If a Poisson variate X satisfies the relation | vedclass

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $P(X=3) = P(X=5)$, we have

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$\frac{50}{3 e^{\sqrt{20}}}$

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $P(X=3) = P(X=5)$, we have $\frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^5}{5!}$.Canceling $e^{-\lambda}$ and $\lambda^3$ from both sides, we get $\frac{1}{6} = \frac{\lambda^2}{120}$.Thus, $\lambda^2 = \frac{120}{6} = 20$, which implies $\lambda = \sqrt{20}$.Now, we need to find $P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!}$.Substituting $\lambda = \sqrt{20}$ and $\lambda^2 = 20$, we get $P(X=4) = \frac{e^{-\sqrt{20}} (20)^2}{24} = \frac{400}{24 e^{\sqrt{20}}} = \frac{50}{3 e^{\sqrt{20}}}$.

If a Poisson variate $X$ satisfies the relation $P(X=3)=P(X=5)$, then $P(X=4)=$

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