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Question
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
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Solution
The p.m.f. of X is given by
P(X = x) = `("e"^-"m" "m"^x)/(x!)`
Given, P(X = 2) = P(X = 3)
∴ `("e"^-"m" "m"^2)/(2!) = ("e"^-"m" "m"^3)/(3!)`
∴ `"m"^2/(2) = "m"^3/(6)`
∴ `1/2 = m/6`
∴ `6/2=m`
∴ m = 3
∴ P(X ≥ 2) = 1 – P(X < 2)
= 1 – P(X = 0 or X = 1)
= 1 – [P(X = 0) + P(X = 1)]
= `1 - [("e"^-3(3)^0)/(0!) + ("e"^-3(3)^1)/(1!)]`
= 1 – [0.0497 + 3 × 0.0497]
= 0.8012
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