A basketball player was an 84% free throw shooter. a. At the moment you turn the...

A basketball player was an 84% free throw shooter.

a. At the moment you turn the game on he is 5 of 7 shooting from the free-throw line. What is the probability that he made 5 of his first 7 shots?
b. What is the probability that he made his 5th shot on his 7th attempt?
c. What is the probability that he made his first shot on his third attempt?

Solutions

Expert Solution

p = 0.84, q=1-0.84=0.16

a) P(5 shots in first 7) =

Hence, the probability that he makes 5 of 7 shooting is 0.2248 (rounded to 4 decimal places)

b) 5th shot in 7th attempt implies there were 4 shots in first 6 and then 5th shot in 7th attempt.

Hence, the probabiluty that he makes 5th shot in 7th attempt is 0.1606

c) First shot in 3rd attempt implies first two attempts unsuccessful.

So, P = 0.16*0.16*0.84 = 0.021504 = 0.0215 (rounded to 4 decimals)

Please comment if any doubt. Thank you.

orchestra answered 1 year ago

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