Question

A human gene carries a certain disease from the mother to the child with a probability rate of 70%. suppose a female carrier of the gene has three children. also assume that the infections of the three children are independent of one another. find the probability that at least one child gets the disease from their mother.

Answer 1

The probability of at least one child inheriting the disease can be given by
P(X ≥ 1) = 1 - P(X = 0).  This is because the only choice other than at least one child inheriting the disease is that no children get it.
Since there is a 70% or 0.7 chance that each child will inherit it, there is a 30% or 0.3 chance that they will not.  This gives us
1 - P(X = 0) = 1 - (0.3)(0.3)(0.3) = 1 - 0.027 = 0.973
There is a 97.3% chance that at least one of the children will get the disease from their mother.