In class we looked at the probability of a 91% free throw shooter taking three f

Question

In class we looked at the probability of a 91% free throw shooter taking three free throws and the odds. Let's look at the same problem with a 70% shooter taking three free throws. We will look at the problem in two ways: making exactly r shots out of n and making at least r shots out of n, which is to say the sum of all the probabilities of exactly k shots where k greaterthanorequalto r. (For example, the probability of at least two shots made is p(exactly 2 made) + p(exactly 3 made). Write the probability as a decimal number with no rounding.

Explanation / Answer

Using binomial distribution formula, ncr*pr*(1-p)(n-r)

                                                        Exactly                                       Atleast
Making 3 of 3 shots              3c3*0.73*0.30 = 0.343                    Same as exactly, 0.343
Making 2 of 3 shots              3c2*0.72*0.31 = 0.441                       0.441+0.343 = 0.784
Making 1 of 3 shots              3c1*0.71*0.32 = 0.189                       0.189+0.441+0.343 = 0.973
Making 0 of 3 shots              3c0*0.70*0.33 = 0.027                       0.027+0.189+0.441+0.343 = 1

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