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
3 answers
not sure but leaning to C to
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
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
the answer is 95%