In a class 21 of students, 10 play an instrument and 6 play a sport. There are 9 students who do not play an instrument or a sport. What is the probability that a student does not play an instrument given that they play a sport?

To find the probability that a student does not play an instrument given that they play a sport, we need to use conditional probability.

The probability of an event A given that event B has occurred is written as P(A|B) and is calculated as:

P(A|B) = P(A ∩ B) / P(B)

where P(A ∩ B) is the probability that both events A and B occur, and P(B) is the probability that event B occurs.

Let's define the following events: A: the event that a student does not play an instrument B: the event that a student plays a sport

We're given the following information: - The total number of students is 21. - The number who play an instrument is 10, which means the number who do not play an instrument, which is event A, is 21 - 10 = 11. - The number who play a sport is 6, which is event B. - The number of students who do not play an instrument or a sport is 9.

Now, we need to find P(A ∩ B), the probability that a student does not play an instrument and plays a sport. Since we know that 9 students do neither, this implies that out of the 6 students who play a sport, 6 - 9 = -3 would not play an instrument, which isn't possible. The issue lies in the fact that the 9 students who do neither are already accounted for within the 21 total students. Therefore, all students who play a sport also play an instrument. This means P(A ∩ B) = 0 (there are no students who play a sport but not an instrument).

Thus, the probability that a student does not play an instrument given that they play a sport is:

P(A|B) = P(A ∩ B) / P(B) = 0 / (6/21) = 0

Therefore, the probability is 0.