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 have:

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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.

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

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

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