Question

in a certain algebra 2 class of 28 students, 11 of them play basketball and 16 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?

Explanation:

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

Total students \(n(T)=28\), students who play neither \(n(N) = 10\).
Number of students who play at least one sport \(n(A\cup B)=n(T)-n(N)=28 - 10=18\).

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

Let \(n(A) = 11\) (basketball players), \(n(B)=16\) (baseball players).
We know \(n(A\cup B)=18\), substituting into \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\) gives \(18=11 + 16-n(A\cap B)\).

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

Rearrange the equation: \(n(A\cap B)=11 + 16-18\).
\(n(A\cap B)=9\).

Step4: Calculate the probability

Probability \(P=\frac{n(A\cap B)}{n(T)}\).
Substitute \(n(A\cap B) = 9\) and \(n(T)=28\), so \(P=\frac{9}{28}\).

Answer:

\(\frac{9}{28}\)