Question

Suppose you have a friend on the school basketball team but, unfortunately, he is not a good free throw shooter. Over the course of his career, he has only made 40% of his shots. Otherwise, he is a good player and gets fouled a lot so he shoots 10 free throws a game, on average. In the biggest game of the season, he does very well and makes 6 out of 10 free throw attempts. ?

A. Assuming that each shot is independent, set up the expression you would use to determine the exact probability that your friend actually makes 6 or more free throws in a given game.

B. Using the set of random numbers shown here, explain how you would set up a simulation to determine the probability of making 6 or more free throws out of 10 attempts if your probability of making any one free throw is .4.

98713 19464 57512 49898 84837 52245 38816 62042 76406 89744 51623 65847 53436 08320 45677 46304 71946 17871 88983 25948 21826 94930 96270 97988 37221 07174 51855 00969 24384 98095 41970 19537 18910 17433 36753 41545 01058 47896 75215 35996 04853 00796 78602 54962 42771 39128 96541 47221 39264 05585

C. Suppose you wanted to do a normal approximation to the binomial for this situation. What are the mean and standard deviation of the binomial distribution? Is the normal approximation justified in this situation? Explain why or why not? Would the normal approximation be justified if n = 100?

D. One of your friend's attempts during a game is a "one-and-one." That is, if he makes one free throw, he is given the opportunity to take a second shot. Assuming each shot is independent, is he most likely to make 0, 1, or 2 points on a "one-and-one" opportunity.

E. What is the probability that the first free throw he makes is his fourth attempt?

Answer 1

A.   To determine the exact probability that your friend actually makes 6 or more free throws in a given game

=

 P(6 out of 10 shots successful) + P(7 out of 10 shots successful) + 

 P(8 out of 10 shots successful) + P(9 out of 10 shots successful) + P(10 out of 10 shots successful)

Now,

P(6 successes followed by 4 misses) = (0.4)^6 * (0.6)^4 =0.134

But there are 10 ways of happening successes and failures in any order. Hence we have the given probability as
= (0.134)^10