Question

At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is the probability that all 5 have a wireless device?

Answer 1

34.87% probability that all 5 have a wireless device

Explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So 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.

81% of students own a wireless device.

This means that
`p = 0.81`

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

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

`P(X = 5) = C_(5,5).(0.81)^(5).(0.19)^(0) = 0.3487`

34.87% probability that all 5 have a wireless device