Category: Class 11

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.1

Question 1.
Find 77% of 580 + 34% of 390.
Solution:

Question 2.
240 candidates appeared for an examination, of which 204 passed. What is the pass percentage?
Solution:
We find the pass percentage using the unitary method

∴ The pass percentage for the examination is 85%.

Question 3.
What percent of 8.4 kg are 168 grams?
Solution:
Let 168 gms be x% of 8.4 kg
i.e., let 168 gms be \(\frac{x}{100}\) of 8400 gms
∴ 168 = \(\frac{x}{100}\) × 8400
∴ x = \(\frac{168}{84}\) = 2
∴ 168 gms is 2% of 8.4 kg.

Question 4.
If the length of a rectangle is decreased by 20%, what should be the increase in the breadth of the rectangle so that the area remains the same?
Solution:
Let x and y represent the length and breadth of the rectangle respectively.
∴ The original area of the rectangle = xy
There is a 20% decrease in length.

Let k % be the required increase in breadth

Given that the new and old areas should be equal.

∴ 100 + k = 125
∴ k = 125 – 100 = 25
∴ Breadth should be increased by 25% so that the area remains same.

Question 5.
The price of rice increased by 20%, as a result, a person can have 5kg rice for ₹ 600. What was the initial price of rice per kg?
Solution:
A person can buy 5 kg of rice for ₹ 600 after the increase in price
∴ New price of rice = \(\frac{600}{5}\) = ₹ 120/kg …..(i)
Let ‘x’ be the initial price per kg of rice.
There is a 20% increase in the price of rice.
Thus the new price of the rice will be given as

∴ The initial price of rice is ₹ 100 per kg

Question 6.
What percent is 3% of 5%?
Solution:
Let 3% be x % of 5%.
Then \(\frac{3}{100}=\frac{x}{100} \times \frac{5}{100}\)
∴ x = \(\frac{3 \times 100}{5}\) = 60
∴ 3% is 60% of 5%.

Question 7.
After availing of two successive discounts of 20% each, Madhavi paid ₹ 64 for a book. If she would have got only one discount of 20%, how much additional amount would she have paid?
Solution:
Let the price of the book be ₹ x.
After the first 20% discount, the price of the book becomes

After another 20% discount, the price of the book becomes

This price = ₹ 64 …..[Given]
∴ \(\frac{16}{25}\)x = 64
∴ x = 4 × 25 = 100
Thus, Amount of the book after one discount = \(\frac{4}{5}\)(100) = 80 …..[from (i)]
∴ The additional amount that Madhavi would have paid = 80 – 64 = ₹ 16

Question 8.
The price of the table is 40% more than the price of a chair. By what percent price of a chair is less than the price of a table?
Solution:
Let ₹ x and ₹ y be the price of a table and chair respectively.
The price of the table is 40% more than the price of a chair
∴ \(\frac{x-y}{y}\) × 100 = 40

We need to find by how much percent is the price of a chair less than that of a table.

∴ The price of a chair is 28.57% less than the price of a table.

Question 9.
A batsman scored 92 runs which includes 4 boundaries 5 sixes. He scored other runs by running between the wickets. What percent of his total score did he make by running between the wickets?
Solution:
Batsman scores 4 fours (boundaries) and 5 sixes in 92 runs.
Number of runs scored by fours and sixes = 4 × 4 + 5 × 6 = 46
∴ 92 – 46 = 46
Let 46 be x% of 92.
Then 46 = \(\frac{x}{100}\) × 92
∴ x = \(\frac{46 \times 100}{92}=\frac{100}{2}\) = 50
∴ 50% of the total runs were scored by running between the wickets.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Miscellaneous Exercise 8

Solve the following system of inequalities graphically.

Question 1.
x ≥ 3, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
2x + y ≥ 6, 3x + 4y < 12
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
x + y ≥ 4, 2x – y ≤ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
2x – y ≥1, x – 2y ≤ -1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
x + y ≤ 6, x + y ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 10.
3x + 4y ≤ 60, x +3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 14.
3x + 2y ≤ 150, x + 4y ≥ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 15.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.3

Find the graphical solution for the following system of linear inequations.

Question 1.
x – y ≤ 0, 2x – y ≥ -2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
2x + 3y ≥ 12, -x + y ≤ 3, x ≤ 4, y ≥ 3
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
3x + 2y ≤ 1800, 2x + 7y ≤ 1400
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
0 ≤ x ≤ 350, 0 ≤ y ≤ 150
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
\(\frac{x}{60}+\frac{y}{90}\) ≤ 1, \(\frac{x}{120}+\frac{y}{75}\) ≤ 1, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
3x + 2y ≤ 24, 3x + y ≥ 15, x ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2

Question 1.
The following table shows the classification of applications for secretarial and for sales positions according to gender. Calculate the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{225 \times 100}{300}\) = 75
E₁₂ = \(\frac{225 \times 200}{300}\) = 150
E₂₁ = \(\frac{75 \times 100}{300}\) = 25
E₂₂ = \(\frac{75 \times 200}{300}\) = 50
Table of expected frequencies.

Question 2.
200 teenagers were asked which takeaway food do they prefer – French fries, burgers, or pizza. The results were-

Compute ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 24}{200}\) = 6
E₁₂ = \(\frac{50 \times 60}{200}\) = 15
E₁₃ = \(\frac{50 \times 116}{200}\) = 29
E₂₁ = \(\frac{150 \times 24}{200}\) = 18
E₂₂ = \(\frac{150 \times 60}{200}\) = 45
E₂₃ = \(\frac{150 \times 116}{200}\) = 87
Table of expected frequencies.

Question 3.
A sample of men and women who had passed their driving test either in 1st attempt or in 2nd attempt
were surveyed. Compute ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{60 \times 40}{80}\) = 30
E₁₂ = \(\frac{60 \times 40}{80}\) = 30
E₂₁ = \(\frac{20 \times 40}{80}\) = 10
E₂₂ = \(\frac{20 \times 40}{80}\) = 10
Table of expected frequencies.

Question 4.
800 people were asked whether they wear glasses for reading with the following results.

Compute the ϰ² square statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{400 \times 600}{800}\) = 300
E₁₂ = \(\frac{400 \times 200}{800}\) = 100
E₂₁ = \(\frac{400 \times 600}{800}\) = 300
E₂₂ = \(\frac{400 \times 200}{800}\) = 100
Table of expected frequencies.

Question 5.
Out of a sample of 120 persons in a village, 80 were administered a new drug for preventing influenza, and out of the 18 were attacked by influenza. Out of those who are not administered the new drug, 10 persons were not attacked by influenza:
(i) Prepare a two-way table showing frequencies.
(ii) Compute the ϰ² square statistic.
Solution:
(i) The given data can be arranged in the following table.

The observed frequency table can be prepared as follows:

(ii) Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{48 \times 80}{120}\) = 32
E₁₂ = \(\frac{48 \times 40}{120}\) = 16
E₂₁ = \(\frac{72 \times 80}{120}\) = 48
E₂₂ = \(\frac{72 \times 40}{120}\) = 24
Table of expected frequencies.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1

Question 1.
The following table gives income (X) and expenditure (Y) of 25 families:

Find
(i) Marginal frequency distributions of income and expenditure.
(ii) Conditional frequency distribution of X when Y is between 300 – 400.
(iii) Conditional frequency distribution of Y when X is between 200 – 300.
(iv) How many families have their income ₹ 300 and more and expenses ₹ 400 and less?
Solution:
The bivariate frequency distribution table for Income (X) and Expenditure (Y) is as follows:

(i) Marginal frequency distribution of income (X):

Marginal frequency distribution of expenditure ( Y):

(ii) Conditional frequency distribution of X when Y is between 300 – 400:

(iii) Conditional frequency distribution of Y when X is between 200 – 300:

(iv) The cells 300 – 400 and 400 – 500 are having income ₹ 300 and more and the cells 200 – 300 and 300 – 400 are having expenditure ₹ 400 and less.
Now, the following table indicates the number of families satisfying the above condition.

∴ There are 17 families with income ₹ 300 and more and expenditure ₹ 400 and less.

Question 2.
Two dice are thrown simultaneously 25 times. The following pairs of observations are obtained.
(2, 3) (2, 5) (5, 5) (4, 5) (6, 4) (3, 2) (5, 2) (4, 1) (2, 5) (6, 1) (3, 1) (3, 3) (4, 3) (4, 5) (2, 5) (3, 4) (2, 5) (3, 4) (2, 5) (4, 3) (5, 2) (4, 5) (4, 3) (2, 3) (4, 1)
Prepare a bivariate frequency distribution table for the above data. Also, obtain the marginal distributions.
Solution:
Let X = Observation on 1st die
Y = Observation on 2nd die
Now, the minimum value of X is 1 and the maximum value is 6.
Also, the minimum value of Y is 1 and the maximum value is 6.
A bivariate frequency distribution can be prepared by taking X as row and Y as a column.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Question 3.
Following data gives the age of husbands (X) and age of wives (Y) in years. Construct a bivariate frequency distribution table and find the marginal distributions.

Find conditional frequency distribution of age of husbands when the age of wife is 23 years.
Solution:
Given, X = Age of Husbands (in years)
Y = Age of Wives (in years)
Now, the minimum value of X is 25 and the maximum value is 29.
Also, the minimum value of Y is 19 and the maximum value is 23.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of X when Y is 23:

Question 4.
Construct a bivariate frequency distribution table of the marks obtained by students in Statistics (X) and English (Y).

Construct a bivariate frequency distribution table for the above data by taking class intervals 20 – 30, 30 – 40, …. etc. for both X and Y. Also find the marginal distributions and conditional frequency distribution of Y when X lies between 30 – 40.
Solution:
Given, X = Marks in Statistics
Y = Marks in English
A bivariate frequency table can be prepared by taking class intervals 20 – 30, 30 – 40,…, etc for both X and Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when X lies between 30 – 40:

Question 5.
Following data gives height in cms (X) and weight in kgs (Y) of 20 boys. Prepare a bivariate frequency table taking class intervals 150 – 154, 155 – 159,…etc. for X and 35 – 39, 40 – 44,…, etc for Y. Also, find
(i) Marginal frequency distributions.
(ii) Conditional frequency distribution of Y when 155 ≤ X ≤ 159.
(152,40) (160,54) (163,52) (150,35) (154,36) (160,49) (166,54) (157,38)
(159,43) (153,48) (152,41) (158,51) (155,44) (156,47) (156,43) (166,53)
(160,50) (151,39) (153,50) (158,46)
Solution:
Given X = Height in cms.
Y = Weight in kgs.
Bivariate frequency table can be prepared by taking class intervals 150 – 154, 155 – 159, …, etc for X and 35 – 39, 40 – 44,…etc for Y.
The bivariate frequency distribution table is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when 155 ≤ X ≤ 159:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Skewness Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Skewness Miscellaneous Exercise 3

Question 1.
For u distribution, mean = 100, mode = 80 and S.D. = 20. Find Pearsonian coefficient of skewness Sk_p.
Solution:
Given, Mean = 100, Mode = 80, S.D. = 20

Question 2.
For a distribution, mean = 60, median = 75 and variance = 900. Find Pearsonian coefficient of skewness Sk_p.
Solution:
Given. Mean = 60, Median = 75, Variance = 900
∴ S.D. = √Variance = √900 = 30

Question 3.
For a distribution, Q₁ = 25, Q₂ = 35 and Q₃ = 50. Find Bowley’s coefficient of skewness Sk_b.
Solution:
Given Q₁ = 25, Q₂ = 35, Q₃ = 50

Question 4.
For a distribution Q₃ – Q₂ = 40, Q₂ – Q₁ = 60. Find Bowlev’s coefficient of skewness Sk_b.
Solution:
Given, Q₃ – Q₂ = 40, Q₂ – Q₁ = 60

Question 5.
For a distribution, Bowley’s coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.
Solution:
Given, Sk_b = 0.6, Q₃ + Q₁ = 100,
Median = Q₂ = 38
Skb = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)
∴ 0.6 = \(\frac{100-2(38)}{Q_{3}-Q_{1}}\)
∴ 0.6(Q₃ – Q₁) = 100 – 76 = 24
∴ Q₃ – Q₁ = 40 ….(i)
Q₃ + Q₁ = 100 …..(ii) (given)
Adding (i) and (ii), we get
2Q₃ = 140
∴ Q₃ = 70
Substituting the value of Q₃ in (ii), we get
70 + Q₁ = 100
∴ Q₁ = 100 – 70 = 30
∴ upper quartile = 70 and lower quartile = 30

Question 6.
For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution.
Solution:
Mean = \(\bar{x}\) = 200
Coefficient of variation, C.V. = 8%, Skp = 0.3
C.V. = \(\frac{\sigma}{\bar{x}} \times 100\), where σ = standard deviation
∴ 8 = \(\frac{\sigma}{200} \times 100\)
∴ σ = \(\frac{8 \times 200}{100}\) = 16
Now, Sk_p = \(\frac{\text { Mean – Mode }}{\text { S.D. }}\)
∴ 0.3 = \(\frac{200-\text { Mode }}{16}\)
∴ 0.3 × 16 = 200 – Mode
∴ Mode = 200 – 4.8 = 195.2
Since, Mean – Mode = 3(Mean – Median)
∴ 200 – 195.2 = 3(200 – Median)
∴ 4.8 = 600 – 3Median
∴ 3Median = 600 – 4.8 = 595.2
∴ Median = 198.4

Question 7.
Calculate Karl Pearsonian’s coefficient of skewness Skp from the follow ing data:

Solution:
The given table is the cumulative frequency table of more than type.
From this table, we have to prepare the frequency distribution table and then calculate the value of Sk_p.
Construct the following table:

From the table, N = 120, Σf_ix_i = 5490 and \(\sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}^{2}\) = 284600
Mean = \(\bar{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{5490}{120}\) = 45.75
Maximum frequency 42 is of the class 50 – 60
∴ Mode lies in the class 50 – 60
∴ L = 50, f₁ = 42, f₀ = 25, f₂ = 13, h = 10

Alternate Method:
Let u = \(\frac{x-45}{10}\)

\(\overline{\mathrm{u}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\mathrm{N}}=\frac{9}{120}\) = 0.075
∴ \(\bar{x}\) = 45 + 10(\(\bar{u}\))
= 45 + 10(0.075)
= 45 + 0.75
= 45.75
Var(u) = \(\sigma_{u}^{2}=\frac{\sum f_{i} u_{i}{ }^{2}}{N}-(\bar{u})^{2}\)
= \(\frac{335}{120}\) – (0.075)²
= 2.7917 – 0.0056
= 2.7861
Var(X) = h² × Var(u)
= 100 × 2.7861
= 278.61
S.D. = √278.61 = 16.6916
Maximum frequency 42 is of the class 50 – 60.
∴ Mode lies in the class 50 – 60.
∴ L = 50, f₁ = 42, f₀ = 25, f₂ = 13, h = 10

Question 8.
Calculate Bowley’s coefficient of skewness Skb from the following data.

Solution:
To calculate Bowley’s coefficient of skewness Skb, we construct the following table:

Here, N = 120
Q₁ class = class containing the \(\left(\frac{N}{4}\right)^{t h}\) observation
∴ \(\frac{N}{4}=\frac{120}{4}\) = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 30-40.
∴ L = 30, h = 10, f = 13, c.f. = 22

Q₂ class = class containing the \(\left(\frac{N}{2}\right)^{t h}\) observation
∴ \(\frac{\mathrm{N}}{2}=\frac{120}{2}\) = 60
Cumulative frequency which is just greater than (or equal to) 60 is 60.
∴ Q₂ lies in the class 40-50.
∴ L = 40, h = 10, f = 25, c.f. = 35

Q₃ class = class containing the \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90
Cumulative frequency which is just greater than (or equal to) 90 is 102.
∴ Q₃ lies in the class 50 – 60
∴ L = 50, h = 10, f = 42, c.f. = 60

Question 9.
Find Sk_p for the following set of observations:
18, 27, 10, 25, 31, 13, 28
Solution:
The given data can be arranged in ascending order as follows:
10, 13, 18, 25, 27, 28, 31
Here, n = 7
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{7+1}{2}\right)^{\text {th }}\) observation
= value of 4th observation
= 25
For finding standard deviation, we construct the following table:

Question 10.
Find Skb for the following set of observations:
18, 27, 10, 25, 31, 13, 28
Solution:
The given data can be arranged in ascending order as follows:
10, 13, 18, 25, 27, 28, 31
Here, n = 7
∴ Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of 2nd observation
∴ Q₁ = 13
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 2)th observation
= value of 4th observation
∴ Q₂ = 25
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 2)th observation
= value of 6th observation
∴ Q₃ = 28
Coefficient of skewness,

∴ Sk_b = -0.6

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.4

Question 1.
Two dice are thrown simultaneously, if at least one of the dice shows a number 5, what is the probability that sum of the numbers on two dice is 9?
Solution:
When two dice are thrown simultaneously, the sample space 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 A be the event that at least one die shows number 5.
∴ A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ n(A) = 11
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{11}{36}\)
Let B be the event that sum of the numbers on two dice is 9.
∴ B = {(3, 6), (4, 5), (5, 4), (6, 3)}
Also, A ∩ B = {(4, 5), (5, 4)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{2}{36}\)
∴ Probability of sum of numbers on two dice is 9, given that one dice shows number 5, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{2}{36}}{\frac{11}{36}}=\frac{2}{11}\)

Question 2.
A pair of dice is thrown. If sum of the numbers is an even number, what is the probability that it is a perfect square?
Solution:
When two dice are thrown simultaneously, the sample space 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 A be the event that sum of the numbers is an even number.
∴ A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
∴ n(A) = 18
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{18}{36}\)
Let B be the event that sum of outcomes is a perfect square.
∴ B = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
Also, A n B= {(1, 3), (2, 2), (3, 1)}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}\)
∴ Probability of sum of the numbers is a perfect square, given that sum of numbers is an even number, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{3}{36}}{\frac{18}{36}}=\frac{3}{18}=\frac{1}{6}\)

Question 3.
A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
Solution:
Two tickets can be drawn from 11 tickets with replacement in 11 × 11 = 121 ways.
∴ n(S) = 121
Let A be the event that the sum of two numbers is even.
The event A occurs, if either both the tickets with odd numbers or both the tickets with even numbers are drawn.
There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11.
∴ n(A) = 6 × 6 + 5 × 5
= 36 + 25
= 61
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{61}{121}\)
Let B be the event that the numbers tickets drawn are odd
∴ n(B) = 6 × 6 = 36
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}\)
Since 6 odd numbers are common between A and B.
∴ n(A ∩ B) = 6 × 6 = 36
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}\)
∴ Probability of both the numbers are odd, given that sum is even, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{36}{121}}{\frac{61}{121}}=\frac{36}{61}\)

Question 4.
A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as
A: a club card is drawn.
B: an ace card is drawn.
Determine whether events A and B are independent or not.
Solution:
One card can be drawn out of 52 cards in ⁵²C₁ ways.
∴ n(S) = ⁵²C₁
Let A be the event that a club card is drawn.
1 club card out of 13 club cards can be drawn in ¹³C₁ ways.
∴ n(A) = ¹³C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}\)
Let B be the event that an ace card is drawn.
An ace card out of 4 aces can be drawn in ⁴C₁ ways.
∴ n(B) = ⁴C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}\)
Since 1 card is common between A and B
∴ n(A ∩ B) = ¹C₁
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{1}{52}\) …….(i)
∴ P(A) × P(B) = \(\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}} \times \frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{13 \times 4}{52 \times 52}=\frac{1}{52}\) …….(ii)
From (i) and (ii), we get
P(A ∩ B) = P(A) × P(B)
∴ A and B are independent events.

Question 5.
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that,
(i) problem is solved?
(ii) problem is not solved?
(iii) exactly two students solve the problem?
Solution:
Let A be the event that student A can solve the problem.
B be the event that student B can solve the problem.
C be the event that student C can solve problem.
∴ P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\), P(C) = \(\frac{1}{5}\)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
P(B’) = 1 – P(B) = 1 – \(\frac{1}{4}\) = \(\frac{5}{4}\)
P(C’) = 1 – P(C) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Since A, B, C are independent events
∴ A’, B’, C’ are also independent events
(i) Let X be the event that problem is solved.
Problem can be solved if at least one of the three students solves the problem.
P(X) = P(at least one student solves the problem)
= 1 – P(no student solved problem)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\)
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)

(ii) Let Y be the event that problem is not solved
∴ P(Y) = P(A’ ∩ B’ ∩ C’)
= P(A’) P(B’) P(C’)
= \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\)
= \(\frac{2}{5}\)

(iii) Let Z be the event that exactly two students solve the problem.
∴ P(Z) = P(A ∩ B ∩ C’) ∪ P(A ∩ B’ ∩ C) ∪ P(A’ ∩ B ∩ C)
= P(A) . P(B) . P(C’) + P(A) . P(B’) . P(C) + P(A’) . P(B) . P(C)
= \(\left(\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}\right)+\left(\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right)\)
= \(\frac{4}{60}+\frac{3}{60}+\frac{2}{60}\)
= \(\frac{3}{20}\)

Question 6.
The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive for the next 10 years?
Solution:
Let A be the event that man will be alive at 60.
∴ P(A) = 0.83
Let B be the event that a woman will be alive at 55.
∴ P(B) = 0.97
A ∩ B = Event that both will be alive.
Also, A and B are independent events
∴ P(both man and his wife will be alive) = P(A ∩ B)
= P(A) . P(B)
= 0.83 × 0.97
= 0.8051

Question 7.
In an examination, 30% of the students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subjects I and subject II. A student is selected at random, what is the probability that the student
(i) has failed in the subject I, if it is known that he is failed in subject II?
(ii) has failed in at least one subject?
(iii) has failed in exactly one subject?
Solution:
Let A be the event that the student failed in Subject I
B be the event that the student failed in Subject II
Then P(A) = 30% = \(\frac{30}{100}\)
P(B) = 20% = \(\frac{20}{100}\)
and P(A ∩ B) = 10% = \(\frac{10}{100}\)
(i) P (student failed in Subject I, given that he has failed in Subject II) = P(A/B)
\(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\left(\frac{10}{100}\right)}{\left(\frac{20}{100}\right)}=\frac{10}{20}=\frac{1}{2}\)

(ii) P(student failed in at least one subject) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{30}{100}+\frac{20}{100}-\frac{10}{100}\)
= 0.40

(iii) P(student failed in exactly one subject) = P(A) + P(B) – 2P(A ∩ B)
= \(\frac{30}{100}+\frac{20}{100}-2\left(\frac{10}{100}\right)\)
= 0.30

Question 8.
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third gun respectively. Assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
Solution:
A be the event that first gun hits the target
B be the event that second gun hits the target
C be the event that third gun hits the target
P(A) = 0.5, P(B) = 0.6, P(C) = 0.8
∴ P(A’) = 1 – P(A) = 1 – 0.5 = 0.5
∴ P(B’) = 1 – P(B) = 1 – 0.6 = 0.4
∴ P(C’) = 1 – P(C) = 1 – 0.8 = 0.2
Now A, B, C are independent events
∴ A’, B’, C are also independent events.
∴ P (at least one hit is registered)
= 1 – P(no hit is registered)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – (0.5) (0.4) (0.2)
= 1 – 0.04
= 0.96

Question 9.
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that
(i) first is white and second is black?
(ii) one is white and the other is black?
Solution:
Total number of balls = 10 + 15 = 25
Let S be an event that two balls are drawn at random without replacement in succession
∴ n(S) = ²⁵C₁ × ²⁴C₁ = 25 × 24
(i) Let A be the event that the first ball is white and the second is black.
First white ball can be drawn from 10 white balls in ¹⁰C₁ ways
and second black ball can be drawn from 15 black balls in ¹⁵C₁ ways.
∴ n(A) = ¹⁰C₁ × ¹⁵C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{10} \mathrm{C}_{1} \times{ }^{15} \mathrm{C}_{1}}{25 \times 24}=\frac{10 \times 15}{25 \times 24}=\frac{1}{4}\)

Question 10.
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, what is the probability that at least one ball is black?
Solution:
Total number of balls in the urn = 4 + 5 + 6 = 15
Two balls can be drawn without replacement in ¹⁵C₂ = \(\frac{15 \times 14}{1 \times 2}\) = 105 ways
∴ n(S) = 105
Let A be the event that at least one ball is black
i.e., 1 black and 1 non-black or 2 black and 0 non-black.
1 black ball can be drawn out of 4 black balls in ⁴C₁ = 4 ways
and 1 non-black ball can be drawn out of remaining 11 non-black balls in ¹¹C₁ = 11 ways
∴ 1 black and 1 non black ball can be drawn in 4 × 11 = 44 ways
Also, 2 black balls can be drawn from 4 black balls in ⁴C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6 ways
∴ n(A) = 44 + 6 = 50
∴ Required probability = P(A) = \(\frac{n(A)}{n(S)}=\frac{50}{105}\) = \(\frac{10}{21}\)

Alternate Solution:
Total number of balls = 15
Required probability = 1 – P(neither of two balls is black)
Balls are drawn without replacement
Probability of first non-black ball drawn = \(\frac{11}{15}\)
Probability of second non-black ball drawn = \(\frac{10}{14}\)
Probability of neither of two balls is black = \(\frac{11}{15} \times \frac{10}{14}=\frac{11}{21}\)
Required probability = 1 – \(\frac{11}{21}\) = \(\frac{10}{21}\)

Question 11.
Two balls are drawn from an urn containing 5 green, 3 blue, 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Solution:
Total number of balls in the urn = 5 + 3 + 7 = 15
Out of these 12 are non-blue balls.
Two balls can be drawn from 15 balls without replacement in ¹⁵C₂
= \(\frac{15 \times 14}{1 \times 2}\)
= 105 ways.
∴ n(S) = 105
Let A be the event that at least one ball is blue,
i.e., 1 blue and other non-blue or both are blue.
∴ n(A) = ³C₁ × ¹²C₁ + ³C₂
= 3 × 12 + 3
= 36 + 3
= 39
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{39}{105}=\frac{13}{35}\)

Alternate solution:
Total number of balls in the urn = 15
Required probability = 1 – P(neither of two balls is blue)
Balls are drawn one by one without replacement.
Probability of first non-blue ball drawn = \(\frac{12}{15}\)
Probability of second non-blue ball drawn = \(\frac{11}{14}\)
Probability of neither of two ball is blue = \(\frac{12}{15} \times \frac{11}{14}=\frac{22}{35}\)
∴ Required probability = 1 – \(\frac{22}{35}\) = \(\frac{13}{35}\)

Question 12.
A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball ¡s drawn from each bag, what is the Probability that two balls are of the same colour?
Solution:
Let A be the event that a blue ball is drawn from each bag.
Probability of drawing one blue ball out of 4 blue balls where there are a total of 9 balls in the first bag and that of drawing one blue ball out of 3 blue balls where there are a total of 10 balls in the second bag is
P(A) = \(\frac{4}{9} \times \frac{3}{10}\)
Let B be the event that a green ball is drawn from each bag.
Probability of drawing one green ball out of 5 green balls where there are a total of 9 balls in the first bag and that of drawing one green ball out of 7 green balls where there are a total of 10 balls in the second bag is
P(B) = \(\frac{5}{9} \times \frac{7}{10}\)
Since both, the events are mutually exclusive and exhaustive events
∴ P(that both the balls are of the same colour) = P(both are of blue colour) or P(both are of green colour)
= P(A) + P(B)
= \(\frac{4}{9} \times \frac{3}{10}+\frac{5}{9} \times \frac{7}{10}\)
= \(\frac{12}{90}+\frac{35}{90}\)
= \(\frac{47}{90}\)

Question 13.
Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards are drawn are face cards?
Solution:
Two cards are drawn from a pack of 52 cards with replacement.
∴ n(S) = 52 × 52
Let A be the event that two cards drawn are face cards.
First card from 12 face cards is drawn with replacement in ¹²C₁ = 12 ways
and second face card is drawn from 12 face card in ¹²C₁ = 12 ways after replacement.
∴ n(A) = 12 × 12
∴ P(that both the cards drawn are face cards) = P(A)
= \(\frac{n(A)}{n(S)}=\frac{12 \times 12}{52 \times 52}=\frac{9}{169}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.3

Question 1.
Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or the number on the second die is greater than or equal to the number on the first die?
Solution:
When two dice are thrown, the sample space 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 A be the event that sum of numbers on two dice is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}\)
Let B be the event that number on second die is greater than or equal to number on first die.
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)}
∴ n(B) = 21
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{21}{36}\)
Now, A ∩ B = {(1, 4), (2, 3)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{36}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{4}{36}+\frac{21}{36}-\frac{2}{36}\)
= \(\frac{23}{36}\)

Question 2.
A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either red or face card?
Solution:
One card can be drawn from the pack of 52 cards in ⁵²C₁ = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that a red card is drawn Red card can be drawn in ²⁶C₁ = 26 ways
∴ n(A) = 26
∴ P(A) = \(\frac{26}{52}\)
Let B be the event that a black card is drawn
∴ Black card can be drawn in ²⁶C₁ = 26 ways.
∴ n(B) = 26
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= \(\frac{26}{52}+\frac{26}{52}\)
= \(\frac{52}{52}\)
= 1

(ii) Let A be the event that a red card is drawn
∴ red card can be drawn in ²⁶C₁ = 26 ways
∴ n(A) = 26
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}\)
Let B be the event that a face card is drawn There are 12 face cards in the pack of 52 cards
∴ 1 face card can be drawn in ¹²C₁ = 12 ways
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}\)
There are 6 red face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{52}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\)
= \(\frac{32}{52}\)
= \(\frac{8}{13}\)

Question 3.
Two cards are drawn from a pack of 52 cards. What is the probability that,
(i) both the cards are of the same colour?
(ii) both the cards are either black or queens?
Solution:
Two cards can be drawn from 52 cards in ⁵²C₂ ways.
∴ n(S) = ⁵²C₂
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that both cards are red.
∴ 2 red cards can be drawn in ²⁶C₂ ways.
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Let B be the event that both cards are black.
∴ 2 black cards can be drawn in ²⁶C₂ ways
∴ n(B) = ²⁶C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= \(\frac{25}{102}+\frac{25}{102}\)
= \(\frac{25}{51}\)

(ii) Let A be the event that both cards are black.
∴ 2 black cards can be drawn in ²⁶C₂ ways.
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Let B be the event that both cards are queens.
There are 4 queens in a pack of 52 cards
∴ 2 queen cards can be drawn in ⁴C₂ ways.
∴ n(B) = ⁴C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
There are two black queen cards.
∴ n(A ∩ B) = ²C₂ = 1

Question 4.
A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is the probability that
(i) number on the ticket is a perfect square or divisible by 4?
(ii) number on the ticket is a prime number or greater than 30?
Solution:
Out of the 50 tickets, a ticket can be drawn in ⁵⁰C₁ = 50 ways.
∴ n(S) = 50
(i) Let A be the event that the number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49}
∴ n(A) = 7
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{50}\)
Let B be the event that the number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{50}\)
Now, A ∩ B = {4, 16, 36}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{50}\)
Required probability = P (A u B)
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{7}{50}+\frac{12}{50}-\frac{3}{50}\)
= \(\frac{8}{25}\)

(ii) Let A be the event that the number on the ticket is a prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
∴ n(A) = 15
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{15}{50}\)
Let B be the event that the number is greater than 30.
∴ B = {31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}
∴ n(B) = 20
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{20}{50}\)
Now, A ∩ B = {31, 37, 41, 43, 47}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{15}{50}+\frac{20}{50}-\frac{5}{50}\)
= \(\frac{15+20-5}{50}\)
= \(\frac{3}{5}\)

Question 5.
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random
(i) passed at least one examination.
(ii) passed in exactly one examination.
(iii) failed in both examinations.
Solution:
Out of hundred students 1 student can be selected in ¹⁰⁰C₁ = 100 ways.
∴ n(S) = 100
Let A be the event that the student passed in the first examination.
Let B be the event that student passed in second examination.
∴ n(A) = 60, n(B) = 50 and n(A ∩ B) = 30

(i) P(student passed in at least one examination) = P(A ∪ B)
= P(A) + P(B) – P (A ∩ B)
= \(\frac{6}{10}+\frac{5}{10}-\frac{3}{10}\)
= \(\frac{4}{5}\)

(ii) P(student passed in exactly one examination) = P(A) + P(B) – 2.P(A ∩ B)
= \(\frac{6}{10}+\frac{5}{10}-2\left(\frac{3}{10}\right)\)
= \(\frac{1}{2}\)

(iii) P(student failed in both examinations) = P(A’ ∩ B’)
= P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – \(\frac{4}{5}\)
= \(\frac{1}{5}\)

Question 6.
If P(A) = \(\frac{1}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{1}{2}\). Find the values of the following probabilities.
(i) P(A ∩ B)
(ii) P(A ∩ B’)
(iii) P(A’ ∩ B)
(iv) P(A’ ∪ B’)
(v) P(A’ ∩ B’)
Solution:
Here, P(A) = \(\frac{1}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{1}{2}\)
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{1}{4}+\frac{2}{5}-\frac{1}{2}\)
= \(\frac{3}{20}\)

(ii) P(A’ ∩ B’) = P(A) – P(A ∩ B)
= \(\frac{1}{4}-\frac{3}{20}\)
= \(\frac{1}{10}\)

(iii) P(A’ ∩ B) = P(B) – P(A ∩ B)
= \(\frac{2}{5}-\frac{3}{20}\)
= \(\frac{1}{4}\)

(iv) P(A’ ∪ B’) = P(A ∩ B)’ …..[De Morgan’s law]
= 1 – P(A ∩ B)
= 1 – \(\frac{3}{20}\)
= \(\frac{17}{20}\)

(v) P(A’ ∩ B’) = P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)

Question 7.
A computer software company is bidding for computer programs A and B. The probability that the company will get software A is \(\frac{3}{5}\), the probability that the company will get software B is \(\frac{1}{3}\) and the probability that company will get both A and B is \(\frac{1}{8}\). What is the probability that the company will get at least one software?
Solution:
Let A be the event that the company will get software A.
∴ P(A) = \(\frac{3}{5}\)
Let B be the event that company will get software B.
∴ P(B) = \(\frac{1}{3}\)
Also, P(A ∩ B) = \(\frac{1}{8}\)
∴ P(the company will get at least one software) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{3}{5}+\frac{1}{3}-\frac{1}{8}\)
= \(\frac{72+40-15}{120}\)
= \(\frac{97}{120}\)

Question 8.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of it being a heart or a queen.
Solution:
One card can be drawn from the pack of 52 cards in ⁵²C₁ = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 13 heart cards and 4 queen cards
Let A be the event that a card drawn is the heart.
A heart card can be drawn from 13 heart cards in ¹³C₁ ways
∴ n(A) = ¹³C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{52}=\frac{13}{52}\)
Let B be the event that a card drawn is queen.
A queen card can be drawn from 4 queen cards in ⁴C₁ ways
∴ n(B) = ⁴C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{52}=\frac{4}{52}\)
There is one queen card out of 4 which is also a heart card
∴ n(A ∩ B) = ¹C₁
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{52}=\frac{1}{52}\)
∴ P(card is a heart or a queen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\)
= \(\frac{13+4-1}{52}\)
= \(\frac{16}{52}\)
∴ P(A ∪ B) = \(\frac{4}{13}\)

Question 9.
In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in ⁷C₄
= \(\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}\)
= 35 ways.
∴ n(S) = 35
Let A be the event that 3 boys and 1 girl are selected.
3 boys can be selected in ³C₃ ways while a girl can be selected in ⁴C₁ ways.
∴ n(A) = ³C₃ × ⁴C₁ = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{35}\)
Let B be the event that 3 girls and 1 boy are selected.
3 girls can be selected in ⁴C3 ways and a boy can be selected in ³C₁ ways.
∴ n(B) = ⁴C₃ × ³C₁ = 12
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{12}{35}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
= P(A) + P(B)
= \(\frac{4}{35}+\frac{12}{35}\)
= \(\frac{16}{35}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.2

Question 1.
A fair die is thrown two times. Find the chance that
(i) product of the numbers on the upper face is 12.
(ii) sum of the numbers on the upper face is 10.
(iii) sum of the numbers on the upper face is at least 10.
(iv) sum of the numbers on the upper face is 4.
(v) the first throw gives an odd number and the second throw gives a multiple of 3.
(vi) both the times die to show the same number (doublet).
Solution:
If a fair die is thrown twice, the sample space 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
(i) Let A be the event that the product of the numbers on uppermost face is 12.
∴ A = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(A) = 4
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{4}{36}=\frac{1}{9}\)

(ii) Let B be the event that sum of the numbers on uppermost face is 10.
∴ B = {(4, 6), (5, 5), (6, 4)}
∴ n(B) = 3
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

(iii) Let C be the event that sum of the numbers on uppermost face is at least 10 (i.e., 10 or more than 10 which are 10 or 11 or 12)
∴ C = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), ( 6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(iv) Let D be the event that sum of the numbers on uppermost face is 4.
∴ D = {(1, 3), (2, 2), (3, 1)}
∴ n(D) = 3
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}=\frac{1}{12}\)

(v) Let E be the event that 1st throw gives an odd number and 2nd throw gives multiple of 3.
∴ E = {(1, 3), (1, 6), (3, 3), (3, 6), (5, 3), (5, 6)}
∴ n(E) = 6
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(vi) Let F be the event that both times die shows same number.
∴ F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(F) = 6
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

Question 2.
Two cards are drawn from a pack of 52 cards. Find the probability that
(i) both are black.
(ii) both are diamonds.
(iii) both are ace cards.
(iv) both are face cards.
(v) one is a spade and the other is a non-spade.
(vi) both are from the same suit.
(vii) both are from the same denomination.
Solution:
Two cards can be drawn from a pack of 52 cards in ⁵²C₂ ways.
∴ n(S) = ⁵²C₂
(i) Let A be the event that both the cards drawn are black.
The pack of 52 cards contains 26 black cards.
∴ 2 cards can be drawn from them in ²⁶C₂ ways
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(ii) Let B be the event that both the cards drawn are diamond.
There are 13 diamond cards in a pack of 52 cards.
∴ 2 diamond cards can be drawn from 13 diamond cards in ¹³C₂ ways
∴ n(B) = ¹³C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(iii) Let C be the event that both the cards drawn are aces.
In a pack of 52 cards, there are 4 ace cards.
∴ 2 ace cards can be drawn from 4 ace cards in ⁴C₂ ways
∴ n(C) = ⁴C₂
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(iv) Let D be the event that both the cards drawn are face cards.
There are 12 face cards in a pack of 52 cards.
∴ 2 face cards can be drawn from 12 face cards in ¹²C₂ ways.
∴ n(D) = ¹²C₂
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(v) Let E be the event that out of the two cards drawn one is a spade and other is non-spade.
There are 13 spade cards and 39 cards are non-spade cards in a pack of 52 cards.
∴ One spade card can be drawn from 13 spade cards in ¹³C₁ ways and one non-spade card can be drawn from 39 non-spade cards in 39C1 ways.
∴ n(E) = ¹³C₁ . ³⁹C₁
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \cdot{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}\)

(vi) Let F be the event that both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
2 cards can be drawn from a suit in ¹³C₂ ways.
A suit can be selected in 4 ways.
∴ n(F) = ¹³C₂ × 4
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(vii) Let G be the event that both the cards drawn are of same denominations.
A pack of cards has 13 denominations and 4 different cards for each denomination
∴ n(G) = 13 × ⁴C₂
∴ P(G) = \(\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}=\frac{13 \times{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

Question 3.
Four cards are drawn from a pack of 52 cards. Find the probability that
(i) 3 are Kings and 1 is Jack.
(ii) all the cards are from different suits.
(iii) at least one heart.
(iv) all cards are club and one of them is a jack.
Solution:
4 cards can be drawn out of 52 cards in ⁵²C₄ ways.
∴ n(S) = ⁵²C₄
(i) Let A be the event that out of the four cards drawn, 3 are kings and 1 is a jack.
There are 4 kings and 4 jacks in a pack of 52 cards.
∴ 3 kings can be drawn from 4 kings in ⁴C₃ ways.
Similarly, 1 jack can be drawn out of 4 jacks in ⁴C₁ ways.
∴ Total number of ways in which 3 kings and 1 jack can be drawn is ⁴C₃ × ⁴C₁
∴ n(A) = ⁴C₃ × ⁴C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

(ii) Let B be the event that all the cards drawn are of different suits.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ A card can be drawn from each suit in ¹³C₁ ways.
∴ 4 cards can be drawn from 4 different suits in ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ ways.
∴ n(B) = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

(iii) Let C be the event that out of the four cards drawn at least one is a heart.
∴ C’ is the event that all 4 cards drawn are non-heart cards.
In a pack of 52 cards, there are 39 non-heart cards.
∴ 4 non-heart cards can be drawn in ³⁹C₄ ways.
∴ n(C’) = ³⁹C₄
∴ P(C’) = \(\frac{\mathrm{n}\left(\mathrm{C}^{\eta}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}\)
∴ P(C) = 1 – P(C’) = 1 – \(\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}\)

(iv) Let D be the event that all the 4 cards drawn are clubs and one of them is a jack.
In a pack of 52 cards, there are 13 club cards having 1 jack card.
∴ 1 jack can be drawn in ¹C₁ way and the other 3 cards can be drawn from remaining 12 club cards in ¹²C₃ ways.
∴ n(D) = ¹²C₃ × ¹C₁
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{3} \times{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

Question 4.
A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is 1/3. The probability of yellow is 1/5.
(i) What is the probability of blackball?
(ii) How many balls are green, black, and yellow?
Solution:
(i) The bag contains 15 balls of three different colours i.e., green (G), black (B) and yellow (Y)
∴ P(G) = \(\frac{1}{3}\) and P(Y) = \(\frac{1}{5}\)
If a ball is drawn from the bag, then it can be any one of the green, black and yellow.
∴ P(G) + P(B) + P(Y) = 1
∴ \(\frac{1}{3}\) + P(B) + \(\frac{1}{5}\) = 1
∴ P(B) + \(\frac{8}{15}\) = 1
∴ P(B) = 1 – \(\frac{8}{15}\) = \(\frac{7}{15}\)
∴ Probability of black ball is \(\frac{7}{15}\)

(ii) Total number of balls = 15 and
P(G) = \(\frac{1}{3}\), P(B) = \(\frac{7}{15}\), P(Y) = \(\frac{1}{5}\)
∴ number of green balls = \(\frac{1}{3}\) × 15 = 5
number of black balls = \(\frac{7}{15}\) × 15 = 7
and number of yellow balls = \(\frac{1}{5}\) × 15 = 3.

Question 5.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that the
(i) number on the ticket is divisible by 6?
(ii) number on the ticket is a perfect square?
(iii) number on the ticket is prime?
(iv) number on the ticket is divisible by 3 and 5?
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in ⁷⁵C₁ = 75 ways.
∴ n(S) = 75
(i) Let A be the event that number on ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12}{75}=\frac{4}{25}\)

(ii) Let B be the event that number on ticket is a perfect square.
∴ B = {1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{8}{75}\)

(iii) Let C be the event that the number on the ticket is a prime number.
∴ C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(C) = 21
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{s})}=\frac{21}{75}=\frac{7}{25}\)

(iv) Let D be the event that number on ticket is divisible by 3 and 5 i.e., divisible by L.C.M. of 3 and 5 i.e., 15
∴ D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{5}{75}=\frac{1}{15}\)

Question 6.
From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains
(i) 3 boys and 2 girls
(ii) at least 3 boys.
Solution:
The group consists of 8 boys and 5 girls i.e., 8 + 5 = 13 persons.
A committee of 5 is to be formed from this group.
∴ 5 persons from 13 persons can be selected in ¹³C₅ ways
∴ n(S) = ¹³C₅
(i) Let A be the event that the committee contains 3 boys and 2 girls.
3 boys from 8 boys can be selected in ⁸C₃ ways and 2 girls from 5 girls can be selected in ⁵C₂ ways
∴ n(A) = ⁸C₃ . ⁵C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}}{{ }^{13} \mathrm{C}_{5}}\)

(ii) Let B be the event that the committee contains at least 3 boys (i.e., 3 boys and 2 girls or 4 boys and 1 girl or 5 boys and no girl)
∴ n(B) = ⁸C₃ . ⁵C₂ + ⁸C₄ . ⁵C₁ + ⁸C₅ . ⁵C₀
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}+{ }^{8} \mathrm{C}_{4} \cdot{ }^{5} \mathrm{C}_{1}+{ }^{8} \mathrm{C}_{5} \cdot{ }^{5} \mathrm{C}_{0}}{{ }^{13} \mathrm{C}_{5}}\)

Question 7.
A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in a socket. What is the probability that the room is lit?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in ¹⁰C₃ ways.
∴ n(S) = ¹⁰C₃
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in ⁶C₃ ways.
∴ n(A’) = ⁶C₃
∴ P(A’) = \(\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}\)
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 8.
The letters of the word LOGARITHM are arranged at random. Find the probability that
(i) Vowels are always together.
(ii) Vowels are never together.
(iii) Exactly 4 letters between G and H
(iv) begins with O and ends with T
(v) Start with a vowel and ends with a consonant.
Solution:
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ⁹P₉ = 9! ways.
∴ n(S) = 9!
(i) Let A be the event that vowels are always together.
The word LOGARITHM consists of 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M).
3 vowels can be arranged among themselves in = ³P₃ = 3! ways.
Considering 3 vowels as one group, 6 consonants and this group (i.e., altogether 7) can be arranged in ⁷P₇ = 7! ways.
∴ n(A) = 3! × 7!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3 ! \times 7 !}{9 !}\)

(ii) Let B be the event that vowels are never together.
Consider the following arrangement
_C_C_C_C_C_C_
6 consonants create 7 gaps.
∴ 3 vowels can be arranged in 7 gaps in ⁷P₃ ways.
Also 6 consonants can be arranged among themselves in ⁶P₆ = 6! ways.
∴ n(B) = 6! × ⁷P₃
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6 ! \times{ }^{7} \mathrm{P}_{3}}{9 !}\)

(iii) Let C be the event that exactly 4 letters are arranged between G and H.
Consider the following arrangement
1 2 3 4 5 6 7 8 9
∴ Out of 9 places, G and H can occupy any one of following 4 positions in 4 ways.
1st and 6th, 2nd and 7th, 3rd and 8th, 4th and 9th
Now, G and H can be arranged among themselves in ²P₂ = 2! =2 ways.
Also, the remaining 7 letters can be arranged in remaining 7 places in ⁷P₇ = 7! ways.
∴ n(C) = 4 × 2 × 7! = 8 × 7! = 8!
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{8 !}{9 !}=\frac{8 !}{9 \times 8 !}=\frac{1}{9}\)

(iv) Let D be the event that word begins with O and ends with T.
Thus first and last letter can be arranged in one way each and the remaining 7 letters can be arranged in remaining 7 places in ⁷P₇ = 7! ways
∴ n(D) = 7! × 1 × 1 = 7!
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{7 !}{9 !}\)

(v) Let E be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in the word LOGARITHM.
∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in ⁷P₇ = 7! ways
∴ n(E) = 3 × 6 × 7!
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{3 \times 6 \times 7 !}{9 !}\)

Question 9.
The letters of the word SAVITA are arranged at random. Find the probability that vowels are always together.
Solution:
The word SAVITA contains 6 letters. Out of 6 letters, 3 are vowels (A, A, I) and 3 are consonants (S, V, T).
6 letters in which A repeats twice can be arranged among themselves in \(\frac{6 !}{2 !}\) ways.
∴ n(S) = \(\frac{6 !}{2 !}\)
Let A be the event that vowels are always together.
3 vowels (A, A, I) can be arranged among themselves in \(\frac{3 !}{2 !}\) ways.
Considering 3 vowels as one group, 3 consonants and this group (i.e. altogether 4) can be arranged in ⁴P₄ = 4! ways.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.1

Question 1.
State the sample space and n(S) for the following random experiments.
(i) A coin is tossed twice. If a second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If a second throw results in a head, a die is thrown, otherwise, a coin is tossed.
Solution:
(i) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
A coin is tossed twice and if the second throw results in a tail, a die is thrown.
Out of the above 4 possibilities, on second throw tail occurs in two cases only i.e., HT, TT.
∴ S = {HH, TH, HT1, HT2, HT3, HT4, HT5, HT6, TT1, TT2, TT3, TT4, TT5, TT6}
∴ n(S) = 14

(ii) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
Let A be the event that the second throw results in the head when a coin is tossed twice followed by a die is thrown.
∴ A = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6}
The remaining outcomes i.e., HT, TT are followed by the tossing of a coin.
Let us consider this as event B.
∴ B = {HTT, HTH, TTT, TTH}
The sample space S of the experiment is A ∪ B.
∴ S = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6, HTT, HTH, TTT, TTH}
∴ n(S) = 16

Question 2.
In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).
Solution:
The bag contains 3 balls out of which one is black (B), one is red (R) and the other one is green (G).
Two balls are drawn one after the other, with replacement, from the bag.
∴ the sample space S is given by
S = {BB, BR, BG, RB, RR, RG, GB, GR, GG}
∴ n(S) = 9

Question 3.
A coin and die are tossed. State sample space of following events.
(i) A: Getting a head and an even number.
(ii) B: Getting a prime number.
(iii) C: Getting a tail and perfect square.
Solution:
When a coin and a die are tossed the sample space S is given by
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(i) A: getting a head and an even number
∴ A = {H2, H4, H6}

(ii) B: getting a prime number
∴ B = {H2, H3, H5, T2, T3, T5}

(iii) C: getting a tail and a perfect square.
∴ C = {T1, T4}

Question 4.
Find the total number of distinct possible outcomes n(S) for each of the following random experiments.
(i) From a box containing 25 lottery tickets and 3 tickets are drawn at random.
(ii) From a group of 4 boys and 3 girls, any two students are selected at random.
(iii) 5 balls are randomly placed into five cells, such that each cell will be occupied.
(iv) 6 students are arranged in a row for photographs.
Solution:
(i) Let S be the event that 3 tickets are drawn at random from 25 tickets
∴ 3 tickets can be selected in ²⁵C₃ ways
∴ n(S) = ²⁵C₃
= \(\frac{25 \times 24 \times 23}{3 \times 2 \times 1}\)
= 2300

(ii) There are 4 boys and 3 girls i.e., 7 students.
2 students can be selected from these 7 students in ⁷C₂ ways.
∴ n(S) = ⁷C₂
= \(\frac{7 \times 6}{2 \times 1}\)
= 21

(iii) 5 balls have to be placed in 5 cells in such a way that each cell is occupied.
∴ The first ball can be placed into one of the 5 cells in 5 ways, the second ball placed into one of the remaining 4 cells in 4 ways.
Similarly, the third, fourth, and fifth balls can be placed in 3 ways, 2 ways, and 1 way, respectively.
∴ a total number of ways of filling 5 cells such that each cell is occupied = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(iv) Six students can be arranged in a row for a photograph in ⁶P₆ = 6! ways.
∴ n(S) = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Question 5.
Two dice are thrown. Write favourable outcomes for the following events.
(i) P: The sum of the numbers on two dice is divisible by 3 or 4.
(ii) Q: sum of the numbers on two dice is 7.
(iii) R: sum of the numbers on two dice is a prime number.
Also, check whether
(a) events P and Q are mutually exclusive and exhaustive.
(b) events Q and R are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, all possible outcomes are
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)}
(i) P: sum of the numbers on two dice is divisible by 3 or 4.
∴ P = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(ii) Q: sum of the numbers on two dice is 7.
∴ Q = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(iii) R: sum of the numbers on two dice is a prime number.
∴ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
(a) P and Q are mutually exclusive events as P ∩ Q = φ and
P ∪ Q = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ P and Q are not exhaustive events as P ∪ Q ≠ S.

(b) Q ∩ R = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ Q ∩ R ≠ φ
Also, Q ∪ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} ≠ S
∴ Q and R are neither mutually exclusive nor exhaustive events.

Question 6.
A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
consideration of suits
(i) is not taken into account.
(ii) is taken into account.
Solution:
(i) If consideration of suits is not taken into account, then one card can be drawn from the pack of 52 playing cards in ⁵²C₁ = 52 ways.
∴ n(S) = 52

(ii) If consideration of suits is taken into account, then one card can be drawn from each suit in ¹³C₁ × ⁴C₁
= 13 × 4
= 52 ways.
∴ n(S) = 52

Question 7.
Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-I; if it turns up tails, a marble is chosen from Box-II. Describe the sample space.
Solution:
Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box II contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
∴ the sample space is
S = {(H, R11), (H, R12), (H, R13), (H, B11), (H, B12), (T, R21), (T, R22), (T, B21), (T, B22), (T, B23), (T, B24)}

Question 8.
Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn
(i) with replacement
(ii) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8.
Two cards are to be drawn from this bag.
(i) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6,8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(ii) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}