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प्रश्न
If X ∼ P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2).
Given e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
∴ `("e"^square"m"^1)/(1!) = ("e"^"-m""m"^2)/square`
∴ m = `square`
∴ P(X = 2) = `("e"^-2. "m"^2)/(2!)` = `square`
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उत्तर
If X ∼ P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2).
Given e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
∴ `("e"^bb(-m)"m"^1)/(1!) = ("e"^"-m""m"^2)/bb(2!)`
∴ m = 2
∴ P(X = 2) = `("e"^-2. 2^2)/2!`
= `bb((0.1353 xx 4)/2)`
= 0.2706
Notes
Since P(X = 1) = P(X = 2)
`(e^-mm^1)/(1!) = bb((e^-mm^2)/(2!))`
∴ m = 2
∴ mean = m = 2
Then P(X = 2) = `bb((e^-mm^2)/(2!)) = bb((e^-2 2^2)/(2!))`
= `bb((0.1353 xx 4)/2)`
= 0.2706
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