Question

A human gene carries a certain disease from the mother to the child with a probability rate of 39%. That is, there is a 39% 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 all four of the children get the disease from their mother. Round to the nearest thousandth.

Answer 1

Answer:

$Probability = 0.023$

Step-by-step explanation:

Given

Represent the given probability with P(Gene)

$P(Gene) = 39\%$

$Children = 4$

Since all 4 children get the disease, the required probability is calculated as thus:

$Probability = P(Gene)^4$

$Probability = 39\%^4$

Convert % to decimal

$Probability = 0.39^4$

$Probability = 0.02313441$

$Probability = 0.023$ Approximated