Question

Assume a college basketball player makes 75% of his free throws and that the outcome of free throw attempts are independent.

a) The basketball player is fouled and is awarded two free throws. Develop a probability distribution for number of points (free throws made) for the two attempts.

b) The basketball player is fouled and is awarded two free throws. What is the expected number of points for the two attempts?

c) The basketball player is fouled and is awarded a "one and one" (if the first shot is successful, he is allowed a second shot, but no second shot is taken if the first is missed). Develop a probability distribution for number of points (free throws made) for the "one and one".

d) The basketball player is fouled and is awarded a "one and one" (if the first shot is successful, he is allowed a second shot, but no second shot is taken if the first is missed). What is the expected number of points for the "one and one".

Answer #1

a) No.of points 0: Probability = 0.25 x 0.25 = 0.0625 (the player fails on both the throws)

No. of points 1: Probability = 0.75 x 0.25 + 0.25 x 0.75= 0.375 (the player succeeds in either of the two throws and fails in the other)

No. of points = 2: Probability = 0.75 x 0.75 = 0.5625 (the player scores on both the throws)

b) Expected number of points = 0 x 0.0625 + 1 x 0.375 + 2 x 0.5625 = 1.5

c)

No.of points 0: Probability = 0.25 (the player fails on the first throw)

No. of points 1: Probability = 0.75 x 0.25 = 0.1875 (the player succeeds in either first throw and fails in the second)

No. of points = 2: Probability = 0.75 x 0.75 = 0.5625 (the player scores on both the throws)

d) Expected number of points = 0 x 0.25 + 1 x 0.1875 + 2 x 0.5625 = 1.3125

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