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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 more than 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(more than five rats)
= P(X > 5)
= 1 – P(X ≤ 5)
= 1 – P[X = 0 or X = 1 or X = 2 or X = 3 or X = 4 or X = 5]
= 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P (X = 4) + P(X = 5)]
= `1 - [("e"^-5 (5)^0)/(0!) + ("e"^-5(5)^1)/(1!) + ("e"^-5 (5)^2)/(2!) + ("e"^-5 (5)^3)/(3!) + ("e"^-5 (5)^4)/(4!) + ("e"^-5(5)^5)/(5!)]`
= `1 - ["e"^-5 (5^0/(0!) + 5^1/(1!) + 5^2/(2!) + 5^3/(3!) + 5^4/(4!) + 5^5/(5!))]`
= `1 - ["e"^-5 (1/1 + 5/1 + 25/2 + 125/6 + 625/24 + 3125/120)]`
= 1 – [e–5 (1 + 5 + 12.5 + 20.83 + 26.04 + 26.04)]
= 1 – [0.0067 (91.417)]
= 1 – 0.6125
= 0.3875
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