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(A) For a Poisson distribution with parameter $\lambda$,the probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.Given $

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$\frac{e^{-1}}{6}$

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(A) For a Poisson distribution with parameter $\lambda$,the probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.Given $P(X=1) = 2P(X=2)$.Substituting the formula:$\frac{\lambda^1 e^{-\lambda}}{1!} = 2 	imes \frac{\lambda^2 e^{-\lambda}}{2!}$$\lambda e^{-\lambda} = 2 	imes \frac{\lambda^2 e^{-\lambda}}{2}$$\lambda e^{-\lambda} = \lambda^2 e^{-\lambda}$Since $e^{-\lambda} 
eq 0$,we divide both sides by $e^{-\lambda}$:$\lambda = \lambda^2$$\lambda^2 - \lambda = 0 \Rightarrow \lambda(\lambda - 1) = 0$.Since $\lambda > 0$ for a Poisson distribution,we have $\lambda = 1$.Now,we calculate $P(X=3)$:$P(X=3) = \frac{\lambda^3 e^{-\lambda}}{3!} = \frac{1^3 e^{-1}}{3 	imes 2 	imes 1} = \frac{e^{-1}}{6}$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:

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