Question

A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received with probability 0.2. Suppose that we want to transmit a message consisting of one binary digit. To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1. If the receiver of the message uses majority" decoding, what the probability that the message will be wrong when decoded? What independence assumptions are you making?

Answer 1

Answer:

The probability that the message will be wrong when decoded is 0.05792

Step-by-step explanation:

Consider the provided information.

To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1.

We have 5 bits, message will be corrupt if at least 3 bits are incorrect for the same block.

The digit transmitted is incorrectly received with probability p = 0.2

The probability of receiving a digit correctly is q = 1 - 0.2 = 0.8

We want the probability that the message will be wrong when decoded.

This can be written as:

$P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =\frac{5!}{3!2!}(0.2)^3(0.8)^{2}+\frac{5!}{4!1!}(0.2)^4(0.8)^{1}+\frac{5!}{5!}(0.2)^5(0.8)^0\\P(X\geq3) =0.05792$

Hence, the probability that the message will be wrong when decoded is 0.05792