Question

If $X$ is a poisson variate such that $P(X=0)=0.1,P(X=2)=0.2$, then the parameter $\lambda$

• A. $2$
• B. $4$
• C. $5$
• D. $3$

Answer 1

The correct option is A $2$

$P(X=0)=0.1$ and $P(X=2)=0.2$

$P$$(0;\mu )=\frac{e^{-\mu }{\mu }^0}{0!}=e^{-\mu }$ = $0.1$

$P$$(2;\mu )=\frac{e^{-\mu }{\mu }^2}{2!}$ $=0.2$

=> $\frac{0.1{\mu }^2}{2}$ $=0.2$

=> $\mu$ $=2$

The parameter is mean ($\mu$) in Poisson distribution, so $=$ $2$