If X is a Poisson variate such that P(X=1)=2 | vedclass

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(A) For a Poisson distribution, the probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.Given $P(X=1) = 2 P(X=2)$.Subst

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$\frac{e^{-1}}{6}$

Vedclass · Answer

(A) For a Poisson distribution, the probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.Given $P(X=1) = 2 P(X=2)$.Substituting the formula:$\frac{\lambda^1 e^{-\lambda}}{1!} = 2 	imes \frac{\lambda^2 e^{-\lambda}}{2!}$$\lambda e^{-\lambda} = 2 	imes \frac{\lambda^2 e^{-\lambda}}{2}$$\lambda e^{-\lambda} = \lambda^2 e^{-\lambda}$Since $\lambda > 0$ and $e^{-\lambda} 
eq 0$, we divide both sides by $\lambda e^{-\lambda}$:$1 = \lambda$Thus, the parameter $\lambda = 1$.Now, we calculate $P(X=3)$:$P(X=3) = \frac{\lambda^3 e^{-\lambda}}{3!} = \frac{1^3 e^{-1}}{6} = \frac{e^{-1}}{6}$.

If $X$ is a Poisson variate such that $P(X=1)=2 P(X=2)$, then $P(X=3)$ is equal to

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