Question

At a local sports club, 11/15

of the members play tennis and racquetball. 17/20

play tennis. If a member is selected at random, and that member plays tennis, find the probability that he/she also plays racquetball.

Question 8 options:

Answer 1

To solve the problem, we will use the concept of conditional probability.

Let's define the events:

- Let \( A \) be the event that a member plays tennis.
- Let \( B \) be the event that a member plays racquetball.

From the problem, we know:

1. \(\P(A) = \frac{17}{20}\) (the probability that a member plays tennis)
2. \(\P(A \cap B) = \frac{11}{15}\) (the probability that a member plays both tennis and racquetball)

We need to find the probability that a member plays racquetball given that they play tennis, i.e., \(\P(B | A)\).

Using the formula for conditional probability:

\[
\P(B | A) = \frac{\P(A \cap B)}{\P(A)}
\]

Substituting the known values:

\[
\P(B | A) = \frac{\P(A \cap B)}{\P(A)} = \frac{\frac{11}{15}}{\frac{17}{20}}
\]

To simplify this, we multiply by the reciprocal of \(\P(A)\):

\[
\P(B | A) = \frac{11}{15} \times \frac{20}{17}
\]

Now, let’s perform the multiplication:

\[
\P(B | A) = \frac{11 \times 20}{15 \times 17} = \frac{220}{255}
\]

Next, we simplify \(\frac{220}{255}\):

The GCD (Greatest Common Divisor) of 220 and 255 is 5.

Dividing both the numerator and denominator by 5:

\[
\P(B | A) = \frac{220 \div 5}{255 \div 5} = \frac{44}{51}
\]

Thus, the final probability that a member plays racquetball given that they play tennis is:

\[
\boxed{\frac{44}{51}}
\]