If X is a Poisson variate such that alpha = P | vedclass

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(B) Given that $X$ is a Poisson variate with parameter $\lambda$, the probability mass function is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given

Vedclass · Accepted Answer

$\frac{\alpha}{3}$

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(B) Given that $X$ is a Poisson variate with parameter $\lambda$, the probability mass function is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $\alpha = P(X=1) = P(X=2)$:$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$$\lambda = \frac{\lambda^2}{2} \Rightarrow \lambda = 2$ (since $\lambda > 0$).Now, calculate $\alpha$:$\alpha = P(X=1) = \frac{e^{-2} 	imes 2^1}{1!} = 2e^{-2}$.We need to find $P(X=4)$:$P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-2} 	imes 2^4}{24} = \frac{16 e^{-2}}{24} = \frac{2}{3} e^{-2}$.Since $\alpha = 2e^{-2}$, we have $e^{-2} = \frac{\alpha}{2}$.Substituting this into the expression for $P(X=4)$:$P(X=4) = \frac{2}{3} 	imes \frac{\alpha}{2} = \frac{\alpha}{3}$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$, then $P(X=4)$ is equal to

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