A human gene carries a certain disease from the mother to the child with a probability rate of 40%. That is, there is a 40% 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.
Let X be a number of the children get the disease from their mother among three children.
Here, X ~ Binomial (n = 3, p = 0.40)
Probability mass function of X is,
P(X = x) = nCx px (1 - p)n-x
We want to find, P(X >= 1)
P(X >= 1)
= 1 - P(X < 1)
= 1 - P(X = 0)
= 1 - [ 3C0 * (0.40)0 * (1 - 0.40)3-1 ]
= 1 - [ 1 * 1 * (0.60)3 ]
= 1 - 0.2160
= 0.7840
=> P(X >= 1) = 0.7840
Therefore, the probability that at least one of the children get the disease from their mother is 0.7840
Get Answers For Free
Most questions answered within 1 hours.