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Question
If a random variable X follows Poisson distribution such that P(X = l) =P(X = 2), then find P(X ≥ 1). [Use e-2 = 0.1353]
Solution
Given X - p(m)
The p.m.f of X is,
P(X = x) = P(x) = `("e"^(-"m") . "m"^2)/("x"!)` , x = 0 , 1 , 2
Given P(X= l)= P(X= 2)
`therefore ("e"^(-"m") . "m"^1)/(1!) = ("e"^(-"m") . "m"^2)/(2!)`
⇒ m = 2
Now P (X ≥ 1) = 1 - P(X = 0)
`= 1 - ("e"^(-"m") . "m"^0)/(0!) = 1 - ("e"^(-2) xx 1)/1`
= 1 - e-2
= 1- 0.1353
= 0.8647
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