in a certain algebra 2 class of 23 students, 15 of them play basketball and 16 of them play baseball. there…

in a certain algebra 2 class of 23 students, 15 of them play basketball and 16 of them play baseball. there are 3 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, 15 of them play basketball and 16 of them play baseball. there are 3 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 - students who play neither sport. So, $23 - 3=20$ students play at least one sport.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball players and $B$ be the set of baseball players. We know $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Here $n(A) = 15$, $n(B)=16$ and $n(A\cup B) = 20$. Substituting gives $20=15 + 16 - n(A\cap B)$.

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

Rearrange the equation: $n(A\cap B)=15 + 16-20=11$.

Step4: Calculate the probability

Probability = $\frac{\text{Number of students who play both}}{\text{Total number of students}}$. So, $P=\frac{11}{23}$.

Answer:

$\frac{11}{23}$