A message consists of a string of bits (0s and 1s). Due to noise in the communications channel, each bit has probability 0.3 of being reversed (i.e., a 1 will be changed to a 0 or a 0 to a 1). To improve the accuracy of the communication, each bit is sent five times, so, for example, 0 is sent as 00000. The receiver assigns the value 0 if three or more of the bits are decoded as 0, and 1 if three or more of the bits are decoded as 1. Assume that errors occur independently.

a. A 0 is sent (as 00000). What is the probability that the receiver assigns the correct value of 0?
b. Assume that each bit is sent n times, where n is an odd number, and that the receiver assigns the value decoded in the majority of the bits. What is the minimum value of n necessary so that the probability that the correct value is assigned is at least 0.90?


(a) Let X be the number of bits that are reversed. Then X~ Bin(5,0.3). The correct value is assigned if X 2.





(b) We need to find the smallest odd value of n so that P(X?(n -1)/2) 0.90 when X~ Bin(n, 0.3). Consulting Table A.1, we find that if n = 3, P(X?1) = 0.784, if n = 5, P(X?2) = 0.837, if n = 7, P(X?3) = 0.874, and if n = 9, P(X?4) = 0.901. The smallest value of n is therefore n = 9.

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