Question

In a town $10$ accidents take place in a span of $50$ days. Assuming that number of accidents follows Poisson distribution, the probability that there will be atleast one accident on a selected day at random is

• A. $\frac{e^{-0.02}{.2}^1}{1!}$
• B. $1-e^{-0.2}$
• C. $e^{-0.2}$
• D. $1-e^{1.2}$

Answer 1

Answer: (B)

$1-e^{-0.2}$

Here Poisson parameter $\lambda =\frac{10}{50}=0.2$
$\therefore P(x\ge 1)=1-P(X=0)=1-\frac{e^{-0.2}.(.2{)}^0}{0!}=1-e^{-0.2}$