Question

in a certain algebra 2 class of 26 students, 10 of them play basketball and 5 of them play baseball. there are 3 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

Explanation:

Step1: Recall the principle of inclusion - exclusion for sets

The formula for the number of elements in the union of two sets \( A \) and \( B \) is \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \), where \( n(A) \) is the number of elements in set \( A \), \( n(B) \) is the number of elements in set \( B \), and \( n(A\cap B) \) is the number of elements in the intersection of \( A \) and \( B \). Let \( A \) be the set of students who play basketball and \( B \) be the set of students who play baseball. We know that \( n(A) = 10 \), \( n(B)=5 \), and \( n(A\cap B) = 3 \).

Step2: Calculate the number of students who play basketball or baseball

Using the inclusion - exclusion formula, we substitute the values: \( n(A\cup B)=10 + 5-3=12 \).

Step3: Calculate the probability

The total number of students in the class is \( N = 26 \). The probability \( P \) that a randomly chosen student plays basketball or baseball is the number of students who play basketball or baseball divided by the total number of students, so \( P=\frac{n(A\cup B)}{N}=\frac{12}{26}=\frac{6}{13} \).

Answer:

\(\frac{6}{13}\)