Question

(3 points) In Illinois, 9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders; that is, they have been arrested previously for a DUI offence. Suppose 28 people arrested for DUI in Illinois are selected at random. You may assume that this is a binomial distribution. (0.5 pts.) a) What is the probability that exactly 3 people are repeat offenders

Answer 1

Answer:

22.60% probability that exactly 3 people are repeat offenders

Step-by-step explanation:

For each driver arrested selected, there are only two possible outcomes. Either they are repeat offenders, or they are not. The probability of an arrested driver being a repeat offender is independent from other arrested drivers. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

In Illinois, 9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders;

This means that $p = 0.09$

Suppose 28 people arrested for DUI in Illinois are selected at random.

This means that $n = 28$

a) What is the probability that exactly 3 people are repeat offenders

This is $P(X = 3)$

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

$P(X = 3) = C_{28,3}.(0.91)^{25}.(0.09)^{3} = 0.2260$

22.60% probability that exactly 3 people are repeat offenders