answer - D
suppose there are k places within n bit string where mismatch has occoured
probability of this occouring is ^{n}C_{k}(prob. of mismatch)^{k}(prob. of match)^{(n - k)} = ^{n}C_{k}(1/2)^{k}(1/2)^{(n-k)} = ^{n}C_{k}(1/2)^{n}
k can range from 1 to n hence required probability sum(^{n}C_{k}(1/2)^{n}) where k ranges from 1 to n
hence (1/2^{n})(2^{n} - 1)
Alternatively
Probability of matching at given place 1/2
there are n places hence probability of matching 1/(2^{n})
hence probability of mismatch 1 - 1/(2^{n})