in a certain algebra 2 class of 28 students, 21 of them play basketball and 20 of them play baseball. there…

in a certain algebra 2 class of 28 students, 21 of them play basketball and 20 of them play baseball. there are 5 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, 21 of them play basketball and 20 of them play baseball. there are 5 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 the number of students who play at least one sport

The total number of students is 28 and 5 play neither sport. So the number of students who play at least one sport is $28 - 5=23$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of students who play basketball ($n(A) = 21$) and $B$ be the set of students who play baseball ($n(B)=20$). The number of students who play at least one sport is $n(A\cup B) = 23$. By the inclusion - exclusion principle $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Substituting the known values, we get $23 = 21+20 - n(A\cap B)$.

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

Rearranging the equation from Step 2 gives $n(A\cap B)=21 + 20-23=18$.

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 $P=\frac{n(A\cap B)}{28}=\frac{18}{28}=\frac{9}{14}$.

Answer:

$\frac{9}{14}$