in a certain algebra 2 class of 30 students, 24 of them play basketball and 16 of them play baseball. there…

in a certain algebra 2 class of 30 students, 24 of them play basketball and 16 of them play baseball. there are 4 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 30 students, 24 of them play basketball and 16 of them play baseball. there are 4 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 = 30, students who play neither = 4. So number of students who play at least one sport is $30 - 4=26$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball - players ($n(A)=24$) and $B$ be the set of baseball - players ($n(B)=16$). The formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. We know $n(A\cup B) = 26$, $n(A)=24$, and $n(B)=16$. Substituting into the formula gives $26=24 + 16−n(A\cap B)$.

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

Rearrange the equation $26=24 + 16−n(A\cap B)$ to $n(A\cap B)=24 + 16-26$. So $n(A\cap B)=14$.

Step4: Calculate the probability

The probability $P$ that a randomly chosen student plays both sports is $P=\frac{n(A\cap B)}{30}=\frac{14}{30}=\frac{7}{15}$.

Answer:

$\frac{7}{15}$