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+1 vote

Hamming code can correct 1 bit error. So if more than one bit gets flipped decoding will result in wrong codeword.

So P(more than 1 bit gets flipped) = 1 - P(0 or 1 bit gets flipped) = 1 - P(no bit gets flipped) - P(1 bit gets flipped)

P(particular bit gets flipped) = 0.2, hence P(particular bit is error free) = 0.8

Assuming error in each bit is independent of error in other bits,

P(0 bit gets flipped) = P(all bits are error free) = $(0.8)^7$

P(1 bit gets flipped) = $\binom{7}{1}*(0.2)*(0.8)^6$

Ans = 0.4233

So P(more than 1 bit gets flipped) = 1 - P(0 or 1 bit gets flipped) = 1 - P(no bit gets flipped) - P(1 bit gets flipped)

P(particular bit gets flipped) = 0.2, hence P(particular bit is error free) = 0.8

Assuming error in each bit is independent of error in other bits,

P(0 bit gets flipped) = P(all bits are error free) = $(0.8)^7$

P(1 bit gets flipped) = $\binom{7}{1}*(0.2)*(0.8)^6$

Ans = 0.4233

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