Question

An insurance company found that 9% of drivers were involved in a car accident last year. If seven drivers are randomly selected, find the probability that exactly two of them were involved in a car accident last year.

Answer 1

Answer: 0.1061

Step-by-step explanation:

Given : An insurance company found that 9% of drivers were involved in a car accident last year.

Thus, the probability of drivers involved in car accident last year = 0.09

The formula of binomial distribution :-

$P(X=x)^nC_xp^x(1-p)^{n-x}$

If seven (n=7) drivers are randomly selected then , the probability that exactly two (x=2) of them were involved in a car accident last year is given by :-

$P(X=2)=^7C_2(0.09)^2(1-0.09)^{7-2}\\\\=\dfrac{7!}{2!5!}(0.09)^2(0.91)^{5}=0.106147867882\approx0.1061$

Hence, the required probability :-0.1061