Question

The mother and the father of five children are carriers (heterozygous) of an autosomal recessive allele that causes cystic fibrosis. The probability of having exactly three normal children among five is ........ (up to two decimal places).

Hint:

When solving binomial probability problems, use the binomial distribution formula \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \) to find the probability for specific outcomes in a given number of trials.

Answer 1

The probability of having a normal child (non-cystic fibrosis) is 3/4, since both parents are heterozygous (carriers). The probability of having a cystic fibrosis child is 1/4.
This is a binomial probability problem, where we need to find the probability of having exactly 3 normal children (out of 5). The formula for binomial probability is: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] Where:
- \( n = 5 \) (total number of children),
- \( k = 3 \) (exactly three normal children),
- \( p = \frac{3}{4} \) (probability of a normal child),
- \( 1 - p = \frac{1}{4} \) (probability of a cystic fibrosis child).
First, we calculate the binomial coefficient \( \binom{5}{3} \):
\[ \binom{5}{3} = \frac{5!}{3!(5 - 3)!} = \frac{5 \times 4}{2 \times 1} = 10 \] Now, we can calculate the probability:
\[ P(X = 3) = 10 \times \left(\frac{3}{4}\right)^3 \times \left(\frac{1}{4}\right)^2 \] \[ P(X = 3) = 10 \times \frac{27}{64} \times \frac{1}{16} = 10 \times \frac{27}{1024} = \frac{270}{1024} \approx 0.263 \] Thus, the probability of having exactly three normal children among five is approximately \( 0.26 \).