Answer - $D$
Suppose there are $k$ places within $n$ bit string where mismatch has occoured
Probability of this occurring is
$^{n}C_{k}.\text{(prob. of mismatch)}^{k}\text{(prob. of match)}^{(n-k)}$
$ = ^{n}C_{k}\left(\dfrac{1}{2}\right)^{k}\left(\dfrac{1}{2}\right)^{(n-k)} $
$= ^{n}C_{k}\left(\dfrac{1}{2}\right)^{n}.$
$k$ can range from $1$ to $n$, hence the required probability $\sum \left(^{n}C_{k}\left(\frac{1}{2}\right)^{n}\right)$ where $k$ ranges from $1$ to $n$
is $\left(\dfrac{1}{2^{n}}\right)\left(2^{n} - 1\right).$
Alternatively
Probability of matching at given place $\dfrac{1}{2}.$
there are n places hence probability of matching $\dfrac{1}{2^{n}}.$
hence probability of mismatch $1 - \dfrac{1}{2^{n}}.$