26 votes 26 votes Two $n$ bit binary strings, $S_1$ and $S_2$ are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to $d$ is $\dfrac{^{n}C_{d}}{2^{n}}$ $\dfrac{^{n}C_{d}}{2^{d}}$ $\dfrac{d}{2^{n}}$ $\dfrac{1}{2^{d}}$ Probability gatecse-2004 probability normal uniform-distribution + – Kathleen asked Sep 18, 2014 edited Dec 3, 2017 by pavan singh Kathleen 7.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Take small values for n and then solve it. Jhaiyam answered Aug 31, 2020 Jhaiyam comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes n binary bits with difference 'd' then no. of favourable cases = nCd Total no. of cases where n positions have any binary bit = 2n The probability of 'd' bits differ = nCd / 2n varunrajarathnam answered Nov 7, 2020 varunrajarathnam comment Share Follow See all 0 reply Please log in or register to add a comment.