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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, what is the probability that the message will be incorrectly decoded? What independence assumptions are you making?

(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 answers: Independence assumptions that each bit is transmitted independently, where each bit is sent as 5 bits for error reduction (0 = 00000), majority decoding.

  • Probability of error with this transmission method will only be if

more than 2 bits were incorrectly transmitted. In which case, a transmission for 0 transmits as 00111 (3 errors), will be incorrectly decoded as a 1.

$q=0.2$, $p = 1-q=0.8$

There fore probability for transmitting incorrectly is given by binomial :

$$\begin{align} P(X>2) &= 1 - P(X \le 2)\\ &= 1 - {P(X=0) + P(X=1) + P(X=2)} \end{align} $$ My doubt here since we are supposed to calculate probability for transmitting incorrectly, we use swap positions for q and p right ?

$$p(X=0)= \binom{5}{0} 0.2^0 0.8^5$$

p(x=1)= 5C1 (0.2)^1 X (0.8)^4

p(x=2)= 5C2 (0.2)^2 X (0.8)^3

{P(x=0) + P(x=1) + P(x=2)} = 0.94208

1 - 0.94208 = 0.05792 is this right ?

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I have an add-on question on top of above Binomial probability question. Assume above question remain the same and at the end it need to find probability for below statement:

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?

How would be the solution direction for this statement since the probability for transmitting incorrectly has been calculated?

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  • $\begingroup$ Welcome to MSE. If you have a new question, then post it as such. $\endgroup$ Oct 25, 2021 at 20:57
  • $\begingroup$ Thank you @José Carlos Santos for the reminder $\endgroup$
    – HaRLoFei
    Oct 26, 2021 at 8:06

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