If X is a Poisson variate such that 4P(X=1)=3P(X=2) , then P(X=0)=

The correct answer is Option (1) → \( e^{-8/3} \)

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

    \( \displaystyle 4\cdot\frac{e^{-\lambda}\lambda}{1!}=3\cdot\frac{e^{-\lambda}\lambda^{2}}{2!} \)

Dividing both sides by \( e^{-\lambda}\lambda \) (\( \lambda\gt 0 \)):

    \( \displaystyle 4=\frac{3\lambda}{2} \)

    \( \displaystyle \lambda=\frac{8}{3} \)

    \( \displaystyle P(X=0)=e^{-\lambda}=e^{-8/3} \)