If a Poisson variate X satisfies P(X=2) = P(X | vedclass

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(B) For a Poisson distribution, the probability mass function is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.Given $P(X=2) = P(X=3)$, we hav

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$\frac{81}{40 e^3}$

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(B) For a Poisson distribution, the probability mass function is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.Given $P(X=2) = P(X=3)$, we have:$\frac{\lambda^2 e^{-\lambda}}{2!} = \frac{\lambda^3 e^{-\lambda}}{3!}$$\frac{\lambda^2}{2} = \frac{\lambda^3}{6}$Since $\lambda > 0$, we can divide by $\lambda^2$:$\frac{1}{2} = \frac{\lambda}{6} \Rightarrow \lambda = 3$.Now, we calculate $P(X=5)$:$P(X=5) = \frac{3^5 e^{-3}}{5!} = \frac{243 e^{-3}}{120} = \frac{81 e^{-3}}{40} = \frac{81}{40 e^3}$.Thus, option $B$ is correct.

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.2$
$\omega_{3}$	$0.3$
$\omega_{4}$	$0.4$
$\omega_{5}$	$0.5$
$\omega_{6}$	$0.6$
$\omega_{7}$	$0.7$

If a Poisson variate $X$ satisfies $P(X=2) = P(X=3)$, then $P(X=5) =$

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