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
