in a certain algebra 2 class of 24 students, 7 of them play basketball and 13 of them play baseball. there…

in a certain algebra 2 class of 24 students, 7 of them play basketball and 13 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 24 students, 7 of them play basketball and 13 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 are 24 and 9 play neither. So number of students who play at least one sport is $24 - 9=15$.

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 ($n(B)=13$). The formula for $n(A\cup B)$ is $n(A)+n(B)-n(A\cap B)$. We know $n(A\cup B) = 15$, $n(A) = 7$ and $n(B)=13$. Substituting into the formula gives $15=7 + 13 - n(A\cap B)$.

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

Rearranging the equation $15=7 + 13 - n(A\cap B)$ gives $n(A\cap B)=7 + 13-15$. So $n(A\cap B)=5$.

Step4: Calculate the probability

The probability $P$ that a randomly chosen student plays both sports is the number of students who play both sports divided by the total number of students. So $P=\frac{5}{24}$.

Answer:

$\frac{5}{24}$