Answer:
Option A
Explanation:
Here,
n=500 ,p=2%, q=98%
Required probability
$P(X\leq1)=P(X=0)+P(X=1)$
$=\frac{\lambda^{0}e^{-\lambda}}{0!}+\frac{\lambda^{}e^{-\lambda}}{1!}$
$P(X\leq1)=e^{-\lambda}(1+\lambda)$
= $e^{-10}(1+10)$ [$\because \lambda=np= 500 \times \frac{2}{100}=10$]
= $\frac{11}{e^{10}}$