Question

Forty-three percent of US teens have heard of a fax machine. You randomly select 12 US teens. Find the probability that the number of these selected teens that have heard of a fax machine is exactly six (first answer listed below). Find the probability that the number is more than 8 (second answer listed below).

A. 0.784, 0.974
B. 0.200, 0.974
C. 0.784, 0.026
D. 0.200, 0.026

Answer 1

D. 0.200, 0.026

Explanation:

For each teen, there are only two possible outcomes. Either they have a fax machine, or they do not. The probability of a teen having a fax machine is independent of other teens. 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.

Forty-three percent of US teens have heard of a fax machine.

This means that
`p = 0.43`

You randomly select 12 US teens.

This means that
`n = 12`

Find the probability that the number of these selected teens that have heard of a fax machine is exactly six (first answer listed below).

This is
`P(X = 6)`. So

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

`P(X = 6) = C_(12,6).(0.43)^(6).(0.57)^(6) = 0.200`

So the answer is either option B or D.

Find the probability that the number is more than 8

`P(X > 8) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)`

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

`P(X = 9) = C_(12,9).(0.43)^(9).(0.57)^(3) = 0.021`

`P(X = 10) = C_(12,10).(0.43)^(10).(0.57)^(2) = 0.005`

`P(X = 11) = C_(12,11).(0.43)^(11).(0.57)^(1) = 0`

`P(X = 12) = C_(12,12).(0.43)^(12).(0.57)^(0) = 0`

`P(X > 8) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.021 + 0.005 = 0.026`

So the correct answer is given by option D.