in a certain algebra 2 class of 23 students, 9 of them play basketball and 12 of them play baseball. there…

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?

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?

Answer

Explanation:

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

Total students = 23, students who play neither = 9. So number of students who play at least one sport is $23 - 9=14$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball players ($n(A)=9$) and $B$ be the set of baseball players ($n(B)=12$). Let $n(A\cap B)$ be the number of students who play both. By the inclusion - exclusion principle $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. We know $n(A\cup B) = 14$, $n(A)=9$ and $n(B)=12$. So $14=9 + 12 - n(A\cap B)$.

Step3: Solve for $n(A\cap B)$

Rearranging the equation $14=9 + 12 - n(A\cap B)$ gives $n(A\cap B)=9 + 12-14=7$.

Step4: Calculate the probability

Probability $P=\frac{n(A\cap B)}{\text{Total number of students}}=\frac{7}{23}$.

Answer:

$\frac{7}{23}$