Question

15. The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among U.S. women ages 20 to 24 is 2791.5 per 100,000. Take a random sample of three U.S. women in this age group. What is the probability that at least one of the women has a Chlamydia infection

Answer 1

Answer:

0.0814 = 8.14% probability that at least one of the women has a Chlamydia infection

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they have a Chlamydia infection, or they do not. The probability of a woman having the infection is independent of any other women. This means that the binomial probability distribution is used 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.

The Centers for Disease Control and Prevention reports that the rate of Chlamydia infections among U.S. women ages 20 to 24 is 2791.5 per 100,000.

This means that:

$p = \frac{2791.5}{100000} = 0.027915$

Take a random sample of three U.S. women in this age group.

This means that $n = 3$

What is the probability that at least one of the women has a Chlamydia infection?

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.027915)^{0}.(1-0.027915)^{3} = 0.9186$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9186 = 0.0814$

0.0814 = 8.14% probability that at least one of the women has a Chlamydia infection