Question

in a class of 29 students, 24 play an instrument and 13 play a sport. there are 3 students who do not play an instrument or a sport. what is the probability that a student who plays an instrument also plays a sport?

Explanation:

Step1: Find number of students who play an instrument or a sport

Total students = 29, students who play neither = 3. So number of students who play an instrument or a sport is $29 - 3=26$.

Step2: Use the inclusion - exclusion principle

Let $A$ be the set of students who play an instrument and $B$ be the set of students who play a sport. We know $n(A)=24$, $n(B)=13$ and $n(A\cup B) = 26$. By the formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$, we can find $n(A\cap B)$. So $26=24 + 13 - n(A\cap B)$. Then $n(A\cap B)=24 + 13-26=11$.

Step3: Calculate the conditional probability

The probability that a student who plays an instrument also plays a sport is $P(B|A)=\frac{n(A\cap B)}{n(A)}$. Substituting $n(A\cap B) = 11$ and $n(A)=24$, we get $P(B|A)=\frac{11}{24}$.

Answer:

$\frac{11}{24}$