To find the probability that a student does not play a sport given that they play an instrument, we can use the following information provided in the problem:
- Total number of students: 22
- Students who play an instrument: 13
- Students who play a sport: 7
- Students who play both an instrument and a sport: 5
Let's denote:
- \( A \): The event that a student plays an instrument.
- \( B \): The event that a student plays a sport.
We're interested in the probability \( P(B^c | A) \), where \( B^c \) is the complement of event \( B \), meaning the student does not play a sport.
Using the formula for conditional probability:
\[ P(B^c | A) = \frac{P(B^c \cap A)}{P(A)} \]
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Calculate \( P(A) \): The probability that a student plays an instrument: \[ P(A) = \frac{13}{22} \]
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Calculate \( P(B^c \cap A) \): This is the number of students who play an instrument but do not play a sport.
- From the 13 students who play an instrument, 5 also play a sport.
- Therefore, the number of students who play an instrument but do not play a sport is: \[ 13 - 5 = 8 \]
- Thus, the probability \( P(B^c \cap A) \) is: \[ P(B^c \cap A) = \frac{8}{22} \]
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Calculate \( P(B^c | A) \): Substitute the values into the conditional probability formula: \[ P(B^c | A) = \frac{P(B^c \cap A)}{P(A)} = \frac{8/22}{13/22} = \frac{8}{13} \]
Therefore, the probability that a student does not play a sport given that they play an instrument is:
\[ \boxed{\frac{8}{13}} \]
If you would prefer this as a decimal, it can be calculated as:
\[ \frac{8}{13} \approx 0.615 \]
Thus, rounded to three decimal places, it's:
\[ \boxed{0.615} \]