Suppose $X$ has Poisson distribution such that $3 P(X=1)=2 P(X=2)$ then $P(X>0)$ is:

$1-e^{-3}$

The correct answer is Option (3) → $1-e^{-3}$

$X\sim \text{Poisson}(\lambda)$

$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$

$P(X=1)=e^{-\lambda}\lambda$

$P(X=2)=\frac{e^{-\lambda}\lambda^2}{2}$

$3P(X=1)=2P(X=2)$

$3(e^{-\lambda}\lambda)=2\left(\frac{e^{-\lambda}\lambda^2}{2}\right)$

$3\lambda=\lambda^2$

$\lambda=3$

$P(X>0)=1-P(X=0)$

$P(X=0)=e^{-3}$

$P(X>0)=1-e^{-3}$

The probability is $1-e^{-3}$.