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

Answer:

D. 0.200, 0.026

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.