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.

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

And
is the probability of X happening.
In this problem
15 drivers are randomly selected, so
.
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
.
What is the probability of getting 3 or more who were involved in a car accident last year?
This is
.
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:


In which





So

Finally

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