Q.40 A disease is inherited by a child with a probability of 1/4. In a family with two children, the
probability that exactly one sibling is affected by this disease is
(A) 1/4 (B) 3/8 (C) 7/16 (D) 9/16
This genetics probability problem models autosomal recessive inheritance, where each child has an independent 1/4 chance of being affected. For two children, the goal is to find the chance that exactly one is affected.
Problem Breakdown
Assume children are independent, with p = 1/4 probability of disease (affected, A) and q = 3/4 of no disease (N). Possible outcomes for child 1 and child 2: AA, AN, NA, NN, each with probabilities 1/16, 3/16, 3/16, 9/16.
Exactly one affected means AN or NA. Probability = 2 × 1/4 × 3/4 = 6/16 = 3/8.
✅ Correct Answer: 3/8 (Option B)
The calculation confirms 3/8 or 0.375, matching binomial probability for n=2, k=1:
Why Other Options Wrong
(A) 1/4: Equals both affected ((1/4)2 = 1/16) or one specific child affected (1/4), ignores the other case.
(C) 7/16: At least one affected is 1 – 9/16 = 7/16, not exactly one.
(D) 9/16: Both unaffected ((3/4)2 = 9/16), opposite of exactly one.
Outcomes Table
| Outcome | Probability | Description |
|---|---|---|
| AA | 1/16 | Both affected |
| AN | 3/16 | First affected, second not |
| NA | 3/16 | First not, second affected |
| NN | 9/16 | Neither affected |
Total for exactly one (AN + NA): 3/8
