Question

in a certain algebra 2 class of 30 students, 7 of them play basketball and 11 of them play baseball. there are 3 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

Explanation:

Step1: Recall the principle of inclusion - exclusion for probability

The formula for \( P(A \cup B) \) (the probability that event \( A \) or event \( B \) occurs) is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \), where \( A \) is the event that a student plays basketball and \( B \) is the event that a student plays baseball.

First, find the probabilities of \( A \), \( B \), and \( A \cap B \). The total number of students \( n = 30 \).

For event \( A \) (plays basketball): \( n(A)=7 \), so \( P(A)=\frac{n(A)}{n}=\frac{7}{30} \).

For event \( B \) (plays baseball): \( n(B) = 11 \), so \( P(B)=\frac{n(B)}{n}=\frac{11}{30} \).

For event \( A\cap B \) (plays both): \( n(A\cap B)=3 \), so \( P(A\cap B)=\frac{n(A\cap B)}{n}=\frac{3}{30} \).

Step2: Calculate \( P(A \cup B) \)

Substitute the values into the formula:
\( P(A \cup B)=\frac{7}{30}+\frac{11}{30}-\frac{3}{30} \)
First, add the first two fractions: \( \frac{7 + 11}{30}=\frac{18}{30} \)
Then subtract the third fraction: \( \frac{18-3}{30}=\frac{15}{30} \)
Simplify \( \frac{15}{30}=\frac{1}{2} \)

Answer:

\(\frac{1}{2}\)