Suppose a women’s college basketball player is fouled in a one-and-one situation (meaning she is awarded two free throws only if she makes her first one...

Question

Suppose a women’s college basketball player is fouled in a one-and-one situation (meaning she is awarded two free throws only if she makes her first one and awarded one point for each shot made). 

 

Knowing that her season average is 85%, what is the probability that she will get one point?

Answer 1

OK, this seemed unclear, but I think I understand the problem after reading about "one-and-one" rule.

Another assumption I am making is that "season average is 85%", this means that the players' free-throw average,

the chance that they will get a given free-throw in, is 85%.

 

So the probability she will get one point is

the probability that she gets her first free-throw in  (85%)

    and  (times)

the probability that she misses the second one. (100-85=15 %)

So we have:

   prob-of-one = 0.85*0.15

                     = 0.1275   <-- the answer

 

For fun and comparison, the prob of getting 0 points (missing first shot) is 0.15,

and the prob of getting 2 points is 0.85*0.85 = 0.7225

So getting one point is the most unlikely result.