Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.

If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e^{-1} = 0.3678.

Solution:

∵ m = 1

∵ X follows Poisson Distribution

= e^{-m} × 1 + e^{-m} × 1

= e^{-1} + e^{-1}

= 2 × e^{-1}

= 2 × 0.3678

= 0.7356

Question 2.

If X ~ P(\(\frac{1}{2}\)), then find P(X = 3) given e^{-0.5} = 0.6065.

Solution:

Question 3.

If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e^{-3} = 0.0497

Solution:

∵ X follows Poisson Distribution

Question 4.

The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e^{-4} = 0.0183.

Solution:

∵ m = 1

∵ X ~ P(m = 4)

∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)

X = No. of complaints recieved

(i) P(Only two complaints on a given day)

(ii) P(Atmost two complaints on a given day)

P(X ≤ 2) = p(0) + p(1) + p(2)

= \(\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}\) + 0.1464

= e^{-4} + e^{-4} × 4 + 0.1464

= e^{-4} [1 + 4] + 0.1464

= 0.0183 × 5 + 0.1464

= 0.0915 + 0.1464

= 0.2379

Question 5.

A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that

(i) no car is used on a given day.

(ii) some demand is refused on a given day, given e^{-1.5} = 0.2231.

Solution:

Let X = No. of demands for a car on any day

∴ No. of cars hired

n = 2

m = 1.5

∵ X ~ P(m = 1.5)

Question 6.

Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e^{-1} = 0.3678.

Solution:

∵ X = No. of defects on a plywood sheet

∵ m = -1

∵ X ~ P(m = -1)

∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)

(i) P(No defect)

P(X = 0) = \(\frac{e^{-1} \times 1^{0}}{0 !}\)

= e^{-1}

= 0.3678

(ii) P(At least one defect)

P(X ≥ 1) = 1 – P(X < 1)

= 1 – p(0)

= 1 – 0.3678

= 0.6322

Question 7.

It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has

(i) exactly 5 rats

(ii) more than 5 rats

(iii) between 5 and 7 rats, inclusive. Given e^{-5} = 0.0067.

Solution:

X = No. of rats

∵ m = 5

∴ X ~ P(m = 5)

∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)

(i) P(Exactly five rats)

(ii) P(More than five rats)

P(X > 5) = 1 – P(X ≤ 5)

(iii) P(between 5 and 7 rats, inclusive)

P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)

= 0.0067 × 3125 × 0.02

= 0.0067 × 62.5

= 0.42