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If X~P(0.5), then find P(X = 3) given e−0.5 = 0.6065. - Mathematics and Statistics

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प्रश्न

If X~P(0.5), then find P(X = 3) given e−0.5 = 0.6065.

योग
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उत्तर

Given, X ~ P(0.5) and e–0.5 = 0.6065
∴ m = 0.5
The p.m.f. of X is given by

P(X = x) = `("e"^-"m""m"^x)/(x!)`

∴ P(X = x) = `("e"^(-0.5) (0.5)^x)/(x!), x` = 0, 1, 2,...

∴ P(X = 3) = `("e"^(-0.5) (0.5)^3)/(3!)`

= `(0.6065 xx 0.125)/(3 xx 2 xx 1)`

= 0.0126

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अध्याय 8: Probability Distributions - Exercise 8.4 [पृष्ठ १५२]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 8 Probability Distributions
Exercise 8.4 | Q 1.02 | पृष्ठ १५२

वीडियो ट्यूटोरियलVIEW ALL [2]

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Solution: X : Follows Poisson distribution

∴ P(X) = `("e"^-"m" "m"^x)/(x!)`, P(X = 1) = 0.4 and P(X = 2) = 0.2

∴ P(X = 1) = `square` P(X = 2).

`("e"^-"m" "m"^x)/(1!) = square ("e"^-"m" "m"^2)/(2!)`,

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∴ m = `square`


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lf X ∼ P(m) with P(X = 1) = P(X = 2) then m = 1.


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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∴ m = `square`

∴ P(X = 2) = `("e"^-2. "m"^2)/(2!)` = `square`


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