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Question
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 exactly 5 rats inclusive. Given e-5 = 0.0067.
Solution
Let X denote the number of rats per bungalow.
Given, m = 5 and e–5 = 0.0067
∴ X ~ P(m) ≡ X ~ P(5)
The p.m.f. of X is given by
P(X = x) = `("e"^-"m" "m"^x)/(x!)`
∴ P(X = x) = `("e"^-5*(5)^x)/(x!), x` = 0, 1, ..., 5
P(exactly five rats)
= P(X = 5)
= `("e"^-5*(5)^5)/(5!)`
= `(0.0067 xx 5^5)/(5 xx 4 xx 3 xx 2 xx1)`
= `(0.0067 xx 625)/(24)`
= `(4.1875)/(24)`
= 0.1745
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