a) From the statement of the problem, we know that among middle school students:
• 62% play a sport, 0.62 in decimals,
,
• 42% play an instrument, 0.42 in decimals,
,
• 34% do both, 0.34 in decimals.
The Venn diagram is:
b) The probability that a randomly selected middle school student plays a sport or plays an instrument is equal to the percent in decimals of the students that do both things:

c) The probability that a randomly selected middle school student plays a sport, but does not play an instrument, is given by the part of the zone of sports without the zone of both, so the probability is:

d) The probability that a randomly selected middle school student plays a sport, given that they play an instrument. Is the condition probability:

Where:
• A = a randomly selected middle school student plays a sport,
,
• B = a randomly selected middle school student plays an instrument,
,
• P(A | B) = probability of A given B,
,
• P(A ∩ B) = probability of both events = 0.34,
,
• P(B) = probability of B = 0.42.
Replacing these values in the formula above, we get:

Answers
• b), 0.34
,
• c), 0.28
,
• d) ,0.81