In an athletic composition, the probability that an athlete wins a 100m race is 1/8 and the probability that…

Analyze the question: We want the probability that the athlete wins exactly one of the two events (100m or high jump), assuming the events are independent.

Concept: P(only one) = P(wins 100m and loses HJ) + P(loses 100m and wins HJ)

Solution:

• P(win 100m) = 1/8, so P(lose 100m) = 7/8
• P(win HJ) = 1/4, so P(lose HJ) = 3/4

Case 1: wins 100m only = (1/8)(3/4) = 3/32

Case 2: wins HJ only = (7/8)(1/4) = 7/32

Total = 3/32 + 7/32 = 10/32 = 5/16

Note that option A (3/32) is only one of the two cases, which is a common trap. Options B and C are greater than 1, so they cannot be probabilities.

Tip: For "only one of two events," use: P(A) + P(B) − 2·P(A and B). Check: 1/8 + 1/4 − 2(1/32) = 3/8 − 1/16 = 5/16. Also, any probability greater than 1 can be eliminated immediately.