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i have been working through some old exam papers and have gotten stuck on this last one. can anyone help?

When a piece of information (a bit) is transmitted over a communications channel, it may be wrongly communicated. One method of improving reliability is to transmit the same piece of information an odd number of times, and use a majority decoder. The information is taken to be the bit received the most (from the odd number). A simple model for noisy communication is to assume that each transmitted bit is independently corrupted during transmission with the same probability $p$ where $0 < p < 1$. Now consider a two-out-of-three majority decoder.

(a) Let $Y$ be the random variable that counts how many (out of the three bits) are correctly communicated. What are the possible values of $Y$ ?

(b) Find the probability mass function of $Y$ , writing it either in a table or as a list.

(c) For which values of $Y$ will the majority decoder correctly interpret the information sent?

(d) In terms of $p$, what is the probability that a piece of information is correctly communicated? Also write this in terms of $F(y)$.

(e) For what values of $p$ is the majority decoder more reliable than transmitting the message only once?

i am assuming the answer to (a) is $Y=0,1,2,3$ but my answers for the rest are not making sense

Thanks!

5xum
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Rachael
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1 Answers1

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(a) You are correct, $Y$ can be either $0,1,2$ or $3$.

(b) You simply need to calculate $P(Y=0), P(Y=1), P(Y=2)$ and $P(Y=3)$. How would you calculate $P(Y=0)$? Remember, $P(Y=0)$ is the probability that no errors occured, which means that it is the probability that no error occured in the first transmission and no error occured in the second transmission and no error occured in the third transmission.

(c) Is the decoder correct if $Y=0$? What about if $Y=1$? If $Y=2$? What about $Y=3$?

(d) After you answer (c), this answer will be easier.

(e) After you answer (d), this answer will be easier.

5xum
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  • how do i calculate P(Y=0), P(Y=1), etc with what i have been given? that's whats really got me? – Rachael Aug 26 '15 at 23:36
  • @Rachael Did you read the part where I said that $P(Y=0)$ is the probability of (something and something else and something else)? Do you remember how you calculate $P(A\text{ and } B$ if $A$ and $B$ are independent? – 5xum Aug 27 '15 at 05:44