Math

Coram is at basketball practice preparing to shoot free throws. The probability of Coram making a free throw is 77%. What is the probability that he makes his first free throw on the second shot?

A. 79%

B. 95%

C. 88%

D. 83%

I think the answer is C

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  1. not sure but leaning to C to

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  2. prob(hit) = .77
    so prob(miss) = .23

    My interpretation of the event: he misses the first shot, but makes the second

    so you want prob(miss, then hit) = (.23)(.77) = .1771 or app 18%

    none of the choices are correct

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