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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

  1. $\dfrac{^{n}C_{d}}{2^{n}}$

  2. $\dfrac{^{n}C_{d}}{2^{d}}$

  3. $\dfrac{d}{2^{n}}$

  4. $\dfrac{1}{2^{d}}$

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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
Answer:

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