For a Poisson variate X, if P(X=2)=3 P(X=3), | vedclass

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(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $\lambda$ is the mean of t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $\lambda$ is the mean of the distribution. Given the condition $P(X=2) = 3 P(X=3)$. Substituting the formula: $\frac{\lambda^2 e^{-\lambda}}{2!} = 3 \cdot \frac{\lambda^3 e^{-\lambda}}{3!}$ Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} 
eq 0$) and $\lambda^2$ (assuming $\lambda > 0$): $\frac{1}{2} = 3 \cdot \frac{\lambda}{3 	imes 2 	imes 1}$ $\frac{1}{2} = \frac{3\lambda}{6}$ $\frac{1}{2} = \frac{\lambda}{2}$ $\lambda = 1$. Thus, the mean of the Poisson distribution is $1$.

For a Poisson variate $X$, if $P(X=2)=3 P(X=3)$, then the mean of $X$ is :

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