Question

question 22 (5 points) ✓ saved
a human gene carries a certain disease from the mother to the child with a
probability rate of 49%. that is, there is a 49% 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 all three of the children get the disease from their mother.
round to the nearest thousandth.

Explanation:

Step1: Convert percentage to decimal

The probability of a child getting the disease is 49%, which is \( 0.49 \) in decimal form.

Step2: Use independent events probability formula

For independent events, the probability that all three events occur is the product of their individual probabilities. So we calculate \( 0.49 \times 0.49 \times 0.49 \).

$$ 0.49\times0.49\times0.49 = 0.49^{3}=0.49\times0.49 = 0.2401; \quad 0.2401\times0.49 = 0.117649 $$

Step3: Round to nearest thousandth

Rounding \( 0.117649 \) to the nearest thousandth (three decimal places), we look at the fourth decimal place which is 6. Since \( 6\geq5 \), we round up the third decimal place. So \( 0.117649\approx0.118 \).

Answer:

\( 0.118 \)