Sum

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.

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#### Solution

Let X denote the number of rats per bunglow.

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

Concept: Poisson Distribution

Is there an error in this question or solution?

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