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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]
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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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∴ P(X = 1) = `square` P(X = 2).
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`"e"^-"m" = square "e"^-"m" "m"/2`, m ≠ 0
∴ m = `square`
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e–3 = 0.0497.
P[X = x] = `square`
Since P[X = 2] = P[X = 3]
`square` = `square`
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∴ m = `square`
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