Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 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
Maharashtra Board Solution
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e-1 = 0.3678.
Solution & Step-by-Step Answer:
∵ 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
Maharashtra Board Solution
If X ~ P(), then find P(X = 3) given e-0.5 = 0.6065.
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
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 & Step-by-Step Answer:
∵ X follows Poisson Distribution
Question 4
Maharashtra Board Solution
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 & Step-by-Step Answer:
∵ m = 1 ∵ X ~ P(m = 4) ∴ p(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)
= + 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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
∵ X = No. of defects on a plywood sheet ∵ m = -1 ∵ X ~ P(m = -1) ∴ p(x) = (i) P(No defect) P(X = 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
Maharashtra Board Solution
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 & Step-by-Step Answer:
X = No. of rats ∵ m = 5 ∴ X ~ P(m = 5) ∴ p(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
