SOLUTION: What is the probability of making exactly 3 of 7 free throws when a player shoots 53%.
Is it N!/(N-r)! or 7! / 4! * .53 = or 80 % chance of making excacly 3 of 7 free throws.
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-> SOLUTION: What is the probability of making exactly 3 of 7 free throws when a player shoots 53%.
Is it N!/(N-r)! or 7! / 4! * .53 = or 80 % chance of making excacly 3 of 7 free throws.
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Question 723991: What is the probability of making exactly 3 of 7 free throws when a player shoots 53%.
Is it N!/(N-r)! or 7! / 4! * .53 = or 80 % chance of making excacly 3 of 7 free throws. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! What is the probability of making exactly 3 of 7 free throws when a player shoots 53%.
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Binomial Problem with n = 7 ; p(score) = 0.53
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P(x = 3) = 7C3*0.53^3*0.47^4 = 35*0.149*0.0488 = 0.2543
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Cheers,
Stan H.
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