Question

in a certain algebra 2 class of 27 students, 17 of them play basketball and 19 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?

Explanation:

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

Total students \(n(T)=27\). Students who play neither \(n(N) = 2\).
So, students who play at least one sport \(n(A\cup B)=n(T)-n(N)=27 - 2=25\).

Step2: Use the formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\)

Here \(n(A) = 17\) (basketball players), \(n(B)=19\) (baseball players), and \(n(A\cup B) = 25\).
Substitute into the formula: \(25=17 + 19 - n(A\cap B)\).

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

Rearrange the equation: \(n(A\cap B)=17 + 19-25\).
\(n(A\cap B)=11\).

Step4: Calculate the probability

Probability \(P=\frac{n(A\cap B)}{n(T)}\).
Substitute \(n(A\cap B) = 11\) and \(n(T)=27\). So \(P=\frac{11}{27}\).

Answer:

\(\frac{11}{27}\)