Question

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers living on one particular street, 3 were involved in a car accident last year. If 15 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?

Answer 1

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$.

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year.  This means that $\pi = 0.08$.

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$.

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863$

$P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734$

$P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113$

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.