A random variable $X$ has Poisson distribution with mean $2$ The $P (X > 1.5)$ equals

0

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2 / e^{2}

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3 / e^{2}

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1 - \frac{3}{e^{2}}

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Solution

The correct option is D $1 - \frac{3}{e^{2}}$
We know $P (X = r) = \frac{e^{- λ} λ^{r}}{r!}, (λ =$ mean $)$

$∴ P (X = r > 1.5) = P (2) + P (3) + . . . . \infty = 1 - [P (0) + P (1)]$

$= 1 - [e^{- 2} + \frac{e^{- 2} \times 2^{2}}{2}] = 1 - \frac{3}{e^{2}}$

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