Question

Assume that when adults with smartphones are randomly​ selected, 58​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that at least 5 of them use their smartphones in meetings or classes.

Answer 1

79.85% probability that at least 5 of them use their smartphones in meetings or classes.

Explanation:

For each adult, there are only two possible outcomes. Either they use their smartphone during meetings or classes, or they do not. The probability of an adult using their smartphone in these situations are independent of other adults. 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.

Assume that when adults with smartphones are randomly​ selected, 58​% use them in meetings or classes.

This means that
`p = 0.58`

10 adults selected.

This means that
`n = 10`

Find the probability that at least 5 of them use their smartphones in meetings or classes.

`P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)`

In which

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

`P(X = 5) = C_(10,5).(0.58)^(5).(0.42)^(5) = 0.2162`

`P(X = 6) = C_(10,6).(0.58)^(6).(0.42)^(4) = 0.2488`

`P(X = 7) = C_(10,7).(0.58)^(7).(0.42)^(3) = 0.1963`

`P(X = 8) = C_(10,8).(0.58)^(8).(0.42)^(2) = 0.1017`

`P(X = 9) = C_(10,9).(0.58)^(9).(0.42)^(1) = 0.0312`

`P(X = 10) = C_(10,10).(0.58)^(10).(0.42)^(0) = 0.0043`

`P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2162 + 0.2488 + 0.1963 + 0.1017 + 0.0312 + 0.0043 = 0.7985`

79.85% probability that at least 5 of them use their smartphones in meetings or classes.