in a certain algebra 2 class of 28 students, 7 of them play basketball and 9 of them play baseball. there…

in a certain algebra 2 class of 28 students, 7 of them play basketball and 9 of them play baseball. there are 14 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 28 students, 7 of them play basketball and 9 of them play baseball. there are 14 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 = 28, students who play neither = 14. So number of students who play at least one sport is $28 - 14=14$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball - players ($n(A)=7$) and $B$ be the set of baseball - players ($n(B)=9$). Let $n(A\cap B)$ be the number of students who play both. We know that $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Since $n(A\cup B) = 14$, we have $14=7 + 9 - n(A\cap B)$.

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

Rearranging the equation $14=7 + 9 - n(A\cap B)$ gives $n(A\cap B)=7 + 9-14 = 2$.

Step4: Calculate the probability

The probability $P$ that a randomly chosen student plays both sports is $P=\frac{n(A\cap B)}{28}=\frac{2}{28}=\frac{1}{14}$.

Answer:

$\frac{1}{14}$