Question

A survey found that 20% of the people in Fiji drive the car without seatbelt. If 3 people are randomly selected, find the probability that at least 2 drive without seat.

Answer 1

0.104 = 10.4% probability that at least 2 drive without a seatbelt.

Explanation:

For each person, there are only two possible outcomes. Either they drive without a seatbelt, or they do not. The probability of a person driving without a seatbelt is independent of any other person. This means that we use the binomial probability distribution to solve this question.

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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

20% of the people in Fiji drive the car without seatbelt.

This means that
`p = 0.2`

3 people are randomly selected

This means that
`n = 3`

Find the probability that at least 2 drive without a seatbelt?

This is:

`P(X \geq 2) = P(X = 2) + P(X = 3)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 2) = C_(3,2).(0.2)^(2).(0.8)^(1) = 0.096`

`P(X = 3) = C_(3,3).(0.2)^(3).(0.8)^(0) = 0.008`

`P(X \geq 2) = P(X = 2) + P(X = 3) = 0.096 + 0.008 = 0.104`

0.104 = 10.4% probability that at least 2 drive without a seatbelt.