Question

According to the National Institute of Allergy and Infectious Diseases, approximately 7% of U.S. children 4 years of age or younger have a food allergy. A day care program has capacity for 8 children in that age range. Assume that the children attending the day care program are independent. Let the random variable X be the number of children in this day care who have a food allergy. What is the probability that one of the eight children has a food allergy

Answer 1

Answer:

0.337 = 33.7% probability that one of the eight children has a food allergy.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they have a food allergy, or they do not. The probability of a child having food allergy is independent of any other child. This means that 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.

7% of U.S. children 4 years of age or younger have a food allergy.

This means that $p = 0.07$

A day care program has capacity for 8 children in that age range.

This means that $n = 8$

What is the probability that one of the eight children has a food allergy?

This is P(X = 1).

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

$P(X = 1) = C_{8,1}.(0.07)^{1}.(0.93)^{7} = 0.337$

0.337 = 33.7% probability that one of the eight children has a food allergy.