Question

About 3% of children in the United States are allergic to peanuts. Choose three children at random and let the random variable X be the number in this sample who are allergic to peanuts. The possible values X can take are 0,1,2,and 3. What is the conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy?Round your answer to six decimal places.

Answer 1

Answer:

0.029991 = 2.9991% conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are allergic to peanuts, or they are not. The probability of a children being allergic to peanut butter is independent of other children. 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.

3% of children in the United States are allergic to peanuts.

So $p = 0.03$

Three children:

So n = 3,

The probabilities are:

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

$P(X = 0) = C_{3,0}.(0.03)^{0}.(0.97)^{3} = 0.912673$

$P(X = 1) = C_{3,1}.(0.03)^{1}.(0.97)^{2} = 0.084681$

$P(X = 2) = C_{3,2}.(0.03)^{2}.(0.97)^{1} = 0.002619$

$P(X = 3) = C_{3,3}.(0.03)^{3}.(0.97)^{0} = 0.000027$

What is the conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy?

This is

$p = \frac{P(X = 2)}{P(X \geq 1)}$

In which

$P(X \geq 1) = 0.084681 + 0.002619 + 0.000027 = 0.087327$

$P(X = 2) = 0.002619$

So

$p = \frac{0.002619}{0.087327} = 0.03$

0.029991 = 2.9991% conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy