The problem states that the probability of an athlete winning a 100m race is $\frac{1}{8}$, and the probability of him winning a high jump is $\frac{1}{4}$. We are asked to find the probability that he wins only one of the events.

1. Problem Description

The problem states that the probability of an athlete winning a 100m race is $\frac{1}{8}$, and the probability of him winning a high jump is $\frac{1}{4}$. We are asked to find the probability that he wins only one of the events.

2. Solution Steps

Let A be the event that the athlete wins the 100m race, and B be the event that the athlete wins the high jump.

We are given $P(A) = \frac{1}{8}$ and $P(B) = \frac{1}{4}$.

We want to find the probability that the athlete wins only one of the events. This means he either wins the 100m race and loses the high jump, or he loses the 100m race and wins the high jump.

We assume that the events A and B are independent.

The probability that he wins the 100m race and loses the high jump is $P(A \cap B^c) = P(A) \times P(B^c)$, where $B^c$ is the complement of B.

Since $P(B) = \frac{1}{4}$, $P(B^c) = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}$.

So, $P(A \cap B^c) = \frac{1}{8} \times \frac{3}{4} = \frac{3}{32}$.

The probability that he loses the 100m race and wins the high jump is $P(A^c \cap B) = P(A^c) \times P(B)$, where $A^c$ is the complement of A.

Since $P(A) = \frac{1}{8}$, $P(A^c) = 1 - P(A) = 1 - \frac{1}{8} = \frac{7}{8}$.

So, $P(A^c \cap B) = \frac{7}{8} \times \frac{1}{4} = \frac{7}{32}$.

The probability that he wins only one of the events is the sum of these probabilities:

$P(\text{wins only one event}) = P(A \cap B^c) + P(A^c \cap B) = \frac{3}{32} + \frac{7}{32} = \frac{10}{32} = \frac{5}{16}$.

3. Final Answer

The probability that he wins only one of the events is $\frac{5}{16}$.

Answer: D. $\frac{5}{16}$