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Chapter 4: Q. 4.43 (page 166)

A communications channel transmits the digits 0and 1. However, due to static, the digit transmitted is incorrectly received with probability 2. Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we transmit 00000 instead of 0 and 11111 instead of 1. If the receiver of the message uses “majority” decoding, what is the probability that the message will be wrong when decoded? What independence assumptions are you making?

Short Answer

Expert verified

The probability that the message is wrong when decoded is0.0579.

Step by step solution

01

Given Information

The probability that the digit transmitted incorrectly is, 0.2.

To reduce the chance of error 5digits are transmitted instead of 1digits.

The message is wrongly received when an3,4or5digits are transmitted incorrectly.

02

Solution of the Problem

The probability that the message will be wrong when decoded is,

P(Wrongmessage)=P(X≥3)

=53(0.2)3(0.8)2+54(0.2)4(0.8)1+55(0.2)5(0.8)0

=0.0512+0.0064+0.0003

We get,

=0.0579.

03

Final Answer

The probability that the message is wrong when decoded is0.0579.

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