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?

Question

A. 79%

B. 95%

C. 88%

D. 83%

I think the answer is C

Answer 1

not sure but leaning to C to

Answer 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

Answer 3

the answer is 95%