Question

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.

Answer #1

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) = ^{n}C_{x} p^{x}
(1 - p)^{n-x}

We want to find, P(X >= 1)

P(X >= 1)

= 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - [ ^{3}C_{0} * (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**

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