If X is a Poisson variate such that P(X=2)=P( | vedclass

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(A) $X$ is a Poisson variate. Given $P(X=2) = P(X=3)$. Using the formula $P(X=r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!}$,we have: $\frac{e^{-\lambd

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$\left(\frac{3}{2}\right)^3$

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(A) $X$ is a Poisson variate. Given $P(X=2) = P(X=3)$. Using the formula $P(X=r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!}$,we have: $\frac{e^{-\lambda} \cdot \lambda^2}{2!} = \frac{e^{-\lambda} \cdot \lambda^3}{3!}$ $\Rightarrow \frac{\lambda^2}{2} = \frac{\lambda^3}{6}$ $\Rightarrow \lambda = \frac{6}{2} = 3$. Now,we calculate $P(X=4)$: $P(X=4) = \frac{e^{-\lambda} \cdot \lambda^4}{4!} = \frac{e^{-3} \cdot 3^4}{4!}$. Finally,we find $e^3 P(X=4)$: $e^3 P(X=4) = e^3 \cdot \frac{e^{-3} \cdot 3^4}{4!} = \frac{3^4}{4!} = \frac{81}{24} = \frac{27}{8} = \left(\frac{3}{2}\right)^3$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x)$	$0.2$	$0.3$	$0.1$	$0.15$	$0.25$

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

If $X$ is a Poisson variate such that $P(X=2)=P(X=3)$,then $e^3 P(X=4)$ is

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