Question

A human gene carries a certain disease from the mother to the child with a probability rate of 34%. That is, there is a 34% chance that the child becomes infected with the disease. Suppose a female carrier of the gene has three children. Assume that the infections of the three children are independent of one another. Find the probability that at least one of the children get the disease from their mother.
Answer the following questions:

State the complement of the event "At least one of the children get the disease from their mother".
Find the probability of the complement. Round your answer to four decimals
Find the probability that at least one of the children get the disease from their mother.

Answer 1

Answer:

The probability that at least one of the children get the disease from their mother is P=0.7125.

Step-by-step explanation:

This can be modeled as a binomial random variable.

The parameter p=0.34 is the probability of a child being infected, and is constant and independent of the other events.

The parameter n=3 is the number of children  (sample size).

Then, we have to calculate the probabilty that at least one of the children get the disease from their mother. That is:

$P(x\geq1)$

The probability of exactly k children being infected can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}$

Then, the easiest way to calculate this probability is using the complement: the value of the probability is 1 (or 100%) less the probability that no children gets infected.

$P(x\geq1)=1-P(x=0)\\\\\\P(x=0) = \dbinom{3}{0} p^{0}(1-p)^{3}=1*1*0.2875=0.2875\\\\\\P(x\geq1)=1-0.2875=0.7125$