The variance of a Poisson variate X is 2. The | vedclass

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(C) For a Poisson distribution,the mean and variance are both equal to $\lambda$. Given variance $\lambda = 2$. The probability mass function is $P(X=

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$\frac{e^2-5}{e^2}$

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(C) For a Poisson distribution,the mean and variance are both equal to $\lambda$. Given variance $\lambda = 2$. The probability mass function is $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$. We need to find $P(X \geq 3) = 1 - \{P(X=0) + P(X=1) + P(X=2)\}$. $P(X=0) = \frac{2^0 e^{-2}}{0!} = e^{-2}$. $P(X=1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2}$. $P(X=2) = \frac{2^2 e^{-2}}{2!} = \frac{4 e^{-2}}{2} = 2e^{-2}$. Summing these,$P(X Therefore,$P(X \geq 3) = 1 - \frac{5}{e^2} = \frac{e^2-5}{e^2}$.

The variance of a Poisson variate $X$ is $2$. Then $P(X \geq 3) = $

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