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The equation I used is this one

http://mathworld.wolfram.com/BinomialDistribution.html

My answer:

p(no error in 7-bits) = (7 choose 0) * (.001)^0 (1 - .001)^7 = .99302

p(single error in 7-bits) = (7 choose 1) * (.001)^1 (1 - .001)^6 = .00696

p(undetected error in 7-bit transmission) = 1 - p(no error in 7-bits) - p(single error in 7-bits)

= 1 - .99302 - .00696

= .00002

Correct? Resonable?

Thanks in advance.