SOLUTION: a human gene carries a certain disease from the mother to the child with a probability rate 42%. That is a 42% chance that the child becomes infected with the disease. Suppose a

Algebra ->  Probability-and-statistics -> SOLUTION: a human gene carries a certain disease from the mother to the child with a probability rate 42%. That is a 42% chance that the child becomes infected with the disease. Suppose a       Log On


   

Question 1056935: a human gene carries a certain disease from the mother to the child with a probability rate 42%. That is a 42% chance that the child becomes infected with the disease. Suppose a female carrier of the gene has four children. Assume that the infections of the four children are independent of one another. Find the probability that at least one of the children get the disease from their mother.
Answer by jorel555(1290) About Me  (Show Source):
You can put this solution on YOUR website!
The probability of the children NOT getting the disease is 1-.42, or .58. So, the probability of NONE of them getting the disease is .58^4, or 0.11316496. So the probability of at least one of them getting the disease is 1-0.11316496, or 0.88683504.