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A college basketball player makes 80% of his free throws. At the end of a game, his team is losing by two points. He
is fouled and is awarded three free throws. Assuming each free throw is independent, what is the probability that he
makes at least two of the three free throws?
Hey lpg0005.
The first thing with the probability problems is to identify if a distribution exists that model the situation. You are looking at two successes of three trials and each trial is the same.
This corresponds to a Binomial distribution. You should look at the assumptions for this but basically they are that each trial has the same probability of a success or failure and all trials are independent where the distribution gives the probability of getting so many successes.
Well we got close to the correct answer using P(x)=q^(x-1)*p
P(3)=q^(3-1)*.8
But now that I am looking at it, I dont know how we arrived at 1.25, which we used as a z-score to get close the correct answer of .896
This distribution is not the Binomial one which fits the assumptions.
Consider finding P(X = 2) for P(X = x) = nCx p^(x) * (1-p)^(n-x). n = 3, p = 0.8 and P(X = 3) and add the two (since they are disjoint).