For a Poisson’s distribution with mean \( \displaystyle\frac{1}{2} \), which of the following statements are correct?
(A) \( P(0)=0 \)
(B) \( P(1)=\displaystyle\frac{1}{2\sqrt{e}} \)
(C) \( P(3)=\displaystyle\frac{1}{48\sqrt{e}} \)
(D) \( P(0)=\sqrt{e} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (C) only
For a Poisson distribution with mean λ,
\( \displaystyle P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \)
Here \( \lambda=\frac{1}{2} \), so \( e^{-\lambda}=e^{-1/2}=\frac{1}{\sqrt{e}} \).
\( \displaystyle P(0)=e^{-1/2}=\frac{1}{\sqrt{e}} \)
\( \displaystyle P(1)=\frac{e^{-1/2}\cdot\frac{1}{2}}{1!}=\frac{1}{2\sqrt{e}} \)
\( \displaystyle P(3)=\frac{e^{-1/2}\left(\frac{1}{2}\right)^3}{3!} \)
\( \displaystyle =\frac{1}{8\times 6\sqrt{e}}=\frac{1}{48\sqrt{e}} \)
So \( P(0)=\frac{1}{\sqrt{e}} \), \( P(1)=\frac{1}{2\sqrt{e}} \) and \( P(3)=\frac{1}{48\sqrt{e}} \).
Checking the statements
(A) \( P(0)=0 \) — Incorrect. P(0) is \( \frac{1}{\sqrt{e}} \), not 0.
(B) \( P(1)=\frac{1}{2\sqrt{e}} \) — Correct. \( e^{-1/2}\times\frac{1}{2}=\frac{1}{2\sqrt{e}} \).
(C) \( P(3)=\frac{1}{48\sqrt{e}} \) — Correct. \( e^{-1/2}\times\frac{1}{8}\div 6=\frac{1}{48\sqrt{e}} \).
(D) \( P(0)=\sqrt{e} \) — Incorrect. P(0) is \( \frac{1}{\sqrt{e}} \); besides, \( \sqrt{e}\gt 1 \) cannot be a probability.