in a certain algebra 2 class of 30 students, 12 of them play basketball and 13 of them play baseball. there…

in a certain algebra 2 class of 30 students, 12 of them play basketball and 13 of them play baseball. there are 10 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

in a certain algebra 2 class of 30 students, 12 of them play basketball and 13 of them play baseball. there are 10 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

Answer

Answer:

$\frac{15}{30}=\frac{1}{2}$

Explanation:

Step1: Use the inclusion - exclusion principle

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$ Here, $n(A) = 12$ (students who play basketball), $n(B)=13$ (students who play baseball), and $n(A\cap B) = 10$ (students who play both). So, $n(A\cup B)=12 + 13-10=15$.

Step2: Calculate the probability

The probability $P$ that a randomly - chosen student plays basketball or baseball is given by $P=\frac{n(A\cup B)}{n(S)}$, where $n(S)=30$ (total number of students in the class). So, $P=\frac{15}{30}=\frac{1}{2}$.