If \( X \) is a Poisson variate such that \( P(X=3)=2P(X=4) \), then \( P(X=2) \) equals:
[given \( e^{-2}=0.14 \)]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.28
For a Poisson variate with mean \( \lambda \), \( P(X=r)=\frac{e^{-\lambda}\lambda^{r}}{r!} \). The given relation is
\( \displaystyle \frac{e^{-\lambda}\lambda^{3}}{3!}=2\cdot\frac{e^{-\lambda}\lambda^{4}}{4!} \)
Cancelling \( e^{-\lambda}\lambda^{3} \),
\( \displaystyle \frac{1}{6}=\frac{2\lambda}{24} \)
\( \displaystyle \lambda=2 \)
Then
\( \displaystyle P(X=2)=\frac{e^{-2}\cdot2^{2}}{2!} \)
\( \displaystyle =2e^{-2} \)
\( \displaystyle =2(0.14) \)
\( \displaystyle =0.28 \)