Question

1. in a music school of 124 students, 82 students play the piano, 71 students play the guitar and 10 students play neither. 7b what is the probability that a student chosen at random plays both the piano and the guitar?

Explanation:

Step1: Find total students who play either instrument

Total students = 124, students who play neither = 10. So students who play either = \(124 - 10 = 114\).

Step2: Use principle of inclusion - exclusion

Let \(P\) be piano players (\(n(P)=82\)), \(G\) be guitar players (\(n(G)=71\)). Let \(n(P\cap G)\) be students who play both. Then \(n(P\cup G)=n(P)+n(G)-n(P\cap G)\). We know \(n(P\cup G)=114\), so \(114 = 82 + 71 - n(P\cap G)\).

Step3: Solve for \(n(P\cap G)\)

\(n(P\cap G)=82 + 71 - 114 = 39\).

Step4: Calculate probability

Probability = \(\frac{\text{Number of students who play both}}{\text{Total number of students}}=\frac{39}{124}\).

Answer:

\(\frac{39}{124}\)