If X is a Poisson variate such that frac{5}{3 | vedclass

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(D) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.Given $P(X=2) = P(X=3)$, we have:

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$\frac{3}{4} k$

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(D) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.Given $P(X=2) = P(X=3)$, we have:$\frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^3}{3!}$$\frac{1}{2} = \frac{\lambda}{6} \implies \lambda = 3$.Now, we are given $\frac{5}{3} k = P(X=2) = \frac{e^{-3} 3^2}{2!} = \frac{9 e^{-3}}{2}$.So, $\frac{5}{3} k = \frac{9 e^{-3}}{2} \implies e^{-3} = \frac{10}{27} k$.We need to find $P(X=5) = \frac{e^{-3} 3^5}{5!} = \frac{e^{-3} 	imes 243}{120} = \frac{81}{40} e^{-3}$.Substituting $e^{-3} = \frac{10}{27} k$:$P(X=5) = \frac{81}{40} 	imes \frac{10}{27} k = \frac{3}{4} k$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

If $X$ is a Poisson variate such that $\frac{5}{3} k = P(X=2) = P(X=3)$, then $P(X=5) =$

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