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If the mean of a poisson distribution is $\frac{1}{2}$, then the ratio of $P(X=3)$ to $P(X=2)$ is
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Verified Answer
The correct answer is:
$1: 6$
Given that $\lambda=\frac{1}{2}$
Now, $P(X=n)=\frac{\lambda^n}{n !} e^\lambda$
$$
\begin{aligned}
& \therefore \quad P(X=3)=\frac{\left(\frac{1}{2}\right)^3}{3 !} e^{1 / 2} \\
& \text { and } P(X=2)=\frac{\left(\frac{1}{2}\right)^2}{2 !} e^{1 / 2} \\
& \therefore \quad \frac{P(X=3)}{P(X=2)}=\frac{\frac{\left(\frac{1}{2}\right)^3 e^{1 / 2}}{3 !}}{\frac{\left(\frac{1}{2}\right)^2 e^{1 / 2}}{2 !}}=\frac{\left(\frac{1}{2}\right)^3 e^{1 / 2} 2 !}{\left(\frac{1}{2}\right)^2 e^{1 / 2} 3 !}=\frac{\frac{1}{2}}{3}=\frac{1}{6} \\
&
\end{aligned}
$$
Now, $P(X=n)=\frac{\lambda^n}{n !} e^\lambda$
$$
\begin{aligned}
& \therefore \quad P(X=3)=\frac{\left(\frac{1}{2}\right)^3}{3 !} e^{1 / 2} \\
& \text { and } P(X=2)=\frac{\left(\frac{1}{2}\right)^2}{2 !} e^{1 / 2} \\
& \therefore \quad \frac{P(X=3)}{P(X=2)}=\frac{\frac{\left(\frac{1}{2}\right)^3 e^{1 / 2}}{3 !}}{\frac{\left(\frac{1}{2}\right)^2 e^{1 / 2}}{2 !}}=\frac{\left(\frac{1}{2}\right)^3 e^{1 / 2} 2 !}{\left(\frac{1}{2}\right)^2 e^{1 / 2} 3 !}=\frac{\frac{1}{2}}{3}=\frac{1}{6} \\
&
\end{aligned}
$$
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