# A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows Poisson distribution with mean 1.5. Find the probability that (i) no car is used on a given day, (ii) - Mathematics and Statistics

Sum

A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows Poisson distribution with mean 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 denote the number of cars hired on a day.
Given, m = 1.5 and e–1.5 = 0.2231
∴ X ~ P(m) = X ~ P(1.5)
The p.m.f. of X is given by

P(X = x) = ("e"^-"m"^x)/(x!)

∴ P(X = x) = ("e"^-1.5 (1.5)^x)/(x!)

i. P(no car is used on a given day)
= P(X = 0)

= ("e"^-1.5 (1.5)^0)/(0!)

= 0.2231

ii. P(some demand is refused on a given day)
= P(X > 2)
= 1 – P(X ≤ 2)
= 1 – P(X = 0 or X = 1 or X = 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

= 1 - [("e"^-1.5 (1.5)^0)/(0!) + ("e"^-1.5 (1.5)^1)/(1!) + ("e"^-1.5 (1.5)^2)/(2!)]

= 1 - [0.2231 + 0.2231 xx 1.5 + (0.2231 xx 2.25)/(2 xx 1)]

= 1 – 0.8087
= 0.1913

Concept: Poisson Distribution
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Probability Distributions
Exercise 8.4 | Q 1.05 | Page 152