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(C) Let $\lambda$ be the mean of the Poisson distribution for the random variable $X$. The probability mass function is given by $P(X=r) = \frac{\lamb

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$\frac{4}{15} e^{-2}$

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(C) Let $\lambda$ be the mean of the Poisson distribution for the random variable $X$. The probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$ for $r = 0, 1, 2, \dots$. Given $P(X=1) = P(X=2)$, we have: $\frac{\lambda^1 e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}$ $\lambda = \frac{\lambda^2}{2}$ Since $\lambda > 0$, we divide by $\lambda$ to get $1 = \frac{\lambda}{2}$, which implies $\lambda = 2$. Now, we calculate $P(X=5)$: $P(X=5) = \frac{2^5 e^{-2}}{5!} = \frac{32 e^{-2}}{120}$. Simplifying the fraction $\frac{32}{120}$ by dividing both numerator and denominator by $8$, we get $\frac{4}{15}$. Thus, $P(X=5) = \frac{4}{15} e^{-2}$.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

Suppose that a random variable $X$ follows a Poisson distribution. If $P(X=1) = P(X=2)$, then $P(X=5)$ is equal to:

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