in a certain algebra 2 class of 23 students, 10 of them play basketball and 11 of them play baseball. there…

in a certain algebra 2 class of 23 students, 10 of them play basketball and 11 of them play baseball. there are 10 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, 10 of them play basketball and 11 of them play baseball. there are 10 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 = 10. So number of students who play at least one sport is $23 - 10=13$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball players ($n(A)=10$) and $B$ be the set of baseball players ($n(B)=11$). Let $x$ be the number of students who play both. Then $n(A\cup B)=n(A)+n(B)-x$. We know $n(A\cup B) = 13$, $n(A)=10$, $n(B)=11$. So $13=10 + 11-x$.

Step3: Solve for $x$

$13=21 - x$, then $x=21 - 13=8$.

Step4: Calculate the probability

Probability = $\frac{\text{Number of students who play both}}{\text{Total number of students}}=\frac{8}{23}$.

Answer:

$\frac{8}{23}$