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 $e^{-3} = \frac{10}{27} k$:$P(X=5) = \frac{81}{40} 	imes \frac{10}{27} k = \frac{3}{4} k$.

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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