Q: The simplest error detection scheme used in data communication is parity-checking. Messages sent consists of characters, each character consisting of a number of bits, 0 or 1. In parity checking, a 0 or a 1 is appended to the end to make the number of 1's even. The receiver checks the number of 1's in every character received and if the result is odd, it signals an error. Suppose that each bit is receveid correctly with probability of 0.999 independently of other bits. What is the probability that a 7bit character is received in error and not dected by the parity check?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Binomial Distrinution - is my answer reasonable?

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