If X is a Poisson variate such that 3P(X = 2) = 2P(X =1), then P(X =…

The correct answer is Option (2) → 0.264

For a Poisson variate with mean λ, \( P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \).

    \( \displaystyle 3\cdot\frac{e^{-\lambda}\lambda^2}{2}=2\cdot e^{-\lambda}\lambda \)

Dividing by \( e^{-\lambda}\lambda \) (λ > 0):

    \( \displaystyle \frac{3\lambda}{2}=2 \)

    \( \displaystyle \lambda=\frac{4}{3} \)

    \( \displaystyle P(X=0)=e^{-\lambda}=e^{-4/3} \)

    \( \displaystyle =0.264 \)