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(C) Given that the mean of the Poisson distribution is $\lambda = \frac{1}{2}$.The probability mass function of a Poisson distribution is given by $P(

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$1: 6$

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(C) Given that the mean of the Poisson distribution is $\lambda = \frac{1}{2}$.The probability mass function of a Poisson distribution is given by $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$.We need to find the ratio $\frac{P(X=3)}{P(X=2)}$.$P(X=3) = \frac{(\frac{1}{2})^3 e^{-1/2}}{3!}$ and $P(X=2) = \frac{(\frac{1}{2})^2 e^{-1/2}}{2!}$.Taking the ratio:$\frac{P(X=3)}{P(X=2)} = \frac{\frac{(\frac{1}{2})^3 e^{-1/2}}{3!}}{\frac{(\frac{1}{2})^2 e^{-1/2}}{2!}}$$= \frac{(\frac{1}{2})^3}{3!} 	imes \frac{2!}{(\frac{1}{2})^2}$$= \frac{1}{2} 	imes \frac{2!}{3!}$$= \frac{1}{2} 	imes \frac{2}{6} = \frac{1}{6}$.Thus,the ratio is $1:6$.

$X=k$	$0$	$1$	$2$	$3$	$4$
$P(X=k)$	$0.1$	$0.4$	$0.3$	$0.2$	$0$

If the mean of a Poisson distribution is $\frac{1}{2}$,then the ratio of $P(X=3)$ to $P(X=2)$ is

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