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If a random variable $\mathrm{X}$ follows a poisson distribution such that $P(X=1)=3 P(X=2)$, then $P(x=3)=$
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Verified Answer
The correct answer is:
$\frac{4}{81} e^{-\frac{2}{3}}$
In Poisson distribution
$$
\begin{aligned}
& P(X=r)=\frac{\lambda^p e^{-\lambda}}{r !} \text {, where } \lambda=n p \\
& P(X=1)=3 P(X=4 \\
& \frac{\lambda e^{-\lambda}}{1 !}=\frac{3 \lambda^2 e^{-\lambda}}{2 !} \\
& \Rightarrow \quad \lambda=2 / 3 \\
& \therefore \quad P(X=3)=\frac{\lambda^3 e^{-\lambda}}{3 !}=\frac{\left(\frac{2}{3}\right)^3 e^{-2 / 3}}{3 !}=\frac{4}{81} e^{-2 / 3} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& P(X=r)=\frac{\lambda^p e^{-\lambda}}{r !} \text {, where } \lambda=n p \\
& P(X=1)=3 P(X=4 \\
& \frac{\lambda e^{-\lambda}}{1 !}=\frac{3 \lambda^2 e^{-\lambda}}{2 !} \\
& \Rightarrow \quad \lambda=2 / 3 \\
& \therefore \quad P(X=3)=\frac{\lambda^3 e^{-\lambda}}{3 !}=\frac{\left(\frac{2}{3}\right)^3 e^{-2 / 3}}{3 !}=\frac{4}{81} e^{-2 / 3} \\
&
\end{aligned}
$$
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