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my question is on below:

Q. 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 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,

a. What is the probability that the message will be incorrectly decoded?

b. A message consisting of 118 bytes (944 bits) is encoded and send over the channel. What is the probability that the message contains error after it is decoded?

(By majority decoding we mean that the message is decoded as “0” if there are at least three zeros in the message received and as “1” otherwise.)

My solution for part a is:

$$\begin{align} P(X>2) &= 1 - P(X \le 2) \\ & = 1 - {P(X=0) + P(X=1) + P(X=2)} \end{align} $$

or

$$\begin{align} P(X\ge 3) &= P(X=3) + P(X=4) + P(X=5) \\ \end{align} $$

Both solutions return answer 0.05792

My confusion for part b is:

Since I already get the probability that the message will be incorrectly decoded, how can I calculate the probability of the message which contains 944 bits that decode error? Does this chains into a conditional probability?

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1 Answer 1

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Each bit is decoded independently.

It contain an error if one of the bit is decoded wrongly.

Hence the probability that there is an error is equal to $1$ subtract the probability that there is no error.

Probability of error for a single bit is $q$.

Probability of a bit is decoded correctly is $1-q$.

The probability that there is no error is $(1-q)^n$ where $n$ is the number of bits.

The probability that there is an error is hence $1-(1-q)^n$.

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  • $\begingroup$ Hello @Siong Thye Goh, thank you very much for your answer. So if I understand correctly, for part b, it is to calculate each bit no error/an error probability rather than bundle to five bit as n trial like part a? $\endgroup$
    – HaRLoFei
    Oct 26, 2021 at 14:04
  • $\begingroup$ We bundle group of $5$ bits together. $\endgroup$ Oct 26, 2021 at 16:06
  • $\begingroup$ Thank you @Siong Thye Goh for your explanation :) $\endgroup$
    – HaRLoFei
    Nov 2, 2021 at 20:46

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