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24 August, 01:51

The best player on a basketball team makes 80 % of all free throws. The second-best player makes 75 % of all free throws. The third-best player makes 70 % of all free throws. Based on their experimental probabilities, estimate the number of free throws each player will make in his or her next 60 attempts.

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  1. 24 August, 02:08
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    Expected number of free throws in 60 attempts:

    Best player = 48

    2nd best player = 45

    3rd best player = 42

    Step-by-step explanation:

    Solution:-

    - The probability that best player makes free throw, p1 = 0.8

    - The probability that second-best player makes free throw, p2 = 0.75

    - The probability that third-best player makes free throw, p3 = 0.70

    - Total number of attempts made in free throws, n = 60.

    - The estimated number of free throws that any player makes is defined by:

    E (Xi) = n*pi

    Where, Xi = Player rank

    pi = Player rank probability

    - Expected value for best player making the free throws would be:

    E (X1) = n*p1

    = 60*0.8

    = 48 free throws

    - Expected value for second-best player making the free throws would be:

    E (X2) = n*p2

    = 60*0.75

    = 45 free throws

    - Expected value for third-best player making the free throws would be:

    E (X3) = n*p3

    = 60*0.70

    = 42 free throws
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