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

Answer:

$P(x \le 3) = 0.9920$

Step-by-step 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