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(C) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.Given the condition $3 P(X=2) = P

Vedclass · Accepted Answer

$\frac{324}{5 e^6}$

Vedclass · Answer

(C) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.Given the condition $3 P(X=2) = P(X=4)$.Substituting the formula, we get $3 \cdot \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^4}{4!}$.Canceling $e^{-\lambda}$ from both sides and simplifying the factorials: $\frac{3 \lambda^2}{2} = \frac{\lambda^4}{24}$.Multiplying both sides by $24$, we get $36 \lambda^2 = \lambda^4$.Since $\lambda > 0$, we have $\lambda^2 = 36$, which implies $\lambda = 6$.Now, we need to find $P(X=6) = \frac{e^{-6} \cdot 6^6}{6!}$.Calculating the value: $P(X=6) = \frac{e^{-6} \cdot 46656}{720} = \frac{46656}{720 e^6} = \frac{324}{5 e^6}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$k$	$\frac{k}{3}$	$\frac{k}{4}$	$\frac{k}{2}$	$\frac{k}{2}$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

If $X$ is a Poisson variate satisfying the condition $3 P(X=2)=P(X=4)$, then find $P(X=6)$.

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