Question

Consider a noisy communication channel, where each bit is flipped with probability p (the probability that a bit is sent in error is p). Assume that n−1 bits, b1,b2,⋯,b(n−1), are going to be sent on this channel. A parity check bit is added to these bits so that the sum b1+b2+⋯+bn is an even number. This way, the receiver can distinguish occurrence of odd number of errors, that is, if one, three, or any odd number of errors occur, the receiver finds it out since the sender forced the sum to be an even number. Find the probability that the received message has errors, but the receiver does not detect it, in terms of n and p. If for p=0.1, this probability is equal to (1+(a^n))/2−b^n, what is the value of a+b?

Answer 1

Binomial (n, p) Let X be the number of errors caught. Then observe that X Hence the number of X can only be even if the error is not caught by the receiver Now observe that n (1-p) +p) =E()*(1 - p)* =* k k-0 L(n)/2 L(n)/2] 2 1)p+1(1- p)"-(4-+1) 2k (2p (1-p)-2k + 2k1 k-0 k-0 = P{X even}P{X odd} Similarly n (1-p)-p)" -()-p)*(1 - p)"-k k-0 L(n)/2 L(n)/2] n (24 (1-p)"-2k 2k12(1- p)"-4+1) 2k+1 k=O k 0 = P{X even} - P{X odd} Therefore the probability that the error did not get detected by the receiver is 1 P{X even} = (1 -p) p) (1p) -p)" 1 (1(2p)" Hence for p ab0.8 (1(0.8)") Therefore a = 0.8,b = 0 i.e. 0.1, P{X even} N IN