Question

Assume that when adults with smartphones are randomly​ selected, 5656​% use them in meetings or classes. If 1212 adult smartphone users are randomly​ selected, find the probability that fewer than 55 of them use their smartphones in meetings or classes.

Answer 1

9.88% probability that fewer than 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 is meetings or classes, or they do not. The probability of an adult using their phones is 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.

56​% use them in meetings or classes.

This means that
`p = 0.56`

12 adult smartphone users are randomly​ selected

This means that
`n = 12`

Probability that fewer than 5 of them use their smartphones in meetings or classes.

`P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)`

In which

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

`P(X = 0) = C_(12,0).(0.56)^(0).(0.44)^(12) = 0.0001`

`P(X = 1) = C_(12,1).(0.56)^(1).(0.44)^(11) = 0.0008`

`P(X = 2) = C_(12,2).(0.56)^(2).(0.44)^(10) = 0.0056`

`P(X = 3) = C_(12,3).(0.56)^(3).(0.44)^(9) = 0.0239`

`P(X = 4) = C_(12,4).(0.56)^(4).(0.44)^(8) = 0.0684`

`P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0001 + 0.0008 + 0.0056 + 0.0239 + 0.0684 = 0.0988`

9.88% probability that fewer than 5 of them use their smartphones in meetings or classes.