Question

A basketball player with an 85 % free throw percentage (average probability of making a free throw) takes 10 independent free throws and records the outcome. (a) What is the probability of making exactly 6 free throws? (b) What is the probability of making at least one free throw? (c) What is the probability of making between 7 and 10 free throws? (d) What is the probability that the first made free throw is the 3rd shot attempt? (e) What is the probability that the last two free throws are missed given that the first two free throws are missed?

Answer 1

d. det X: No of throws made by the player. of follows binomial dist where b=0.85 m-10 - P(X=x): (?)(0.85)(1–0.85)1023 @ P(X+6)=(8)(0.85)*(0.15)9=0.0401] © P(x21): 1- P(X=0): 1 – 0.15)) © P175X310) = 3 (0,855*0.15)0? 10.9500] @P(45+ success on goo attempt): P(fail fail success) 0.15* 0.15-0.85=10.019|| @ : Throws are independent. So result of last throw doesnot depend on first throw and vice versa. .. Read prob=P(last two throws missed, X 225)