If X has a Poisson distribution such that 2P(X=1)=P(X=2) , then which…

The correct answer is Option (1) → A and C only

For a Poisson distribution with mean λ, \( P(X=r)=\frac{e^{-\lambda}\lambda^{r}}{r!} \). The given condition is

    \( \displaystyle 2\cdot\frac{e^{-\lambda}\lambda}{1!}=\frac{e^{-\lambda}\lambda^{2}}{2!} \)

    \( \displaystyle 2\lambda=\frac{\lambda^{2}}{2} \)

    \( \lambda^{2}=4\lambda \)

Since λ > 0, λ = 4. Then

    \( \displaystyle P(X=2)=\frac{e^{-4}\cdot 4^{2}}{2!} \)

    \( \displaystyle =\frac{8}{e^{4}} \)

    \( \displaystyle P(X=3)=\frac{e^{-4}\cdot 4^{3}}{3!} \)

    \( \displaystyle =\frac{32}{3e^{4}} \)

So the mean is 4, \( P(X=2)=\frac{8}{e^{4}} \) and \( P(X=3)=\frac{32}{3e^{4}} \).

Checking the statements

(A) The mean of X is 4 — Correct. The mean is λ = 4.

(B) The mean of X is 2 — Incorrect. The mean is 4, not 2.

(C) \( P(X=2)=\frac{8}{e^{4}} \) — Correct. \( \frac{16}{2!\,e^{4}}=\frac{8}{e^{4}} \).

(D) \( P(X=3)=\frac{32}{e^{4}} \) — Incorrect. \( \frac{64}{3!\,e^{4}}=\frac{32}{3e^{4}} \), not \( \frac{32}{e^{4}} \).