Question

in a certain algebra 2 class of 27 students, 22 of them play basketball and 7 of them play baseball. there are 3 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 \(n(N) = 3\).
So, students who play at least one sport \(n(A\cup B)=n(T)-n(N)=27 - 3=24\).

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

Let \(n(A) = 22\) (basketball players), \(n(B)=7\) (baseball players).
We know \(n(A\cup B) = 24\).
Substitute into the formula: \(24=22 + 7-n(A\cap B)\).

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

Rearrange the equation: \(n(A\cap B)=22 + 7-24\).
\(n(A\cap B)=5\).

Step4: Calculate the probability

Probability \(P=\frac{n(A\cap B)}{n(T)}\).
\(P=\frac{5}{27}\).

Answer:

\(\frac{5}{27}\)