Question

A car insurance company has determined that6% of all drivers were involved in a car accident last year. If14drivers are randomly selected, what is the probability of getting at most 3 who were involved in a car accidentlast year

Answer 1

`P(x \le 3) = 0.9920`

Explanation:

Given

`p = 6\%` --- proportion of drivers that had accident

`n = 14` -- selected drivers

Required

`P(x \le 3)`

The question is an illustration of binomial probability, and it is calculated using:

`P(x ) = ^nC_x * p^x * (1 - p)^(n-x)`

So, we have:

`P(x \le 3) = P(x = 0) +P(x = 1) +P(x = 2) +P(x = 3)`

`P(x=0 ) = ^(14)C_0 * (6\%)^0 * (1 - 6\%)^(14-0) = 0.42052319017`

`P(x=1 ) = ^(14)C_1 * (6\%)^1 * (1 - 6\%)^(14-1) = 0.37578668057`

`P(x=2 ) = ^(14)C_2 * (6\%)^2 * (1 - 6\%)^(14-2) = 0.15591149513`

`P(x=3 ) = ^(14)C_3 * (6\%)^3 * (1 - 6\%)^(14-3) = 0.03980719024`

So, we have:

`P(x \le 3) = 0.42052319017+0.37578668057+0.15591149513+0.03980719024`

`P(x \le 3) = 0.99202855611`

`P(x \le 3) = 0.9920` -- approximated