Question

During the 2015–16 NBA season, Stephen Curry of the Golden State Warriors had a free throw shooting percentage of 0.908. Assume that the probability Stephen Curry makes any given free throw is fixed at 0.908, and that free throws are independent.

(a) If Stephen Curry shoots two free throws, what is the probability that he makes both of them?
(b) If Stephen Curry shoots two free throws, what is the probability that he misses both of them?
(c) If Stephen Curry shoots two free throws, what is the probability that he makes exactly one of them?

Answer 1

P( Stephen makes a given freee throw) = 0.908

So, P( Stephen misses a given freee throw) = 1 - P( Stephen makes a given freee throw) = 1 - 0.908 = 0.092

Since the free throws are independent of each other so the probabilities can be multiplied to get the required result.

(a) For the Probability that he makes both the shots we multiply the prob that he makes a given shot twice. So,

P( Stephen makes both free throws) = 0.908 x 0.908 = 0.824464

(b) For the Probability that he misses both the shots we multiply the prob that he misses a given shot twice. So,

P( Stephen misses both free throws) = 0.092 x 0.092 = 0.008464

(c) For the Probability that he makes exactly one of the shots we multiply the prob that he makes a given shot by the prob that he misses a given shot (So as to ensure the "exactly one" part). So, we have 0.908 x 0.092 = 0.083536. But note that this can be achieved in two ways. Specifically if he makes the first shot and misses the second or if he misses the first shot and makes the second. So this above expression of probability of missing one shot and making one shot needs to be multiplied by 2 as that event can occur in two ways.

So P( Stephen makes exactly one free throw) =2 x 0.908 x 0.092 = 0.167072

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