Question 126467
 
If 2% of students are basketball players and 31% of basketball players have an 'A' average what is the probability that a student chosen at random will be a basketball player with an 'A' average?
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Assuming that being a basketball player and getting an A are independent
events you get:
P(b and A) = P(b)*P(A) = 0.02*0.31 = 0.0062
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Cheers,
Stan H.