Question

g The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent. What is the probability that at least one of the births results in a defect

Answer 1

Answer:

0.1426 = 14.26% probability that at least one of the births results in a defect.

Step-by-step explanation:

For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability that a birth results in a defect is independent of any other birth. 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 proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).

This means that $p = \frac{1}{33}$

A local hospital randomly selects five births.

This means that $n = 5$

What is the probability that at least one of the births results in a defect?

This is:

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

In which

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

$P(X = 0) = C_{5,0}.(\frac{1}{33})^{0}.(\frac{32}{33})^{5} = 0.8574$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8574 = 0.1426$

0.1426 = 14.26% probability that at least one of the births results in a defect.