in a certain algebra 2 class of 24 students, 18 of them play basketball and 9 of them play baseball. there…

in a certain algebra 2 class of 24 students, 18 of them play basketball and 9 of them play baseball. there are 2 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, 18 of them play basketball and 9 of them play baseball. there are 2 students who play neither sport. what is the probability that a student chosen randomly from the class plays both basketball and baseball?

Answer

Answer:

$\frac{5}{24}$

Explanation:

Step1: Find number of students who play at least one sport

$24 - 2=22$

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball players and $B$ be the set of baseball players. We know $n(A)=18$, $n(B)=9$ and $n(A\cup B) = 22$. By the formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$, we have $22 = 18+9 - n(A\cap B)$.

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

$n(A\cap B)=18 + 9-22=5$

Step4: Calculate the probability

The probability $P$ that a randomly chosen student plays both sports is $P=\frac{n(A\cap B)}{24}=\frac{5}{24}$