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In a class of 23 students, 12 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 who plays a sport also plays an instrument?



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To determine the probability that a student who plays a sport also plays an instrument, we need to calculate the conditional probability. Conditional probability is the probability of event A occurring given that event B has already occurred.

In this case, event A is playing an instrument and event B is playing a sport. We are given the following information:

* Total number of students (n(S)) = 23

* Number of students who play an instrument (n(I)) = 12

* Number of students who play a sport (n(Sp)) = 6

* Number of students who do not play an instrument or a sport (n(NI)) = 9

We can use the following formula to calculate the conditional probability:

P(A | B) = n(A ∩ B) / n(B)

where:

* P(A | B) is the probability of event A occurring given that event B has already occurred

* n(A ∩ B) is the number of students who play both an instrument and a sport

* n(B) is the number of students who play a sport

First, we need to find the number of students who play both an instrument and a sport (n(A ∩ B)). We can do this by subtracting the number of students who play either an instrument or a sport (n(I ∪ Sp)) from the total number of students (n(S)):

n(A ∩ B) = n(S) - n(I ∪ Sp)

We know that n(S) = 23, n(I) = 12, n(Sp) = 6, and n(NI) = 9. We can use the formula for the union of two sets to find n(I ∪ Sp):

n(I ∪ Sp) = n(I) + n(Sp) - n(I ∩ Sp)

Substituting the values we know, we get:

n(I ∪ Sp) = 12 + 6 - n(I ∩ Sp)

We also know that n(NI) = n(S) - n(I ∪ Sp), so we can substitute this to eliminate n(I ∪ Sp):

n(I ∩ Sp) = n(S) - n(NI) = 23 - 9 = 14

Now we can find n(A ∩ B):

n(A ∩ B) = n(S) - n(I ∪ Sp) = 23 - 14 = 9

Finally, we can calculate the conditional probability:

P(A | B) = n(A ∩ B) / n(B) = 9 / 6 = 1.5

Therefore, the probability that a student who plays a sport also plays an instrument is 1.5.

User Tasin Ishmam
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