If a random variable X follows a Poisson dist | vedclass

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Question

(A) For a Poisson distribution, the probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$, where $\lambda$ is the paramet

Vedclass · Accepted Answer

$\frac{4}{81} e^{-\frac{2}{3}}$

Vedclass · Answer

(A) For a Poisson distribution, the probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$, where $\lambda$ is the parameter of the distribution. Given that $P(X=1) = 3P(X=2)$. Substituting the formula: $\frac{\lambda^1 e^{-\lambda}}{1!} = 3 	imes \frac{\lambda^2 e^{-\lambda}}{2!}$ $\lambda = 3 	imes \frac{\lambda^2}{2}$ Since $\lambda 
eq 0$, we divide by $\lambda$: $1 = \frac{3\lambda}{2} \implies \lambda = \frac{2}{3}$. Now, we calculate $P(X=3)$: $P(X=3) = \frac{\lambda^3 e^{-\lambda}}{3!} = \frac{(\frac{2}{3})^3 e^{-\frac{2}{3}}}{6}$ $P(X=3) = \frac{\frac{8}{27} e^{-\frac{2}{3}}}{6} = \frac{8}{27 	imes 6} e^{-\frac{2}{3}} = \frac{4}{81} e^{-\frac{2}{3}}$.

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$k$	$2k$	$3k$	$4k$	$4k$	$3k$	$2k$	$k$	$k$

If a random variable $X$ follows a Poisson distribution such that $P(X=1) = 3P(X=2)$, then $P(X=3) =$

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