Perhaps
Suppose there are total 'x' shots.
B shoots 2 out of 2 times. means out of x shots (x>2) B shots 2 out of 2 ----> B shots at least 2 out of x.
therefore, 2/x = 0.25 (given)
the target will be missed by B immediately after such two shots : this means he can shot just twice ...the third shot cannot happen which means he missed (x-2) shots.
thus, the probabilty of missing just after 2 shots is (x-2)/x.
(x-2)/x = 1 - 2/x
= 1 - 0.25
= 0.75
Answer : C
Please correct me.
Frankly I don't understand your reasoning above. Anyway:
The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.
Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.
Answer: B.