4 > 1 + 1
When x = 1 and y = 6, 6 > 1 + 1
When x = 2 and y = 2, 2 ≯  2 + 1
When x = 2 and y = 4, 4 > 2 + 1
When x = 2 and y = 6, 6 > 2 + 1
∴ R₄ = {(1, 4), (1, 6), (2, 4), (2, 6)}
Domain (R₄) = {1, 2}
Range (R₄) = {4, 6}

(v) R₅ = {{x, y) / x + y = 3, x, y ∈ (0, 1, 2, 3)}
Here, x + y = 3
When x = 0, y = 3
When x = 1, y = 2
When x = 2, y = 1
When x = 3, y = 0
∴ R₅ = {(0, 3), (1, 2), (2, 1), (3, 0)}
Domain (R₅) = {0, 1, 2, 3}
Range (R₅) = {3, 2, 1, 0}

(vi) R₆ = {(a, b)/ a ∈ N, a < 6 and b = 4}
a ∈ N and a < 6
∴ a = 1, 2, 3, 4, 5 and b = 4
R₆ = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}
Domain (R₆) = {1, 2, 3, 4, 5}
Range (R₆) = {4}

(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
Here, a + b = 6
When a = 1, b = 5
When a = 2, b = 4
When a = 3, b = 3
When a = 4, b = 2
When a = 5, b = 1
∴ R₇ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Domain (R₇) = {1, 2, 3, 4, 5}
Range (R₇) = {5, 4, 3, 2, 1}

(viii) R₈ = {(a, b) / b = a + 2, a ∈ Z, 0 < a < 5}
Here, b = a + 2
When a = 1, b = 3
When a = 2, b = 4
When a = 3, b = 5
When a = 4, b = 6
∴ R₈ = {(1, 3), (2, 4), (3, 5), (4, 6)}
Domain (R₈) = {1, 2, 3, 4}
Range (R₈) = {3, 4, 5, 6}

Question 9.
Identify which of the following relations are reflexive, symmetric, and transitive.

Solution:

(i) Given, R = {(a, b): a, b ∈ Z, a – b is an integer}
Let a ∈ Z, then a – a ∈ Z
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a – b ∈ Z
∴ -(a – b) ∈ Z, i.e., b – a ∈ Z
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a – b ∈ Z and b – c ∈ Z
∴ (a – b) + (b – c) ∈ Z
∴ a – c ∈ Z
∴ (a, c) ∈ R
∴ R is transitive.

(ii) Given, R = {(a, b) : a, b ∈ N, a + b is even}
Let a ∈ N, then a + a = 2a, which is even.
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a + b is even
∴ b + a is even
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a + b and b + c is even
Let a + b = 2x and b + c = 2y for x, y ∈ N
∴ (a + b) + (b + c) = 2x + 2y
∴ a + 2b + c = 2(x + y)
∴ a + c = 2(x + y) – 2b = 2(x + y – b)
∴ a + c is even ……..[∵ x, y, b ∈ N, x + y – b ∈ N]
∴ (a, c) ∈ R
∴ R is transitive.

(iii) Given, R = {(a, b) : a, b ∈ N, a divides b}
Let a ∈ N, then a divides a.
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 2 and b = 8, then 2 divides 8
∴ (a, b) ∈ R
But 8 does not divide 2.
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a divides b and b divides c.
Let b = ax and c = by for x, y ∈ N.
∴ c = (ax) y = a(xy)
i.e., a divides c.
∴ (a, c) ∈ R
∴ R is transitive.

(iv) Given, R = {(a, b) : a, b ∈ N, a² – 4ab + 3b² = 0}
Let a ∈ N, then a² – 4aa + 3a² = a² – 4a² + 3a² = 0
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 3 and b = 1,
then a² – 4ab + 3b² = 9 – 12 + 3 = 0
∴ (a, b) ∈ R
Consider, b² – 4ba + 3a² = 1 – 12 + 9 = -2 ≠ 0
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 3, b = 1 and c = \(\frac{1}{3}\),
then a² – 4ab + 3b² = 9 – 12 + 3 = 0 and
b² – 4bc + 3c² = 1 – \(\frac{4}{3}\) + \(\frac{1}{3}\) = 1 – 1 = 0
∴ we get (a, b) and (b, c) ∈ R.
Consider, a² – 4ac + 3c² = 9 – 4 + \(\frac{1}{3}\) = \(\frac{16}{3}\) ≠ 0
∴ (a, c) ∉ R
∴ R is not transitive.

(v) Given, R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}
Let a ∈ G, then ‘a’ cannot be a sister of herself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ ‘a’ is a sister of ‘b’.
∴ ‘b’ is a sister of ‘a’.
∴ (b, c) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ ‘a’ is a sister of ‘b’ and ‘b’ is a sister of ‘c’
∴ ‘a’ is a sister of ‘c’.
∴ (a, c) ∈ R
∴ R is transitive.

(vi) Given, R = {(a, b) : Line a is perpendicular to line b in a plane}
Let a be any line in the plane, then a cannot be perpendicular to itself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ a is perpendicular to b.
∴ b is perpendicular to a.
∴ (b, a) ∈ R.
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R.
∴ a is perpendicular to b and b is perpendicular to c.
∴ a is parallel to c.
∴ (a, c) ∉ R
∴ R is not transitive.

(vii) Given, R = {(a, b) : a, b ∈ R, a < b}
Let a ∈ R, then a ≮ a.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 1 and b = 2, then 1 < 2
∴ (a, b) ∈ R
But 2 ≮ 1
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a < b and b < c
∴ a < c
∴ (a, c) ∈ R
∴ R is transitive.

(viii) Given, R = {(a, b) : a, b ∈ R, a ≤ b³}
Let a = -3, then a³ = -27.
Here, a ≮ a
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 2 and b = 9, then b³ = 729
Here, a < b³
∴ (a, b) ∈ R
Consider, a3 = 8
Here, b ≮ a³
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 10, b = 3, c = 2,
then b³ = 27 and c³ = 8
Here, a < b³ and b < c³.
∴ (a, b) and (b, c) ∈ R
But a ≮ c³
∴ (a, c) ∉ R.
∴ R is not transitive.

State Board 11th Maths Book Answers 

• Sets and Relations Ex 5.1 Class 11 Maths Solutions
• Sets and Relations Ex 5.2 Class 11 Maths Solutions
• Sets and Relations Miscellaneous Exercise 5 Class 11 Maths Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.1 Questions and Answers.

11th Maths Part 2 Sets and Relations Exercise 5.1 Questions And Answers Maharashtra Board

Question 1.
Describe the following sets in Roster form:
(i) A = {x/x is a letter of the word ‘MOVEMENT’}
(ii) B = {x/x is an integer, –\(\frac{3}{2}\) < x < \(\frac{9}{2}\)>
(iii) C = {x/x = 2n + 1, n ∈ N}
Solution:
(i) A = {M, O, V, E, N, T}
(ii) B = {-1, 0, 1, 2, 3, 4}
(iii) C = {3, 5, 7, 9, … }

Question 2.
Describe the following sets in Set-Builder form:
(i) {0}
(ii) {0, ±1, ±2, ±3}
(iii) \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
(iv) {0, -1, 2, -3, 4, -5, 6,…}
Solution:
(i) Let A = {0}
0 is a whole number but it is not a natural number.
∴ A = {x / x ∈ W, x ∉ N}

(ii) Let B = {0, ±1, ±2, ±3}
B is the set of elements which belongs to Z from -3 to 3.
∴ B = {x /x ∈ Z, -3 ≤ x ≤ 3}

(iii) Let C = \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
∴ C = {x / x = \(\frac{n}{n^{2}+1}\), n ∈ N, n ≤ 7}

(iv) Let D = {0, -1, 2, -3, 4, -5, 6, …}
∴ D = {x/x = (-1)n-1 × (n – 1), n ∈ N}

Question 3.
If A = {x / 6x² + x – 15 = 0}, B = {x / 2x² – 5x – 3 = 0}, C = {x / 2x² – x – 3 = 0}, then find
(i) (A ∪ B ∪ C)
(ii) (A ∩ B ∩ C)
Solution:
A = [x/6x² + x – 15 = 0)
6x² + x – 15 = 0
6x² + 10x – 9x – 15 = 0
2x(3x + 5) – 3(3x + 5) = 0
(3x + 5) (2x – 3) = 0
3x + 5 = 0 or 2x – 3 = 0
x = \(\frac{-5}{3}\) or x = \(\frac{3}{2}\)
A = {\(\frac{-5}{3}\), \(\frac{3}{2}\)}

B = {x/2x² – 5x – 3 = 0}
2x² – 5x – 3 = 0
2x² – 6x + x – 3 = 0
2x(x – 3) + 1(x – 3) = 0
(x – 3)(2x + 1) = 0
x – 3 = 0 or 2x + 1 = 0
x = 3 or x = \(\frac{-1}{2}\)
B = (\(\frac{-1}{2}\), 3)

C = {x/2x² – x – 3 = 0}
2x² – x – 3 = 0
2x² – 3x + 2x – 3 = 0
x(2x – 3) + 1(2x – 3) = 0
(2x – 3) (x + 1) = 0
2x – 3 = 0 or x + 1 = 0
x = \(\frac{3}{2}\) or x = -1
C = {-1, \(\frac{3}{2}\)}

(i) A ∪ B ∪ C = \(\left\{-\frac{5}{3}, \frac{3}{2}\right\} \cup\left\{\frac{-1}{2}, 3\right\} \cup\left\{-1, \frac{3}{2}\right\}\) = \(\left\{\frac{-5}{3},-1, \frac{-1}{2}, \frac{3}{2}, 3\right\}\)

(ii) A ∩ B ∩ C = { }

Question 4.
If A, B, C are the sets for the letters in the words ‘college’, ‘marriage’ and ‘luggage’ respectively, then verify that [A – (B ∪ C)] = [(A – B) ∩ (A – C)].
Solution:
A = {c, o, l, g, e}
B = {m, a, r, i, g, e}
C = {l, u, g, a, e}
B ∪ C = {m, a, r, i, g, e, l, u}
A – (B ∪ C) = {c, o}
A – B = {c, o, l}
A – C = {c, o}
∴ [(A – B) ∩ (A – C)] = {c, o} = A – (B ∪ C)
∴ [A -( B ∪ C)] = [(A – B) ∩ (A – C)]

Question 5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify the following:
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(iii) (A ∪ B)’ = A’ ∩ B’
(iv) (A ∩ B)’ = A’ ∪ B’
(v) A = (A ∩ B) ∪ (A ∩ B’)
(vi) B = (A ∩ B) ∪ (A’ ∩ B)
(vii) (A ∪ B) = (A – B) ∪ (A ∩ B) ∪ (B – A)
(viii) A ∩ (B ∆ C) = (A ∩ B) ∆ (A ∩ C)
(ix) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(x) n(B) = n (A’ ∩ B) + n (A ∩ B)
Solution:
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8},
X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(i) B ∩ C = {4, 5, 6}
∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6} …..(i)
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} …….(ii)
From (i) and (ii), we get
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) B ∪ C = {3, 4, 5, 6, 7, 8}
∴ A ∩ (B ∪ C) = {3, 4} ………(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) ∪ (A∩ C) = {3, 4} ……..(ii)
From (i) and (ii), we get
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) A ∪ B = {1, 2, 3, 4, 5, 6}
∴ (A ∪ B)’ = {7, 8, 9, 10} ………(i)
A’ = {5, 6, 7, 8, 9, 10},
B’ = {1, 2, 7, 8, 9,10}
∴ A’ ∩ B’ = {7, 8, 9, 10} …….(ii)
From (i) and (ii), we get
(A ∪ B)’ = A’ ∩ B’

(iv) A ∩ B = {3, 4}
(A ∩ B)’= {1, 2, 5, 6, 7, 8, 9, 10} …….(i)
A’ = {5, 6, 7, 8, 9, 10}
B’ = {1, 2, 7, 8, 9, 10}
∴ A’ ∪ B’ = {1, 2, 5, 6, 7, 8, 9, 10} …….(ii)
From (i) and (ii), we get
(A ∩ B)’ = A’ ∪ B’

(v) A = {1, 2, 3, 4} ……(i)
A ∩ B = {3, 4}
B’ = {1, 2, 7, 8, 9, 10}
A ∩ B’ = {1, 2}
∴ (A ∩ B) ∪ (A ∩ B’) = {1, 2, 3, 4} …..(ii)
From (i) and (ii), we get
A = (A ∩ B) ∪ (A ∩ B’)

(vi) B = {3, 4, 5, 6} …..(i)
A ∩ B = {3, 4}
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ (A ∩ B) ∪ (A’ ∩ B) = {3, 4, 5, 6} …..(ii)
From (i) and (ii), we get
B = (A ∩ B) ∪ (A’ ∩ B)

(vii) A ∪ B = {1, 2, 3, 4, 5, 6} …….(i)
A – B = {1, 2}
A ∩ B = {3, 4}
B – A = {5, 6}
∴ (A – B) ∪ (A ∩ B) ∪ (B – A) = {1, 2, 3, 4, 5, 6} ……(ii)
From (i) and (ii), we get
A ∪ B = (A – B) ∪ (A ∩ B) ∪ (B – A)

(viii) B – C = {3}
C – B = {7, 8}
B Δ C = (B – C) ∪ (C – B) = {3, 7, 8}
∴ A ∩ (B Δ C) = {3} ……(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) Δ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)] = {3} …..(ii)
From (i) and (ii), we get
A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

(ix) A = {1, 2, 3, 4}, B = {3, 4, 5, 6}
A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6}
∴ n(A) = 4, n(B) = 4,
n(A ∩ B) = 2, n(A ∪ B) = 6 ……(i)
∴ n(A) + n(B) – n(A ∩ B) = 4 + 4 – 2
∴ n(A) + n(B) – n(A ∩ B) = 6 …..(ii)
From (i) and (ii), we get
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

(x) B = {3, 4, 5, 6}
∴ n(B) = 4 …..(i)
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ n(A’ ∩ B) = 2
A ∩ B = {3, 4}
∴ n(A ∩ B) = 2
∴ n(A’ ∩ B) + n(A ∩ B) = 2 + 2 = 4 …..(ii)
From (i) and (ii), we get
n(B) = n(A’ ∩ B) + n (A ∩ B)

Question 6.
If A and B are subsets of the universal set X and n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5, find
(i) n(A ∪ B)
(ii) n(A ∩ B)
(iii) n(A’ ∩ B)
(iv) n(A ∩ B’)
Solution:
n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5
(i) n(A ∪ B) = n(X) – [n(A ∪ B)’]
= n(X) – n(A’ ∩ B’)
= 50 – 5
= 45

(ii) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 35 + 20 – 45
= 10

(iii) n(A’ ∩ B) = n(B) – n(A ∩ B)
= 20 – 10
= 10

(iv) n(A ∩ B’) = n(A) – n(A ∩ B)
= 35 – 10
= 25

Question 7.
In a class of 200 students who appeared in certain examinations, 35 students faded in CET, 40 in NEET and 40 in JEE, 20 faded in CET and NEET, 17 in NEET and JEE, 15 in CET and JEE and 5 faded in ad three examinations. Find how many students
(i) did not fail in any examination.
(ii) faded in NEET or JEE entrance.
Solution:
Let A = set of students who failed in CET
B = set of students who failed in NEET
C = set of students who failed in JEE
X = set of all students
∴ n(X) = 200, n(A) = 35, n(B) = 40, n(C) = 40, n(A ∩ B) = 20, n(B ∩ C) = 17, n(A ∩ C) = 15, n(A ∩ B ∩ C) = 5

(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 35 + 40 + 40 – 20 – 17 – 15 + 5
= 68
∴ No. of students who did not fail in any exam = n(X) – n(A ∪ B ∪ C)
= 200 – 68
= 132

(ii) No. of students who failed in NEET or JEE entrance = n(B ∪ C)
= n(B) + n(C) – n(B ∩ C)
= 40 + 40 – 17
= 63

Question 8.
From amongst 2000 Uterate individuals of a town, 70% read Marathi newspapers, 50% read English newspapers and 32.5% read both Marathi and English newspapers. Find the number of individuals who read
(i) at least one of the newspapers.
(ii) neither Marathi nor English newspaper.
(iii) only one of the newspapers.
Solution:
Let M = set of individuals who read Marathi newspapers
E = set of individuals who read English newspapers
X = set of all literate individuals
∴ n(X) = 2000,
n(M) = \(\frac{70}{100}\) × 2000 = 1400
n(E) = \(\frac{50}{100}\) × 2000 = 1000
n(M ∩ E) = \(\frac{32.5}{100}\) × 2000 = 650
(i) n(M ∪ E) = n(M) + n(E) – n(M ∩ E)
= 1400 + 1000 – 650
= 1750

No. of individuals who read at least one of the newspapers = n(M ∪ E) = 1750.

(ii) No. of individuals who read neither Marathi nor English newspaper = n(M’ ∩ E’)
= n(M ∪ E)’
= n(X) – n(M ∪ E)
= 2000 – 1750
= 250

(iii) No. of individuals who read only one of the newspapers = n(M ∩ E’) + n(M’ ∩ E)
= n(M ∪ E) – n(M ∩ E)
= 1750 – 650
= 1100

Question 9.
In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take tea and milk both and everyone takes atleast one beverage, find the total number of students in the hostel.
Solution:
Let T = set of students who take tea
C = set of students who take coffee
M = set of students who take milk
∴ n(T) = 25, n(C) = 20, n(M) = 15, n(T ∩ C) = 10, n(M ∩ C) = 8, n(T ∩ M) = 0, n(T ∩ M ∩ C) = 0

∴ Total number of students in the hostel = n(T ∪ C ∪ M)
= n(T) + n(C) + n(M) – n(T ∩ C) – n(M ∩ C) – n(T ∩ M) + n(T ∩ M ∩ C)
= 25 + 20 + 15 – 10 – 8 – 0 + 0
= 42

Question 10.
There are 260 persons with skin disorders. If 150 had been exposed to the chemical A, 74 to the chemical B, and 36 to both chemicals A and B, find the number of persons exposed to
(i) Chemical A but not Chemical B
(ii) Chemical B but not Chemical A
(iii) Chemical A or Chemical B
Solution:
Let A = set of persons exposed to chemical A
B = set of persons exposed to chemical B
X = set of all persons
∴ n(X) = 260, n(A) = 150, n(B) = 74, n(A ∩ B) = 36

(i) No. of persons exposed to chemical A but not to chemical B = n(A ∩ B’)
= n(A) – n(A ∩ B)
= 150 – 36
= 114

(ii) No. of persons exposed to chemical B but not to chemical A = n(A’ ∩ B)
= n(B) – n(A ∩ B)
= 74 – 36
= 38

(iii) No. of persons exposed to chemical A or chemical B = n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 150 + 74 – 36
= 188

Question 11.
Write down the power set of A = {1, 2, 3}.
Solution:
A = {1, 2, 3}
The power set of A is given by
P(A) = {{Φ}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

Question 12.
Write the following intervals in Set-Builder form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, ∞)
(iv) (-∞, 5]
(v) (2, 5]
(vi) [-3, 4)
Solution:
(i) (-3, 0) = {x / x ∈ R, -3 < x < 0}

(ii) [6, 12] = {x / x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, ∞) = {x / x ∈ R, x > 6}

(iv) (-∞, 5] = {x / x ∈ R, x ≤ 5}

(v) (2, 5] = {x / x ∈ R, 2 < x ≤ 5}

(vi) [-3, 4) = {x / x ∈ R, -3 ≤ x < 4}

Question 13.
A college awarded 38 medals in volleyball, 15 in football, and 20 in basketball. The medals were awarded to a total of 58 players and only 3 players got medals in all three sports. How many received medals in exactly two of the three sports?
Solution:
Let A = Set of students who received medals in volleyball
B = Set of students who received medals in football
C = Set of students who received medals in basketball
n(A) = 38, n(B) = 15, n(C) = 20, n(A ∪ B ∪ C) = 58, n(A ∩ B ∩ C) = 3
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
58 = 38 + 15 + 20 – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + 3
∴ n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = 18 ……(i)
Number of players who got exactly two medals = p + q + r
Here, s = n(A ∩ B ∩ C) = 3

n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = 18 …..[From (i)]
∴ p + s + s + r + q + s = 18
∴ p + q + r + 3s = 18
∴ p + q + r + 3(3) = 18
∴ p + q + r = 18 – 9 = 9
∴ Number of players who received exactly two medals = 9.

Question 14.
Solve the following inequalities and write the solution set using interval notation.
(i) -9 < 2x + 7 ≤ 19
(ii) x² – x > 20
(iii) \(\frac{2 x}{x-4}\) ≤ 5
(iv) 6x² + 1 ≤ 5x
Solution:
(i) -9 < 2x + 7 ≤ 19
∴ -16 < 2x ≤ 12
∴ -8< x ≤ 6
∴ x ∈ (-8, 6]

(ii) x² – x > 20
∴ x² – x – 20 > 0
∴ x² – 5x + 4x – 20 > 0
∴ (x – 5) (x + 4) > 0
∴ either x – 5 > 0 and x + 4 > 0 or x – 5 < 0 and x + 4 < 0

Case I: x – 5 > 0 and x + 4 > 0
∴ x > 5 and x > -4
∴ x > 5 ….(i)

Case II:
x – 5 < 0 and x + 4 < 0
∴ x < 5 and x < -4
∴ x < -4 …..(ii)
From (i) and (ii), we get
x ∈ (-∞, – 4) ∪ (5, ∞)

(iii) \(\frac{2 x}{x-4}\) ≤ 5
∴ \(\frac{2 x}{x-4}\) – 5 ≤ 0
∴ \(\frac{2 x-5 x+20}{x-4}\) ≤ 0
∴ \(\frac{20-3 x}{x-4}\) ≤ 0
When \(\frac{a}{b}\) ≤ 0,
a ≥ 0 and b < 0 or a ≤ 0 and b > 0
∴ either 20 – 3x ≥ 0 and x – 4 < 0 or 20 – 3x ≤ 0 and x – 4 > 0
Case I:
20 – 3x ≥ 0 and x – 4 < 0
∴ x ≤ \(\frac{20}{3}\) and x < 4
∴ x < 4 ……(I)

Case II: 20 – 3x ≤ 0 and x – 4 > 0
∴ x ≥ \(\frac{20}{3}\) and x > 4
∴ x ≥ \(\frac{20}{3}\) ……(ii)
From (i) and (ii), we get
x ∈ (-∞, 4) ∪ [\(\frac{20}{3}\), ∞)

(iv) 6x² + 1 ≤ 5x
6x² – 5x + 1 ≤ 0
6x² – 3x – 2x + 1 ≤ 0
(3x – 1) (2x – 1) ≤ 0
either 3x – 1 ≤ 0 and 2x – 1 ≥ 0 or 3x – 1 ≥ 0 and 2x – 1 ≤ 0
Case I:
3x – 1 ≤ 0 and 2x – 1 ≥ 0
∴ x ≤ \(\frac{1}{3}\) and x ≥ \(\frac{1}{2}\), which is not possible.

Case II:
3x – 1 ≥ 0 and 2x – 1 ≤ 0
∴ x ≥ \(\frac{1}{3}\) and x ≤ \(\frac{1}{2}\)
∴ x ∈ [\(\frac{1}{3}\), \(\frac{1}{2}\)]

Question 15.
If A = (-7, 3], B = [2, 6] and C = [4, 9], then find
(i) A ∪ B
(ii) B ∪ C
(iii) A ∪ C
(iv) A ∩ B
(v) B ∩ C
(vi) A ∩ C
(vii) A’ ∩ B
(viii) B’ ∩ C’
(ix) B – C
(x) A – B
Solution:
A = (-7, 3], B = [2, 6], C = [4, 9]
(i) A ∪ B = (-7, 6]

(ii) B ∪ C = [2, 9]

(iii) A ∪ C = (-7, 3] ∪ [4, 9]

(iv) A ∩ B = [2, 3]

(v) B ∩ C = [4, 6]

(vi) A ∩ C = { }

(vii) A’ = (-∞, – 7] ∪ (3, ∞)
∴ A’ ∩ B = (3, 6]

(viii) B’ = (-∞, 2) ∪ (6, ∞)
C’ = (-∞, 4) ∪ (9, ∞)
∴ B’ ∩ C’ = (-∞, 2) ∪ (9, ∞)

(ix) B – C = [2, 4)

(x) A – B = (-7, 2)

State Board 11th Maths Book Answers 

• Sets and Relations Ex 5.1 Class 11 Maths Solutions
• Sets and Relations Ex 5.2 Class 11 Maths Solutions
• Sets and Relations Miscellaneous Exercise 5 Class 11 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 8 Binomial Distribution Miscellaneous Exercise 8 Questions and Answers.

12th Maths Part 2 Binomial Distribution Miscellaneous Exercise 8 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
(a) √50
(b) 5
(c) 25
(d) 10
Answer:
(b) 5

Question 2.
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probablity of 2 successes is
(a) \(\frac{128}{256}\)
(b) \(\frac{219}{256}\)
(c) \(\frac{37}{256}\)
(d) \(\frac{28}{256}\)
Answer:
(d) \(\frac{28}{256}\)

Question 3.
For a binomial distribution, n = 5. If P(X = 4) = P(X = 3) then p = ___________
(a) \(\frac{1}{3}\)
(b) \(\frac{3}{4}\)
(c) 1
(d) \(\frac{2}{3}\)
Answer:
(d) \(\frac{2}{3}\)

Question 4.
In a binomial distribution, n = 4. If 2 P(X = 3) = 3 P(X = 2) then p = ___________
(a) \(\frac{4}{13}\)
(b) \(\frac{5}{13}\)
(c) \(\frac{9}{13}\)
(d) \(\frac{6}{13}\)
Answer:
(c) \(\frac{9}{13}\)

Question 5.
If X ~ B (4, p) and P (X = 0) = \(\frac{16}{81}\), then P (X = 4) = ___________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{81}\)
(c) \(\frac{1}{27}\)
(d) \(\frac{1}{8}\)
Answer:
(b) \(\frac{1}{81}\)

Question 6.
The probability of a shooter hitting a target is \(\frac{3}{4}\). How many minimum numbers of times must he fie so that the probability of hitting the target at least once is more than 0·99?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 7.
If the mean and variance of a binomial distribution are 18 and 12 respectively, then n = ___________
(a) 36
(b) 54
(c) 18
(d) 27
Answer:
(b) 54

(II) Solve the following:

Question 1.
Let X ~ B(10, 0.2). Find
(i) P(X = 1)
(ii) P(X ≥ 1)
(iii) P(X ≤ 8).
Solution:
X ~ B(10, 0.2)
∴ n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
The p,m.f. of X is given by

Question 2.
Let X ~ B(n, p).
(i) If n = 10, E(X) = 5, find p and Var(X).
(ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) Given: n = 10 and E(X) = 5
But E(X) = np
∴ np = 5.
∴ 10p = 5
∴ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Var(X) = npq = 10(\(\frac{1}{2}\))(\(\frac{1}{2}\)) = 2.5.
Hence, p = \(\frac{1}{2}\) and Var(X) = 2.5

(ii) Given: E(X) = 5 and Var(X) = 2.5
∴ np = 5 and npq = 2.5
∴ \(\frac{n p q}{n p}=\frac{2.5}{5}\)
∴ q = 0.5 = \(\frac{5}{10}=\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Substituting p = \(\frac{1}{2}\) in np = 5, we get
n(\(\frac{1}{2}\)) = 5
∴ n = 10
Hence, n = 10 and p = \(\frac{1}{2}\)

Question 3.
If a fair coin is tossed 10 times and the probability that it shows heads (i) 5 times (ii) in the first four tosses and tail in the last six tosses.
Solution:
Let X = number of heads.
p = probability that coin tossed shows a head
∴ p = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given: n = 10
∴ X ~ B(10, \(\frac{1}{2}\))
The p.m.f. of X is given by
P(X = x) = \({ }^{n} C_{x} P^{x} q^{n-x}\)

(i) P(coin shows heads 5 times) = P[X = 5]

Hence, the probability that can shows heads exactly 5 times = \(\frac{63}{256}\)

(ii) P(getting heads in first four tosses and tails in last six tosses) = P(X = 4)

Hence, the probability that getting heads in first four tosses and tails in last six tosses = \(\frac{105}{512}\).

Question 4.
The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly 2 will miss the target.
Solution:
Let X = the number of bombs hitting the target.
p = probability that bomb will hit the target
∴ p = 0.8 = \(\frac{8}{10}=\frac{4}{5}\)
∴ q = 1 – p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Given: n = 10
∴ X ~ B(10, \(\frac{4}{5}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e.p(x) = \({ }^{10} \mathrm{C}_{x}\left(\frac{4}{5}\right)^{x}\left(\frac{1}{5}\right)^{10-x}\)
P(exactly 2 bombs will miss the target) = P(exactly 8 bombs will hit the target)
= P[X = 8]
= p(8)

Hence, the probability that exactly 2 bombs will miss the target = 45\(\left(\frac{2^{16}}{5^{10}}\right)\)

Question 5.
The probability that a mountain bike travelling along a certain track will have a tire burst is 0.05. Find the probability that among 17 riders:
(i) exactly one has a burst tyre
(ii) at most three have a burst tyre
(iii) two or more have burst tyres.
Solution:
Let X = number of burst tyres.
p = probability that a mountain bike travelling along a certain track will have a tyre burst.
∴ p = 0.05
∴ q = 1 – p = 1 – 0.05 = 0.95
Given: n = 17
∴ X ~ B(17, 0.05)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} P^{x} q^{n-x}\)
i.e.(x) = \({ }^{17} \mathrm{C}_{x}(0.05)^{x}(0.95)^{17-x}\), x = 0, 1, 2, ……, 17
(i) P(exactly one has a burst tyre)
P(X = 1) = p(1) = \({ }^{17} \mathrm{C}_{1}\) (0.05)¹ (0.95)^17-1
= 17(0.05) (0.95)¹⁶
= 0.85(0.95)¹⁶
Hence, the probability that riders has exactly one burst tyre = (0.85)(0.95)¹⁶

(ii) P(at most three have a burst tyre) = P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= p(0) + p(1) + p(2) + p(3)


Hence, the probability that at most three riders have burst tyre = (2.0325)(0.95)¹⁴.

(iii) P(two or more have tyre burst) = P(X ≥ 2)
= 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]
= 1 – [p(0) + p(1)]
= 1 – [\({ }^{17} \mathrm{C}_{0}\) (0.05)⁰ (0.95)¹⁷ + \({ }^{17} \mathrm{C}_{1}\) (0.05)(0.95)¹⁶]
= 1 – [1(1)(0.95)¹⁷ + 17(0.05)(0.95)¹⁶]
= 1 – (0.95)¹⁶[0.95 + 0.85]
= 1 – (1.80)(0.95)¹⁶
= 1 – (1.8)(0.95)¹⁶
Hence, the probability that two or more riders have tyre burst = 1 – (1.8)(0.95)¹⁶.

Question 6.
The probability that a lamp in a classroom will be burnt out is 0.3. Six such lamps are fitted in the classroom. If it is known that the classroom is unusable if the number of lamps burning in it is less than four, find the probability that the classroom cannot be used on a random occasion.
Solution:
Let X = number of lamps burnt out in the classroom.
p = probability of a lamp in a classroom will be burnt
∴ p = 0.3 = \(\frac{3}{10}\)
∴ q = 1 – p = 1 – \(\frac{3}{10}\) = \(\frac{7}{10}\)
Given: n = 6
∴ X ~ B(6, \(\frac{3}{10}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{6} \mathrm{C}_{x}\left(\frac{3}{10}\right)^{x}\left(\frac{7}{10}\right)^{6-x}\)
Since the classroom is unusable if the number of lamps burning in it is less than four, therefore
P(classroom cannot be used) = P[X < 4]
= P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]
= p(0) + p(1) + p(2) + p(3)

Hence, the probability that the classroom cannot be used on a random occasion is 0.92953.

Question 7.
A lot of 100 items contain 10 defective items. Five items are selected at random from the lot and sent to the retail store. What is the probability that the store will receive at most one defective item?
Solution:
Let X = number of defective items.
p = probability that item is defective
∴ p = \(\frac{10}{100}=\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{10}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{5} C_{x}\left(\frac{1}{10}\right)^{x}\left(\frac{9}{10}\right)^{5-x}\)
P (store will receive at most one defective item) = P[X ≤ 1]
=P[X = 0] + P[X = 1]
= p(0) + p(1)

Hence, the probability that the store will receive at most one defective item is (1.4)(0.9)⁴.

Question 8.
A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. The inspector of the retailer picks 20 items from a shipment. What is the probability that the store will receive at most one defective item?
Solution:
Let X = number of defective electronic devices.
p = probability that device is defective
∴ p = 3% = \(\frac{3}{100}\)
∴ q = 1 – p = 1 – \(\frac{3}{100}\) = \(\frac{97}{100}\)
Given: n = 20
∴ X ~ B(20, \(\frac{3}{100}\))
The p.m.f. of X is given as:

Hence, the probability that the store will receive at most one defective item = (1.57)(0.97)¹⁹.

Question 9.
The probability that a certain kind of component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 tested components tested survive.
Solution:
Let X = number of tested components survive.
p = probability that the component survives the check test

Hence, the probability that exactly 2 of the 4 tested components survive is 0.3456.

Question 10.
An examination consists of 10 multiple choice questions, in each of which a candidate has to deduce which one of five suggested answers is correct. A completely unprepared student guesses each answer completely randomly. What is the probability that this student gets 8 or more questions correct? Draw the appropriate moral.
Solution:
Let X = number of correct answers.
p = probability that student gets correct answer
∴ p = \(\frac{1}{5}\)
∴ q = 1 – p = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Given: n = 10 (number of total questions)
∴ X ~ B(10, \(\frac{1}{5}\))
The p.m.f. of X is given by

Hence, the probability that student gets 8 or more questions correct = \(\frac{30.44}{5^{8}}\)

Question 11.
The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that (i) all 8 machines (ii) 7 or 8 machines (iii) at most 6 machines will produce all bolts within specification.
Solution:
Let X = number of machines which produce the bolts within specification.
p = probability that a machine produce bolts within specification
p = 0.998 and q = 1 – p = 1 – 0.998 = 0.002
Given: n = 8
∴ X ~ B(8, 0.998)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{8} \mathrm{C}_{x}(0.998)^{x}(0.002)^{8-x}\), x = 0, 1, 2, …, 8
(i) P(all 8 machines will produce all bolts within specification) = P[X = 8]
= p(8)
= \({ }^{8} \mathrm{C}_{8}\) (0.998)⁸ (0.002)^8-8
= 1(0.998)⁸ . (1)
= (0.998)⁸
Hence, the probability that all 8 machines produce all bolts with specification = (0.998)⁸.

(ii) P(7 or 8 machines will produce all bolts within i specification) = P (X = 7) + P (X = 8)

Hence, the probability that 7 or 8 machines produce all bolts within specification = (1.014)(0.998)⁷.

(iii) P(at most 6 machines will produce all bolts with specification) = P[X ≤ 6]
= 1 – P[x > 6]
= 1 – [P(X = 7) + P(X = 8)]
= 1 – [P(7) + P(8)]
= 1 – (1.014)(0.998)⁷
Hence, the probability that at most 6 machines will produce all bolts with specification = 1 – (1.014)(0.998)⁷.

Question 12.
The probability that a machine develops a fault within the first 3 years of use is 0.003. If 40 machines are selected at random, calculate the probability that 38 or more will develop any faults within the first 3 years of use.
Solution:
Let X = the number of machines who develop a fault.
p = probability that a machine develops a fait within the first 3 years of use
∴ p = 0.003 and q = 1 – p = 1 – 0.003 = 0.997
Given: n = 40
∴ X ~ B(40, 0.003)
The p.m.f. of X is given by

Hence, the probability that 38 or more machines will develop the fault within 3 years of use = (775.44)(0.003)³⁸.

Question 13.
A computer installation has 10 terminals. Independently, the probability that anyone terminal will require attention during a week is 0.1. Find the probabilities that (i) 0 (ii) 1 (iii) 2 (iv) 3 or more, terminals will require attention during the next week.
Solution:
Let X = number of terminals which required attention during a week.
p = probability that any terminal will require attention during a week
∴ p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9
Given: n = 10
∴ X ~ B(10, 0.1)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{10} C_{x}(0.1)^{x}(0.9)^{10-x}\), x = 0, 1, 2, …, 10
(i) P(no terminal will require attention) = P(X = 0)

Hence, the probability that no terminal requires attention = (0.9)¹⁰

(ii) P(1 terminal will require attention)

Hence, the probability that 1 terminal requires attention = (0.9)⁹.

(iii) P(2 terminals will require attention)

Hence, the probability that 2 terminals require attention = (0.45)(0.9)⁸.

(iv) P(3 or more terminals will require attention)

Hence, the probability that 3 or more terminals require attention = 1 – (2.16) × (0.9)⁸.

Question 14.
In a large school, 80% of the pupil like Mathematics. A visitor to the school asks each of 4 pupils, chosen at random, whether they like Mathematics.
(i) Calculate the probabilities of obtaining an answer yes from 0, 1, 2, 3, 4 of the pupils.
(ii) Find the probability that the visitor obtains answer yes from at least 2 pupils:
(a) when the number of pupils questioned remains at 4.
(b) when the number of pupils questioned is increased to 8.
Solution:
Let X = number of pupils like Mathematics.
p = probability that pupils like Mathematics

(i) The probabilities of obtaining an answer yes from 0, 1, 2, 3, 4 of pupils are P(X = 0), P(X = 1), P(X = 2), P(X = 3) and P(X = 4) respectively

(ii) (a) P(visitor obtains the answer yes from at least 2 pupils when the number of pupils questioned remains at 4) = P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4)

(b) P(the visitor obtains the answer yes from at least 2 pupils when number of pupils questioned is increased to 8)

Question 15.
It is observed that it rains 12 days out of 30 days. Find the probability that
(i) it rains exactly 3 days of the week.
(ii) it will rain at least 2 days of a given week.
Solution:
Let X = the number of days it rains in a week.
p = probability that it rains

(i) P(it rains exactly 3 days of week) = P(X = 3)

Hence, the probability that it rains exactly 3 days of week = 0.2903.

(ii) P(it will rain at least 2 days of the given week)

Hence, the probability that it rains at least 2 days of a given week = 0.8414

Question 16.
If the probability of success in a single trial is 0.01. How many trials are required in order to have a probability greater than 0.5 of getting at least one success?
Solution:
Let X = number of successes.
p = probability of success in a single trial
∴ p = 0.01
and q = 1 – p = 1 – 0.01 = 0.99
∴ X ~ B(n, 0.01)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)

Hence, the number of trials required in order to have a probability greater than 0.5 of getting at least one success is \(\frac{\log 0.5}{\log 0.99}\) or 68.

Question 17.
In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.
Solution:
Given: X ~ B(n = 5, p)
The probability of X success is


Hence, the probability of success is \(\frac{1}{5}\).

12th Maharashtra State Board Maths Solutions Pdf Part 2

• Probability Distributions Ex 7.1 Class 12 Maths Solutions
• Probability Distributions Ex 7.2 Class 12 Maths Solutions
• Probability Distributions Miscellaneous Exercise 7 Class 12 Maths Solutions
• Binomial Distribution Ex 8.1 Class 12 Maths Solutions
• Binomial Distribution Miscellaneous Exercise 8 Class 12 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 8 Binomial Distribution Ex 8.1 Questions and Answers.

12th Maths Part 2 Binomial Distribution Exercise 8.1 Questions And Answers Maharashtra Board

Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of
(i) 5 successes
(ii) at least 5 successes
(iii) at most 5 successes.
Solution:
Let X = number of successes, i.e. number of odd numbers.
p = probability of getting an odd number in a single throw of a die
∴ p = \(\frac{3}{6}=\frac{1}{2}\) and q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given: n = 6
∴ X ~ B(6, \(\frac{1}{2}\))
The p.m.f. of X is given by

Hence, the probability of 5 successes is \(\frac{3}{32}\).

(ii) P(at least 5 successes) = P[X ≥ 5]
= p(5) + p(6)

Hence, the probability of at least 5 successes is \(\frac{7}{64}\).

(iii) P(at most 5 successes) = P[X ≤ 5]
= 1 – P[X > 5]

Hence, the probability of at most 5 successes is \(\frac{63}{64}\).

Question 2.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
Let X = number of doublets.
p = probability of getting a doublet when a pair of dice is thrown
∴ p = \(\frac{6}{36}=\frac{1}{6}\) and
q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Given: n = 4
∴ X ~ B(4, \(\frac{1}{6}\))
The p.m.f. of X is given by

Hence, the probability of two successes is \(\frac{25}{216}\).

Question 3.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Solution:
Let X = number of defective items.
p = probability of defective item
∴ p = 5% = \(\frac{5}{100}=\frac{1}{20}\)
and q = 1 – p = 1 – \(\frac{1}{20}\) = \(\frac{19}{20}\)
∴ X ~ B(10, \(\frac{1}{20}\))
The p.m.f. of X is given by

P(sample of 10 items will include not more than one defective item) = P[X ≤ 1]

Hence, the probability that a sample of 10 items will include not more than one defective item = 29\(\left(\frac{19^{9}}{20^{10}}\right)\).

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that
(i) all the five cards are spades
(ii) only 3 cards are spades
(iii) none is a spade.
Solution:
Let X = number of spade cards.
p = probability of drawing a spade card from a pack of 52 cards.
Since there are 13 spade cards in the pack of 52 cards.
∴ p = \(\frac{13}{52}=\frac{1}{4}\) and
q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{4}\))
The p.m.f. of X is given by

(i) P(all five cards are spade)

Hence, the probability of all the five cards are spades = \(\frac{1}{1024}\)

(ii) P(only 3 cards are spade) = P[X = 3]

Hence, the probability of only 3 cards are spades = \(\frac{45}{512}\)

(iii) P(none of cards is spade) = P[X = 0]

Hence, the probability of none of the cards is a spade = \(\frac{243}{1024}\)

Question 5.
The probability of a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one, will fuse after 150 days of use.
Solution:
Let X = number of fuse bulbs.
p = probability of a bulb produced by a factory will fuse after 150 days of use.
∴ p = 0.05
and q = 1 – p = 1 – 0.05 = 0.95
Given: n = 5
∴ X ~ B(5, 0.05)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{5} C_{x}(0.05)^{x}(0.95)^{5-x}\), x = 0, 1, 2, 3, 4, 5
(i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0]
= p(0)
= \({ }^{5} \mathrm{C}_{0}(0.05)^{0}(0.95)^{5-0}\)
= 1 × 1 × (0.95)⁵
= (0.95)⁵
Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)⁵.

(ii) P(not more than one bulb will fuse after 150 days of j use) = P[X ≤ 1]
= p(0) + p(1)
= \({ }^{5} \mathrm{C}_{0} \cdot(0.05)^{0}(0.95)^{5-0}+{ }^{5} \mathrm{C}_{1}(0.05)^{1}(0.95)^{4}\)
= 1 × 1 × (0.95)⁵ + 5 × (0.05) × (0.95)⁴
= (0.95)⁴ [0.95 + 5(0.05)]
= (0.95)⁴ (0.95 + 0.25)
= (0.95)⁴ (1.20)
= (1.2) (0.95)⁴
Hence, the probability that not more than one bulb will fuse after 150 days = (1.2)(0.95)⁴.

(iii) P(more than one bulb fuse after 150 days)
= P[X > 1]
= 1 – P[X ≤ 1]
= 1 – (1.2)(0.95)⁴
Hence, the probability that more than one bulb fuse after 150 days = 1 – (1.2)(0.95)⁴.

(iv) P(at least one bulb fuse after 150 days)
= P[X ≥ 1]
= 1 – P[X = 0]
= 1 – p(0)
= 1 – \({ }^{5} C_{0}(0.05)^{0}(0.95)^{5-0}\)
= 1 – 1 × 1 × (0.95)⁵
= 1 – (0.95)⁵
Hence, the probability that at least one bulb fuses after 150 days = 1 – (0.95)⁵.

Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Solution:
Let X = number of balls marked with digit 0.
p = probability of drawing a ball from 10 balls marked with the digit 0.
∴ p = \(\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
The p.m.f. of X is given by

P(none of the ball marked with digit 0) = P(X = 0)

Hence, the probability that none of the bulb marked with digit 0 is \(\left(\frac{9}{10}\right)^{4}\)

Question 7.
On a multiple-choice examination with three possible answers for each of the five questions. What is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
Let X = number of correct answers.
p = probability that a candidate gets a correct answer from three possible answers.
∴ p = \(\frac{1}{3}\) and q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{3}\))
The p.m.f. of X is given by

P(four or more correct answers) = P[X ≥ 4]
= p(4) + p(5)

Hence, the probability of getting four or more correct answers = \(\frac{11}{243}\).

Question 8.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac{1}{100}\), find the probability that he will win a prize
(i) at least once
(ii) exactly once
(iii) at least twice.
Solution:
Let X = number of winning prizes.
p = probability of winning a prize
∴ p = \(\frac{1}{100}\)
and q = 1 – p = 1 – \(\frac{1}{100}\) = \(\frac{99}{100}\)
Given: n = 50
∴ X ~ B(50, \(\frac{1}{100}\))
The p.m.f. of X is given by
\(P(X=x)={ }^{n} C_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{50} \mathrm{C}_{x}\left(\frac{1}{100}\right)^{x}\left(\frac{99}{100}\right)^{50-x}\), x = 0, 1, 2,… 50
(i) P(a person wins a prize at least once)

Hence, probability of winning a prize at least once = 1 – \(\left(\frac{99}{100}\right)^{50}\)

(ii) P(a person wins exactly one prize) = P[X = 1] = p(1)

Hence, probability of winning a prize exactly once = \(\frac{1}{2}\left(\frac{99}{100}\right)^{49}\)

(iii) P(a persons wins the prize at least twice) = P[X ≥ 2]
= 1 – P[X < 2]
= 1 – [p(0) + p(1)]

Hence, the probability of winning the prize at least twice = 1 – 149\(\left(\frac{99^{49}}{100^{50}}\right)\).

Question 9.
In a box of floppy discs, it is known that 95% will work. A sample of three of the discs is selected at random. Find the probability that (i) none (ii) 1 (iii) 2 (iv) all 3 of the sample will work.
Solution:
Let X = number of working discs.
p = probability that a floppy disc works
∴ p = 95% = \(\frac{95}{100}=\frac{19}{20}\)
and q = 1 – p = 1 – \(\frac{19}{20}\) = \(\frac{1}{20}\)
Given: n = 3
∴ X ~ B(3, \(\frac{19}{20}\))
The p.m.f. of X is given by

(i) P(none of the floppy discs work) = P(X = 0)

Hence, the probability that none of the floppy disc will work = \(\frac{1}{20^{3}}\).

(ii) P(exactly one floppy disc works) = P(X = 1)

Hence, the probability that exactly one floppy disc works = 3\(\left(\frac{19}{20^{3}}\right)\)

(iii) P(exactly two floppy discs work) = P(X = 2)

Hence, the probability that exactly 2 floppy discs work = 3\(\left(\frac{19^{2}}{20^{3}}\right)\)

(iv) P(all 3 floppy discs work) = P(X = 3)

Hence, the probability that all 3 floppy discs work = \(\left(\frac{19}{20}\right)^{3}\).

Question 10.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
Let X = number of sixes.
p = probability that a die shows six in a single throw
∴ p = \(\frac{1}{6}\)
and q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Given: n = 6
∴ X ~ B(6, \(\frac{1}{6}\))
The p.m.f. of X is given by


Hence, probability of throwing at most 2 sixes = \(\frac{7}{3}\left(\frac{5}{6}\right)^{5}\).

Question 11.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?
Solution:
Let X = number of defective articles.
p = probability of defective articles.
∴ p = 10% = \(\frac{10}{100}=\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
Given: n = 12
∴ X ~ B(12, \(\frac{1}{10}\))
The p.m.f. of X is given by

Hence, the probability of getting 9 defective articles = \(22\left(\frac{9^{3}}{10^{11}}\right)\)

Question 12.
Given X ~ B(n, P)
(i) If n = 10 and p = 0.4, find E(x) and Var(X).
(ii) If p = 0.6 and E(X) = 6, find n and Var(X).
(iii) If n = 25, E(X) = 10, find p and SD(X).
(iv) If n = 10, E(X) = 8, find Var(X).
Solution:
(i) Given: n = 10 and p = 0.4
∴ q = 1 – p = 1 – 0.4 = 0.6
∴ E(X) = np = 10(0.4) = 4
Var(X) = npq = 10(0.4)(0.6) = 2.4
Hence, E(X) = 4, Var(X) = 2.4.

(ii) Given: p = 0.6, E (X) = 6
E(X) = np
6 = n(0.6)
n = \(\frac{6}{0.6}\) = 10
Now, q = 1 – p = 1 – 0.6 = 0.4
∴ Var(X) = npq = 10(0.6)(0.4) = 2.4
Hence, n = 10 and Var(X) = 2.4.

(iii) Given: n = 25, E(X) = 10
E(X) = np
10 = 25p
p = \(\frac{10}{25}=\frac{2}{5}\)
∴ q = 1 – p = 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
Var(X) = npq = \(25 \times \frac{2}{5} \times \frac{3}{5}\) = 6
∴ SD(X) = √Var(X) = √6
Hence, p = \(\frac{2}{5}\) and S.D.(X) = √6.

(iv) Given: n = 10, E(X) = 8
E(X) = np
8 = 10p
p = \(\frac{8}{10}=\frac{4}{5}\)
q = 1 – p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Var(X) = npq = \(10\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)=\frac{8}{5}\)
Hence, Var(X) = \(\frac{8}{5}\).

12th Maharashtra State Board Maths Solutions Pdf Part 2

• Probability Distributions Ex 7.1 Class 12 Maths Solutions
• Probability Distributions Ex 7.2 Class 12 Maths Solutions
• Probability Distributions Miscellaneous Exercise 7 Class 12 Maths Solutions
• Binomial Distribution Ex 8.1 Class 12 Maths Solutions
• Binomial Distribution Miscellaneous Exercise 8 Class 12 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Miscellaneous Exercise 7 Questions and Answers.

12th Maths Part 2 Probability Distributions Miscellaneous Exercise 7 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a =
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 2.
If the p.d.f. of a c.r.v. X is f(x) = 3(1 – 2x²), for 0 < x < 1 and = 0, otherwise (elsewhere), then the c.d.f. of X is F(x) =
(a) 2x – 3x²
(b) 3x – 4x³
(c) 3x – 2x³
(d) 2x³ – 3x
Answer:
(c) 3x – 2x³

Question 3.
If the p.d.f. of a c.r.v. X is f(x) = \(\frac{x^{2}}{18}\), for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) =
(a) \(\frac{1}{27}\)
(b) \(\frac{1}{28}\)
(c) \(\frac{1}{29}\)
(d) \(\frac{1}{26}\)
Answer:
(a) \(\frac{1}{27}\)

Question 4.
If p.m.f. of a d.r.v. X takes values 0, 1, 2, 3, … which probability P(X = x) = k(x +1) . 5-x, where k is a constant, then P(X = 0) =
(a) \(\frac{7}{25}\)
(b) \(\frac{16}{25}\)
(c) \(\frac{18}{25}\)
(d) \(\frac{19}{25}\)
Answer:
(b) \(\frac{16}{25}\)

Question 5.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{\left({ }^{5} \mathrm{C}_{x}\right)}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. If a = P(X ≤ 2) and b = P(X ≥ 3), then
(a) a < b
(b) a > b
(c) a = b
(d) a + b
Answer:
(c) a = b

Question 6.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{x}{n(n+1)}\), for x = 1, 2, 3, ……, n and = 0, otherwise, then E(X) =
(a) \(\frac{n}{1}+\frac{1}{2}\)
(b) \(\frac{n}{3}+\frac{1}{6}\)
(c) \(\frac{n}{2}+\frac{1}{5}\)
(d) \(\frac{n}{1}+\frac{1}{3}\)
Answer:
(b) \(\frac{n}{3}+\frac{1}{6}\)

Question 7.
If p.m.f. of a d.r.v. X is P(x) = \(\frac{c}{x^{3}}\), for x = 1, 2, 3 and = 0, otherwise (elsewhere), then E(X) =
(a) \(\frac{343}{297}\)
(b) \(\frac{294}{251}\)
(c) \(\frac{297}{294}\)
(d) \(\frac{294}{297}\)
Answer:
(b) \(\frac{294}{251}\)

Question 8.
If the d.r.v. X has the following probability distribution:

then P(X = -1) =
(a) \(\frac{1}{10}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{4}{10}\)
Answer:
(a) \(\frac{1}{10}\)

Question 9.
If the d.r.v. X has the following probability distribution:

then k =
(a) \(\frac{1}{7}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{10}\)
Answer:
(d) \(\frac{1}{10}\)

Question 10.
Find the expected value of X for the following p.m.f.

(a) 0.85
(b) -0.35
(c) 0.15
(d) -0.15
Answer:
(b) -0.35

(II) Solve the following:

Question 1.
Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values:
(i) An economist is interested in the number of unemployed graduates in the town of population 1 lakh.
(ii) Amount of syrup prescribed by a physician.
(iii) The person on a high protein diet is interesting to gain weight in a week.
(iv) 20 white rats are available for an experiment. Twelve rats are males. A scientist randomly selects 5 rats, the number of female rats selected on a specific day.
(v) A highway-safety group is interested in studying the speed (in km/hr) of a car at a checkpoint.
Solution:
(i) Let X = number of unemployed graduates in a town.
Since the population of the town is 1 lakh, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, …, 99999, 100000}.

(ii) Let X = amount of syrup prescribed by a physician.
Then X takes uncountable infinite values.
∴ random variable X is continuous.

(iii) Let X = gain of weight in a week
Then X takes uncountable infinite values
∴ random variable X is continuous.

(iv) Let X = number of female rats selected on a specific day.
Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr.
Then X takes uncountable infinite values
∴ random variable X is continuous.

Question 2.
The probability distribution of discrete r.v. X is as follows:

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:

Question 3.
The following is the probability distribution of X:

Find the probability that
(i) X is positive
(ii) X is non-negative
(iii) X is odd
(iv) X is even.
Solution:
(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.25 + 0.15 + 0.1
= 0.50

(ii) P(X is non-negative)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.20 + 0.25 + 0.15 + 0.1
= 0.70

(iii) P(X is odd)
= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.1
= 0.55

(iv) P(X is even)
= P(X = -2) + P(X = 0) + P(X = 2)
= 0.10 + 0.20 + 0.15
= 0.45.

Question 4.
The p.m.f. of a r.v. X is given by P(X = x) = x = \(\frac{{ }^{5} \mathrm{C}_{\mathrm{x}}}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).
Solution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}\)
= \(\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}\) ………[latex]{ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}[/latex]
= P(X = 5) + P(X = 4) + P(X = 3)
= P(X ≥ 3)
∴ P(X ≤ 2) = P(X ≥ 3).

Question 5.
In the p.m.f. of r.v. X

Find a and obtain c.d.f. of X.
Solution:
For p.m.f. of a r.v. X
\(\sum_{i=1}^{5} P(X=x)=1\)
∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

Let F(x) be the c.d.f. of X.
Then F(x) = P(X ≤ x)
∴ F(1) = P(X ≤ 1) = P(X = 1) = \(\frac{1}{20}\)
F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2)
\(=\frac{1}{20}+\frac{3}{20}=\frac{4}{20}=\frac{1}{5}\)
P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}=\frac{9}{20}\)
F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}=\frac{19}{20}\)
F(5) = P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}+\frac{1}{20}=\frac{20}{20}=1\)
Hence, the c.d.f. of the random variable X is as follows:

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.
Solution:
When a fair coin is tossed 4 times then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n (X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{0}}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{1}}{16}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{{ }^{4} \mathrm{C}_{2}}{16}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{3}}{16}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{4}}{16}\)
∴ the probability distribution of X is as follows:

Also, the formula for p.m.f. of X is
P(x) = \(\frac{{ }^{4} \mathrm{C}_{x}}{16}\), x = 0, 1, 2, 3, 4 and = 0, otherwise.

Question 7.
Find the probability distribution of the number of successes in two tosses of a die, where success is defined as
(i) number greater than 4
(ii) six appear on at least one die.
Solution:
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
Here, success refers to the number greater than 4.
P(X = 0) = P(number less than or equal to 4 on both the tosses)
= \(\frac{4}{6} \times \frac{4}{6}=\frac{16}{36}=\frac{4}{9}\)
P(X = 1) = P(number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss)
= \(\frac{4}{6} \times \frac{2}{6}+\frac{4}{6} \times \frac{2}{6}\)
= \(\frac{8}{36}+\frac{8}{36}\)
= \(\frac{16}{36}\)
= \(\frac{4}{9}\)
P(X = 2) = P(number greater than 4 on both the tosses)
= \(\frac{2}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
Thus, the probability distribution is as follows:

(ii) Here, success means six appears on at least one die.
P(Y = 0) = P(six appears on none of the dice) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(Y = 1) = P(six appears on none of the dice x six appears on at least one of the dice ) + P(six appears on none of the dice x six appears on at least one of the dice)
= \(\frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}\)
P(Y = 2) = P(six appears on at least one of the dice) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
Thus, the required probability distribution is as follows:

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X > 6)
(iii) P(0 < X < 3).

Question 9.
The following is the c.d.f. of a r.v. X:

Find
(i) p.m.f. of X
(ii) P( -1 ≤ X ≤ 2)
(iii) P(X ≤ X > 0).
Solution:
(i) From the given table
F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5
F(0) = 0.65, f(1) = 0.75, F(2) = 0.85
F(3) = 0.9, F(4) = 1
P(X = -3) = F(-3) = 0.1
P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2
P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.1
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ the p.m.f of X is as follows:

(ii) P(-1 ≤ X ≤ 2) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55

(iii) (X ≤ 3) ∩ (X > 0)
= { -3, -2, -1, 0, 1, 2, 3} n {1, 2, 3, 4}
= {1, 2, 3}

Question 10.
Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:

Solution:
(i) We construct the following table to find the expected value, variance, and standard deviation:

(ii) We construct the following table to find the expected value, variance, and standard deviation:



(iii) We construct the following table to find the expected value, variance, and standard deviation:



(iv) We construct the following table to find the expected value, variance, and standard deviation:

Question 11.
A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.
Solution:
When a coin is tossed twice, the sample space is
S = {HH, HT, TH, HH}
Let X denote the amount he wins.
Then X takes values 10, 5, 2.
P(X = 10) = P(2 heads appear) = \(\frac{1}{4}\)
P(X = 5) = P(1 head appears) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(X = 2) = P(no head appears) = \(\frac{1}{4}\)
We construct the following table to calculate the mean and the variance of X:

From the table Σx_i . P(x_i) = 5.5, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 38.5
E(X) = Σx_i . P(x_i) = 5.5
Var(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 38.5 – (5.5)²
= 38.5 – 30.25
= 8.25
∴ Hence, expected winning amount = ₹ 5.5 and variance of winning amount = ₹ 8.25.

Question 12.
Let the p.m.f. of r.v. X be P(x) = \(\frac{3-x}{10}\), for x = -1, 0, 1, 2 and = 0, otherwise.
Calculate E(X) and Var(X).
Solution:
P(X) = \(\frac{3-x}{10}\)
X takes values -1, 0, 1, 2
P(X = -1) = P(-1) = \(\frac{3+1}{10}=\frac{4}{10}\)
P(X = 0) = P(0) = \(\frac{3-0}{10}=\frac{3}{10}\)
P(X = 1) = P(1) = \(\frac{3-1}{10}=\frac{2}{10}\)
P(X = 2) = P(2) = \(\frac{3-2}{10}=\frac{1}{10}\)
We construct the following table to calculate the mean and variance of X:

From the table
Σx_iP(x_i) = 0 and \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) = 1
E(X) = Σx_iP(x_i) = 0
Var(X) = \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 1 – 0
= 1
Hence, E(X) = 0, Var (X) = 1.

Question 13.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and = 0 otherwise.
Compute
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(X < -0.5 or X > 0.5).
Solution:
(i) P(X > 0)


(ii) P(-1 < X < 1)


(iii) P(X < -0.5 or X > 0.5)

Question 14.
The p.d.f. of a continuous r.v. X is given by f(x) = \(\frac{1}{2 a}\), for 0 < x < 2a and = 0, otherwise. Show that P( X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
The p.d.f. of r.v. X is given by f(x) = \(\frac{k}{\sqrt{x}}\), for 0 < x < 4 and = 0, otherwise. Determine k. Determine c.d.f. of X and hence find P(X ≤ 2) and P(X ≤ 1).
Solution:
Since f is p.d.f. of the r.v. X,

12th Maharashtra State Board Maths Solutions Pdf Part 2

• Probability Distributions Ex 7.1 Class 12 Maths Solutions
• Probability Distributions Ex 7.2 Class 12 Maths Solutions
• Probability Distributions Miscellaneous Exercise 7 Class 12 Maths Solutions
• Binomial Distribution Ex 8.1 Class 12 Maths Solutions
• Binomial Distribution Miscellaneous Exercise 8 Class 12 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.2 Questions and Answers.

12th Maths Part 2 Probability Distributions Exercise 7.2 Questions And Answers Maharashtra Board

Question 1.
Verify which of the following is p.d.f. of r.v. X:
(i) f(x) = sin x, for 0 ≤ x ≤ \(\frac{\pi}{2}\)
(ii) f(x) = x, for 0 ≤ x ≤ 1 and -2 – x for 1 < x < 2
(iii) fix) = 2, for 0 ≤ x ≤ 1.
Solution:
f(x) is the p.d.f. of r.v. X if
(a) f(x) ≥ 0 for all x ∈ R and

Hence, f(x) is the p.d.f. of X.

(ii) f(x) = x ≥ 0 if 0 ≤ x ≤ 1
For 1 < x < 2, -2 < -x < -1
-2 – 2 < -2 – x < -2 – 1
i.e. -4 < f(x) < -3 if 1 < x < 2
Hence, f(x) is not p.d.f. of X.

(iii) (a) f(x) = 2 ≥ 0 for 0 ≤ x ≤ 1

Hence, f(x) is not p.d.f. of X.

Question 2.
The following is the p.d.f. of r.v. X:
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
Find
(a) P(x < 1.5)
(b) P(1 < x < 2) (c) P(x > 2).
Solution:

Question 3.
It is known that error in measurement of reaction temperature (in 0°C) in a certain experiment is continuous r.v. given by
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0. otherwise.
(i) Verify whether f(x) is p.d.f. of r.v. X
(ii) Find P(0 < x ≤ 1)
(iii) Find the probability that X is negative.
Solution:

Question 4.
Find k if the following function represents p.d.f. of r.v. X
(i) f(x) = kx. for 0 < x < 2 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{3}{2}\)).
(ii) f(x) = kx(1 – x), for 0 < x < 1 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{1}{2}\)), P(x < \(\frac{1}{2}\)).
Solution:
(i) Since, the function f is p.d.f. of X

(ii) Since, the function f is the p.d.f. of X,

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected students and suppose X has p.d.f.
f(x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate:
(i) P(x ≤ 1)
(ii) P(0.5 ≤ x ≤ 1.5)
(iii) P(x ≥ 1.5).
Solution:
(i) P(x ≤ 1)

(ii) P(0.5 ≤ x ≤ 1.5)

(iii) P(x ≥ 1.5)

Question 6.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = \(\frac{1}{5}\), for 0 ≤ x ≤ 5 and = 0 otherwise. Find the probability that
(i) waiting time is between 1 and 3
(ii) waiting time is more than 4 minutes.
Solution:
(i) Required probability = P(1 < X < 3)

(ii) Required probability = P(X > 4)

Question 7.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and 0 otherwise.
Compute:
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(-0.5 < X or X > 0.5).
Solution:
Since, f is the p.d.f. of X,

Question 8.
The following is the p.d.f. of continuous r.v. X
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
(i) Find expression for c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7 and 5.
Solution:
(i) Let F(x) be the c.d.f. of X

Question 9.
Given the p.d.f. of a continuous random r.v. X, f(x) = \(\frac{x^{2}}{3}\), for -1 < x < 2 and = 0 otherwise. Determine c.d.f. of X and hence find P(X < 1); P(X < -2), P(X > 0), P(1 < X < 2).
Solution:

Question 10.
If a r.v. X has p.d.f.
f(x) = \(\frac{c}{x}\) for 1 < x < 3, c > 0. Find c, E(X), Var (X).
Solution:
Since f(x) is p.d.f of r.v. X

12th Maharashtra State Board Maths Solutions Pdf Part 2

• Probability Distributions Ex 7.1 Class 12 Maths Solutions
• Probability Distributions Ex 7.2 Class 12 Maths Solutions
• Probability Distributions Miscellaneous Exercise 7 Class 12 Maths Solutions
• Binomial Distribution Ex 8.1 Class 12 Maths Solutions
• Binomial Distribution Miscellaneous Exercise 8 Class 12 Maths Solutions

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions And Answers Maharashtra Board

(I) Select the correct answers from the given alternatives.

Question 1.
The total number of terms in the expression of (x + y)¹⁰⁰ + (x – y)¹⁰⁰ after simplification is:
(A) 50
(B) 51
(C) 100
(D) 202
Answer:
(B) 51
Hint:

Question 2.
The middle term in the expansion of (1 + x)²ⁿ will be:
(A) (n – 1)^th
(B) n^th
(C) (n + 1)^th
(D) (n + 2)^th
Answer:
(C) (n + 1)^th
Hint:
(1 + x)²ⁿ has (2n + 1) terms.
∴ (n + 1 )^th term is the middle term.

Question 3.
In the expansion of (x² – 2x)¹⁰, the coefficient of x¹⁶ is
(A) -1680
(B) 1680
(C) 3360
(D) 6720
Answer:
(C) 3360
Hint:
(x² – 2x)¹⁰ = x¹⁰ (x – 2)¹⁰
To get the coefficient of x¹⁶ in (x² – 2x)¹⁰,
we need to check coefficient of x⁶ in (x – 2)¹⁰
∴ Required coefficient = ¹⁰C₆ (-2)⁴
= 210 × 16
= 3360

Question 4.
The term not containing x in expansion of \((1-x)^{2}\left(x+\frac{1}{x}\right)^{10}\) is
(A) ¹¹C₅
(B) ¹⁰C₅
(C) ¹⁰C₄
(D) ¹⁰C₇
Answer:
(A) ¹¹C₅
Hint:

Question 5.
The number of terms in expansion of (4y + x)⁸ – (4y – x)⁸ is
(A) 4
(B) 5
(C) 8
(D) 9
Answer:
(A) 4
Hint:

Question 6.
The value of ¹⁴C₁ + ¹⁴C₃ + ¹⁴C₅ + …. + ¹⁴C₁₁ is
(A) 2¹⁴ – 1
(B) 2¹⁴ – 14
(C) 2¹²
(D) 2¹³ – 14
Answer:
(D) 2¹³ – 14
Hint:

Question 7.
The value of ¹¹C₂ + ¹¹C₄ + ¹¹C₆ + ¹¹C₈ is equal to
(A) 2¹⁰ – 1
(B) 2¹⁰ – 11
(C) 2¹⁰ + 12
(D) 2¹⁰ – 12
Answer:
(D) 2¹⁰ – 12
Hint:

Question 8.
In the expansion of (3x + 2)⁴, the coefficient of the middle term is
(A) 36
(B) 54
(C) 81
(D) 216
Answer:
(D) 216
Hint:
(3x + 2)⁴ has 5 terms.
∴ (3x + 2)⁴ has 3rd term as the middle term.
The coefficient of the middle term

= 6 × 9 × 4
= 216

Question 9.
The coefficient of the 8th term in the expansion of (1 + x)¹⁰ is:
(A) 7
(B) 120
(C) ¹⁰C₈
(D) 210
Answer:
(B) 120
Hint:
r = 7
t₈ = ¹⁰C₇ x⁷ = ¹⁰C₃ x⁷
∴ Coefficient of 8th term = ¹⁰C₃ = 120

Question 10.
If the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are the same, then the value of a is
(A) \(-\frac{7}{9}\)
(B) \(-\frac{9}{7}\)
(C) \(\frac{7}{9}\)
(D) \(\frac{9}{7}\)
Answer:
(D) \(\frac{9}{7}\)
Hint:

(II) Answer the following.

Question 1.
Prove by the method of induction, for all n ∈ N.
(i) 8 + 17 + 26 + ….. + (9n – 1) = \(\frac{n}{2}\) (9n + 7)
Solution:
Let P(n) ≡ 8 + 17 + 26 +…..+(9n – 1) = \(\frac{n}{2}\) (9n + 7), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 8
R.H.S. = \(\frac{1}{2}\) [9(1) + 7] = 8
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 8 + 17 + 26 +…..+ (9k – 1) = \(\frac{k}{2}\) (9k + 7) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., 8 + 17 + 26 + …… + [9(k + 1) – 1]

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 8 + 17 + 26 +…..+ (9n – 1) = \(\frac{n}{2}\) (9n + 7) for all n ∈ N.

(ii) 1² + 4² + 7² + …… + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1)
Solution:
Let P(n) = 1² + 4² + 7² + ….. + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S.= 1² = 1
R.H.S.= \(\frac{1}{2}\) [6(1)² – 3(1) – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 4² + 7² +…..+ (3k – 2)² = \(\frac{k}{2}\) (6k² – 3k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 4² + 7² + … + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1) for all n ∈ N.

(iii) 2 + 3.2 + 4.2² + …… + (n + 1) 2^n-1 = n . 2ⁿ
Solution:
Let P(n) ≡ 2 + 3.2 + 4.2² +…..+ (n + 1) 2^n-1 = n.2ⁿ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(2¹) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 3.2 + 4.2² + ….. + (k + 1) 2^k-1 = k.2^k …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 3.2 + 4.2² +….+ (k + 2) 2^k = (k + 1) 2^k+1

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 3.2 + 4.2² +……+ (n + 1) 2^n-1 = n.2ⁿ for all n ∈ N.

(iv) \(\frac{1}{3.4 .5}+\frac{2}{4.5 .6}+\frac{3}{5.6 .7}+\ldots+\frac{n}{(n+2)(n+3)(n+4)}\) = \(\frac{n(n+1)}{6(n+3)(n+4)}\)
Solution:

Question 2.
Given that t_n+1 = 5t_n – 8, t₁ = 3, prove by method of induction that t_n = 5^n-1 + 2.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation t_n+1 = 5t_n – 8, t₁ = 3
and R.H.S. a general statement t_n = 5^n-1 + 2.
Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 5^1-1 + 2 = 1 + 2 = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S = t₂ = 5t₁ – 8 = 5(3) – 8 = 7
R.H.S. = t₂ = 5^2-1 + 2 = 5 + 2 = 7
∴ L.H.S. = R.H.S.
∴ P(n) is tme for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1 = 5t_k – 8 and t_k = 5^k-1 + 2

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
t_k+1 = 5^k+1-1 + 2 = 5^k + 2
t_k+1 = 5t_k – 8 and t_k = 5^k-1 + 2 ……[From Step II]
∴ t_k+1 = 5(5^k-1 + 2) – 8 = 5^k + 2
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n = 5^n-1 + 2, for all n ∈ N.

Question 3.
Prove by method of induction
\(\left(\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right)^{n}=\left(\begin{array}{cc}
2 n+1 & -4 n \\
n & -2 n+1
\end{array}\right)\), ∀ n ∈ N.
Solution:


Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\left(\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right)^{n}=\left(\begin{array}{cc}
2 n+1 & -4 n \\
n & -2 n+1
\end{array}\right)\), ∀ n ∈ N.

Question 4.
Expand (3x² + 2y)⁵
Solution:
Here, a = 3x², b = 2y, n = 5.
Using binomial theorem,

Question 5.
Expand \(\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{4}\)
Solution:

Question 6.
Find third term in the expansion of \(\left(9 x^{2}-\frac{y^{3}}{6}\right)^{4}\)
Solution:

Question 7.
Find tenth term in the expansion of \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\)
Solution:

Question 8.
Find the middle term(s) in the expansion of
(i) \(\left(\frac{2 a}{3}-\frac{3}{2 a}\right)^{6}\)
Solution:
Here, a = \(\frac{2 a}{3}\), b = \(\frac{-3}{2 a}\), n = 6.
Now, n is even.
∴ \(\frac{\mathrm{n}+2}{2}=\frac{6+2}{2}=4\)
∴ Middle term is t4, for which r = 3.

∴ The Middle term is -20.

(ii) \(\left(x-\frac{1}{2 y}\right)^{10}\)
Solution:
Here, a = x, b = \(-\frac{1}{2 y}\), n = 10.
Now, n is even.
∴ \(\frac{\mathrm{n}+2}{2}=\frac{10+2}{2}=6\)
∴ Middle term is t6, for which r = 5

(iii) (x² + 2y²)⁷
Solution:
Here, a = x², b = 2y², n = 7.
Now, n is odd.
∴ \(\frac{\mathrm{n}+1}{2}=\frac{7+1}{2}=4, \frac{\mathrm{n}+3}{2}=\frac{7+3}{2}=5\)
∴ Middle terms are t₄ and t₅, for which r = 3 and r = 4 respectively.

∴ Middle terms are 280x⁸y⁶ and 560x⁶y⁸.

(iv) \(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)
Solution:

Question 9.
Find the coefficients of
(i) x⁶ in the expantion of \(\left(3 x^{2}-\frac{1}{3 x}\right)^{9}\)
Solution:

(ii) x⁶⁰ in the expansion of \(\left(\frac{1}{x^{2}}+x^{4}\right)^{18}\)
Solution:

Question 10.
Find the constant term in the expansion of
(i) \(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Solution:

(ii) \(\left(2 x^{2}-\frac{1}{x}\right)^{12}\)
Solution:

Question 11.
Prove by method of induction
(i) log_a xⁿ = n log_a x, x > 0, n ∈ N
Solution:

(ii) 15^2n-1 + 1 is divisible by 16, for all n ∈ N.
Solution:
15^2n-1 + 1 is divisible by 16, if and only if (15^2n-1 + 1) is is a multiple of 16.
Let P(n) ≡ 15^2n-1 + 1 = 16m, where m ∈ N.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 15^2n-1 + 1 is divisible by 16, for all n ∈ N.

(iii) 5²ⁿ – 2²ⁿ is divisible by 3, for all n ∈ N.
Solution:

Question 12.
If the coefficient of x¹⁶ in the expansion of (x² + ax)¹⁰ is 3360, find a.
Solution:

Question 13.
If the middle term in the expansion of \(\left(x+\frac{b}{x}\right)^{6}\) is 160, find b.
Solution:

∴ 160 = \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times b^{3}\)
∴ 160 = 20b³
∴ 8 = b³
∴ b = 2

Question 14.
If the coefficients of x² and x³ in theexpansion of (3 + kx)⁹ are equal, find k.
Solution:

Question 15.
If the constant term in the expansion of \(\left(x^{3}+\frac{\mathrm{k}}{x^{8}}\right)^{11}\) is 1320, find k.
Solution:

Question 16.
Show that there is no term containing x⁶ in the expansion of \(\left(x^{2}-\frac{3}{x}\right)^{11}\).
Solution:

Question 17.
Show that there is no constant term in the expansion of \(\left(2 x-\frac{x^{2}}{4}\right)^{9}\)
Solution:

Question 18.
State, first four terms in the expansion of \(\left(1-\frac{2 x}{3}\right)^{-1 / 2}\)
Solution:

Question 19.
State, first four terms in the expansion of \((1-x)^{-1 / 4}\).
Solution:

Question 20.
State, first three terms in the expansion of \((5+4 x)^{-1 / 2}\)
Solution:

Question 21.
Using the binomial theorem, find the value of \(\sqrt[3]{995}\) upto four places of decimals.
Solution:

Question 22.
Find approximate value of \(\frac{1}{4.08}\) upto four places of decimals.
Solution:

Question 23.
Find the term independent of x in the expansion of (1 – x²) \(\left(x+\frac{2}{x}\right)^{6}\).
Solution:

Question 24.
(a + bx) (1 – x)⁶ = 3 – 20x + cx² + …, then find a, b, c.
Solution:

Question 25.
The 3rd term of (1 + x)ⁿ is 36x². Find 5th term.
Solution:

Question 26.
Suppose (1 + kx)ⁿ = 1 – 12x + 60x² – …… find k and n.
Solution:

Maths Solutions for Class 11 State Board

• Methods of Induction and Binomial Theorem Ex 4.1 Class 11 Maths Solutions
• Methods of Induction and Binomial Theorem Ex 4.2 Class 11 Maths Solutions
• Methods of Induction and Binomial Theorem Ex 4.3 Class 11 Maths Solutions
• Methods of Induction and Binomial Theorem Ex 4.4 Class 11 Maths Solutions
• Methods of Induction and Binomial Theorem Ex 4.5 Class 11 Maths Solutions
• Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Class 11 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.1 Questions and Answers.

12th Maths Part 2 Probability Distributions Exercise 7.1 Questions And Answers Maharashtra Board

Question 1.
Let X represent the difference between a number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution:
When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6.
The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0.
∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0
i.e. -6, -4, -2, 0, 2, 4, 6.
∴ X = {-6, -4, -2, 0, 2, 4, 6}.

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
The urn contains 5 red and 2 black balls.
If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.
X can take values 0, 1, 2.
∴ X = {0, 1, 2}.

Question 3.
State which of the following are not the probability mass function of a random variable. Give reasons for your answer.

Solution:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 ≤ p_i ≤ 1
(b) Σp_i = 1.

(i)

(a) Here 0 ≤ p_i ≤ 1
(b) Σp_i = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(ii)

P(X = 3) = -0.1, i.e. P_i < 0 which does not satisfy 0 ≤ P_i ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

(iii)

(a) Here 0 ≤ p_i ≤ 1
(b) ∑p_i = 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(iv)

Here ∑p_i = 0.3 + 0.2 + 0.4 + 0 + 0.05 = 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.

(v)

Here ∑p_i = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.

(vi)

(a) Here 0 ≤ p_i ≤ 1
(b) ∑p_i = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Solution:
(i) For two tosses of a coin the sample space is {HH, HT, TH, TT}
Let X denote the number of heads in two tosses of a coin.
Then X can take values 0, 1, 2.
∴ P[X = 0] = P(0) = \(\frac{1}{4}\)
P[X = 1] = P(1) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P[X = 2] = P(2) = \(\frac{1}{4}\)
∴ the required probability distribution is

(ii) When three coins are tossed simultaneously, then the sample space is
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let X denotes the number of tails.
Then X can take the value 0, 1, 2, 3.
∴ P[X = 0] = P(0) = \(\frac{1}{8}\)
P[X = 1] = P(1) = \(\frac{3}{8}\)
P[X = 2] = P(2) = \(\frac{3}{8}\)
P[X = 3] = P(3) = \(\frac{1}{8}\)
∴ the required probability distribution is

(iii) When a fair coin is tossed 4 times, then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
Let X denotes the number of heads.
Then X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n(X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
∴ the probability distribution of X is as follows:

Question 5.
Find the probability distribution of a number of successes in two tosses of a die, where success is defined as a number greater than 4 appearing on at least one die.
Solution:
When a die is tossed twice, the sample space s has 6 × 6 = 36 sample points.
∴ n(S) = 36
The trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears.
Then X can take values 0, 1, 2.
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ n(X) = 16.
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ n(X) = 16
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 2, i.e. 5 or 6 appear on both of the dice, then
X = {(5, 5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)
∴ the required probability distribution is

Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution:
Here, the number of defective bulbs is the random variable.
Let the number of defective bulbs be denoted by X.
∴ X can take the value 0, 1, 2, 3, 4.
Since the draws are done with replacement, therefore the four draws are independent experiments.
Total number of bulbs is 30 which include 6 defectives.
∴ P(X = 0) = P(0) = P(all 4 non-defective bulbs)
= \(\frac{24}{30} \times \frac{24}{30} \times \frac{24}{30} \times \frac{24}{30}\)
= \(\frac{256}{625}\)
P(X = 1) = P (1) = P (1 defective and 3 non-defective bulbs)

P(X = 2) = P(2) = P(2 defective and 2 non-defective)

P(X = 3) = P(3) = P(3 defectives and 1 non-defective)

P(X = 4) = P(4) = P(all 4 defectives)
= \(\frac{6}{30} \times \frac{6}{30} \times \frac{6}{30} \times \frac{6}{30}\)
= \(\frac{1}{625}\)
∴ the required probability distribution is

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. If the coin is tossed twice. Find the probability distribution of a number of tails.
Solution:
Given a biased coin such that heads is 3 times as likely as tails.
∴ P(H) = \(\frac{3}{4}\) and P(T) = \(\frac{1}{4}\)
The coin is tossed twice.
Let X can be the random variable for the number of tails.
Then X can take the value 0, 1, 2.
∴ P(X = 0) = P(HH) = \(\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)
P(X = 1) = P(HT, TH) = \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}=\frac{6}{16}=\frac{3}{8}\)
P(X = 2) = P(TT) = \(\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
∴ the required probability distribution is

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X < 3) (iii) P(X > 4)
Solution:
(i) Since P (x) is a probability distribution of x,
\(\sum_{x=0}^{7} P(x)=1\)
⇒ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(10k – 1) = 0
⇒ 10k – 1 = 0 ……..[∵ k ≠ -1]
⇒ k = \(\frac{1}{10}\)

(ii) P(X< 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

(iii) P(0 < X < 3) = P (1) + P (2)
= k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

Question 9.
Find expected value and variance of X for the following p.m.f.:

Solution:
We construct the following table to calculate E(X) and V(X):

From the table,
Σx_ip_i = -0.05 and \(\Sigma x_{i}^{2} \cdot p_{i}\) = 2.25
∴ E(X) = Σx_ip_i = -0.05
and V(X) = \(\Sigma x_{i}^{2}+p_{i}-\left(\sum x_{i}+p_{i}\right)^{2}\)
= 2.25 – (-0.05)²
= 2.25 – 0.0025
= 2.2475
Hence, E(X) = -0.05 and V(X) = 2.2475.

Question 10.
Find expected value and variance of X, where X is the number obtained on the uppermost face when a fair die is thrown.
Solution:
If a die is tossed, then the sample space for the random variable X is
S = {1, 2, 3, 4, 5, 6}
∴ P(X) = \(\frac{1}{6}\); X = 1, 2, 3, 4, 5, 6.

Hence, E(X) = 3.5 and V(X) = 2.9167.

Question 11.
Find the mean number of heads in three tosses of a fair coin.
Solution:
When three coins are tossed the sample space is {HHH, HHT, THH, HTH, HTT, THT, TTH, TTT}
∴ n(S) = 8
Let X denote the number of heads when three coins are tossed.
Then X can take values 0, 1, 2, 3
P(X = 0) = P(0) = \(\frac{1}{8}\)
P(X = 1) = P(1) = \(\frac{3}{8}\)
P(X = 2) = P(2) = \(\frac{3}{8}\)
P(X = 3) = P(3) = \(\frac{1}{8}\)
∴ mean = E(X) = Σx_iP(x_i)
= \(0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}\)
= \(0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\)
= \(\frac{12}{8}\)
= 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
When two dice are thrown, the sample space S has 6 × 6 = 36 sample points.
∴ n(S) = 36
Let X denote the number of sixes when two dice are thrown.
Then X can take values 0, 1, 2
When X = 0, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(X) = 25
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{25}{36}\)
When X = 1, then
X = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(X) = 10
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{10}{36}\)
When X = 2, then X = {(6, 6)}
∴ n(X) = 1
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{1}{36}\)
∴ E(X) = ΣxiP(xi)
= \(0 \times \frac{25}{36}+1 \times \frac{10}{36}+2 \times \frac{1}{36}\)
= \(0+\frac{10}{36}+\frac{2}{36}\)
= \(\frac{1}{3}\)

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
Two numbers are chosen from the first 6 positive integers.
∴ n(S) = \({ }^{6} C_{2}=\frac{6 \times 5}{1 \times 2}\) = 15
Let X denote the larger of the two numbers.
Then X can take values 2, 3, 4, 5, 6.
When X = 2, the other positive number which is less than 2 is 1.
∴ n(X) = 1
∴ P(X = 2) = P(2) = \(\frac{n(X)}{n(S)}=\frac{1}{15}\)
When X = 3, the other positive number less than 3 can be 1 or 2 and hence can be chosen in 2 ways.
∴ n(X) = 2
P(X = 3) = P(3) = \(\frac{n(X)}{n(S)}=\frac{2}{15}\)
Similarly, P(X = 4) = P(4) = \(\frac{3}{15}\)
P(X = 5) = P(5) = \(\frac{4}{15}\)
P(X = 6) = P(6) = \(\frac{5}{15}\)
∴ E(X) = Σx_iP(x_i)
= \(2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}\)
= \(\frac{2+6+12+20+30}{15}\)
= \(\frac{70}{15}\)
= \(\frac{14}{3}\)

Question 14.
Let X denote the sum of numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution:
If two fair dice are rolled then the sample space S of this experiment is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let X denote the sum of the numbers on uppermost faces.
Then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

∴ the probability distribution of X is given by

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the student is recorded. What is the probability distribution of the random variable X? Find mean, variance, and standard deviation of X.
Solution:
Let X denote the age of the chosen student. Then X can take values 14, 15, 16, 17, 18, 19, 20, 21.
We make a frequency table to find the number of students with age X:

The chances of any student selected are equally likely.
If there are m students with age X, then P(X) = \(\frac{m}{15}\)
Using this, the following is the probability distribution of X:


Variance = V(X) = \(\Sigma x_{i}^{2}\) . P(x_i) – [E(X)]²
= 312.2 – (17.53)²
= 312.2 – 307.3
= 4.9
Standard deviation = √V(X) = √4.9 = 2.21
Hence, mean = 17.53, variance = 4.9 and standard deviation = 2.21.

Question 16.
In a meeting, 70% of the member’s favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
X takes values 0 and 1.
It is given that
P(X = 0) = P(0) = 30% = \(\frac{30}{100}\) = 0.3
P(X = 1) = P(1) = 70% = \(\frac{70}{100}\) = 0.7
∴ E(X) = Σx_i . P(x_i) = 0 × 0.3 + 1 × 0.7 = 0.7
Also, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 0 × 0.3 + 1 × 0.7 = 0.7
∴ Variance = V(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)-[E(X)]^{2}\)
= 0.7 – (0.7)²
= 0.7 – 0.49
= 0.21
Hence, E(X) = 0.7 and Var(X) = 0.21.

12th Maharashtra State Board Maths Solutions Pdf Part 2

• Probability Distributions Ex 7.1 Class 12 Maths Solutions
• Probability Distributions Ex 7.2 Class 12 Maths Solutions
• Probability Distributions Miscellaneous Exercise 7 Class 12 Maths Solutions
• Binomial Distribution Ex 8.1 Class 12 Maths Solutions
• Binomial Distribution Miscellaneous Exercise 8 Class 12 Maths Solutions