I have a problem and I'm wondering if my logic is correct:
Say I have a code word of six bits (e.g. $000000$ or $110110$. For this problem, there are a total of 8 code words that can be computed from a generator matrix, but I don't believe that's relevant here). These bits are transmitted in the form of a code word across a communications link with a bit error rate of $1E-6$. In the error-detecting code scheme used, only 1 error can be detected given any of the code words. The question asked is what is the probability of an error that is undetected in a code word?
My logic was to follow the flow of a binomial distribution. That if our random variable, X means to have an undetected error in the code word $$P(X=1) = \binom61 (1\times10^{-6})^{1}(1-(1\times10^{-6}))^{6-1}$$
However, because the answer comes out to $5.99997E-6$ I'm not sure if this is correctly computed. Would someone verify this? Thanks so much for your help in advanced!
Single-error detection codes are parity-check codes. They check if the total number of bits is even or odd. By looking at your two given codewords, it is an even-parity-check code which means the number of $1$s is always even ($0$ is also even). Now when the number of errors is even, the code cannot detect it since the codeword in error can still be a valid codeword.
When the number of bit errors is even, a single-error detection code fails to detect it. The probability of undetected codeword-error is calculated by taking into account the number of combinations that $2$, $4$, and $6$ bits are in error: $$\Pr(\text{undetected})=\binom{6}{2}\varepsilon^2(1-\varepsilon)^{(6-2)}+\binom{6}{4}\varepsilon^4(1-\varepsilon)^{(6-4)}+\binom{6}{6}\varepsilon^6(1-\varepsilon)^{(6-6)}$$
where $\varepsilon=10^{-6}$ is the error probability of channel.