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Question
If a random variable X follows Poisson distribution sucli that P(X = 1) = P(X = 2), find:
(a) the mean and
(b) P(X = 0). [Given: e-2 = 0.1353)
Solution
P(X = x) = P(x) = `("e"^(-"m") . "m"^2)/("x"!)` ,
Given P(X= l)= P(X= 2)
`therefore ("e"^(-"m") . "m"^1)/(1!) = ("e"^(-"m") . "m"^2)/(2!)`
⇒ m = 2
P[X = 0] = `("e"^(-2) . 2^2)/(0!)`
= `("e"^(-2) . 1)/(1)`
= 0.1353
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