Question

in a certain algebra 2 class of 23 students, 9 of them play basketball and 12 of them play baseball. there are 9 students who play neither sport. what is the probability that a student chosen randomly from the class plays both basketball and baseball?

Explanation:

Step1: Find the number of students who play at least one sport

Total students = 23, students who play neither = 9. So students who play at least one sport = \(23 - 9 = 14\).

Step2: Use the principle of inclusion - exclusion

Let \(B\) be the set of basketball players and \(Base\) be the set of baseball players. We know \(|B| = 9\), \(|Base| = 12\), and \(|B\cup Base| = 14\). The formula is \(|B\cup Base|=|B| + |Base|-|B\cap Base|\). Substituting values: \(14 = 9 + 12 - |B\cap Base|\).

Step3: Solve for \(|B\cap Base|\)

Simplify the equation: \(14 = 21 - |B\cap Base|\). Then \(|B\cap Base| = 21 - 14 = 7\).

Step4: Calculate the probability

Probability = \(\frac{\text{Number of students who play both}}{\text{Total number of students}}=\frac{7}{23}\).

Answer:

\(\frac{7}{23}\)