in a certain algebra 2 class of 29 students, 13 of them play basketball and 14 of them play baseball. there…

in a certain algebra 2 class of 29 students, 13 of them play basketball and 14 of them play baseball. there are 10 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 29 students, 13 of them play basketball and 14 of them play baseball. there are 10 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 - students who play neither sport. So, $29 - 10=19$ students play at least one sport.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of basketball - players and $B$ be the set of baseball - players. We know that $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Here, $n(A\cup B) = 19$, $n(A)=13$, $n(B)=14$. Substituting these values into the formula: $19 = 13+14 - n(A\cap B)$.

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

First, simplify the right - hand side of the equation: $13 + 14=27$. Then, we have $19=27 - n(A\cap B)$. Rearranging gives $n(A\cap B)=27 - 19 = 8$.

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, $P=\frac{8}{29}$.

Answer:

$\frac{8}{29}$