A man and his wife appear for an interview for two posts; their selections are…
2016
A man and his wife appear for an interview for two posts; their selections are independent of each other. The probability of selection of the husband is 1 by 7 and that of the wife is 1 by 5. What is the probability that only one of them is selected?
- A.
1/7
- B.
2/7
- C.
2/5
- D.
1/5
Attempted by 4 students.
Show answer & explanation
Correct answer: B
For two independent events A and B, the probability that exactly one of them occurs is P(A)·P(not B) + P(not A)·P(B). This holds because "A occurs, B does not" and "A does not occur, B occurs" are mutually exclusive cases, so once each joint probability is found using independence (multiplication), the two cases are combined using addition.
Let H = event that the husband is selected, with P(H) = 1/7, so P(not H) = 1 − 1/7 = 6/7.
Let W = event that the wife is selected, with P(W) = 1/5, so P(not W) = 1 − 1/5 = 4/5.
Case 1 — husband selected, wife not selected: P(H) × P(not W) = 1/7 × 4/5 = 4/35.
Case 2 — husband not selected, wife selected: P(not H) × P(W) = 6/7 × 1/5 = 6/35.
These two cases cannot happen together, so add them: P(exactly one selected) = 4/35 + 6/35 = 10/35.
Simplify 10/35 by dividing numerator and denominator by 5: 10/35 = 2/7.
Cross-check with the alternative identity for independent events, P(exactly one) = P(H) + P(W) − 2·P(H)·P(W): 1/7 + 1/5 − 2×(1/7×1/5) = 12/35 − 2/35 = 10/35 = 2/7, the same result.
So the probability that only one of the husband and wife is selected is 2/7.