A human gene carries a certain disease from the mother to the child with a probability rate of 51%. That is,there is a 51% chance that the child becomes infected with the disease. Suppose a female carrier of the gene hasfive children. Assume that the infections of the five children are independent of one another. Find theprobability that at least one of the children get the disease from their mother. Round to the nearest thousandth.

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Quizplus · Accepted Answer

The probability that at least one child gets the disease is 1 minus the probability that none of the children get the disease. The probability that a child does not get the disease is 49% (100% - 51%). For five children, this is $0.49^5$. Thus, the probability at least one gets the disease is $1 - 0.49^5 \approx 0.972$.

A Human Gene Carries a Certain Disease from the Mother