If X is a Poisson variate such that P(X=1)=P( | vedclass

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(C) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $P(X=1) = P(X=2)$,we have:

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$\frac{2}{3 e^2}$

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(C) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $P(X=1) = P(X=2)$,we have:$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$Dividing both sides by $e^{-\lambda} \lambda$ (assuming $\lambda 
eq 0$):$1 = \frac{\lambda}{2}$$\lambda = 2$Now,we need to find $P(X=4)$:$P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-2} (2)^4}{24}$$P(X=4) = \frac{16}{24 e^2} = \frac{2}{3 e^2}$

If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$,then $P(X=4)$ is equal to

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