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Consider a binary communication system that consists of a transmitter, a receiver and a Chanel that transfers bits from the transmitter to the receiver. The nature of the channel is such that it occasionally drops bits, so that when a zero or a one is transmitted, it is possible that nothing is received. to simplify the formulation of the model we denote the event of not receiving as $R_e$. We define the following events:

$R_e$ = error in transmission

$T_o$ = transmission of a zero

$T_1$ = transmission of a one $$ \begin{eqnarray} P\left(R_o|T_o\right) &=& 0.9,\\ P\left(R_e|T_o\right) &=& 0.1,\\ P\left(R_1|T_1\right) &=& 0.8,\\ P\left(R_e|T_1\right) &=& 0.2. \end{eqnarray} $$ Question: Calculate $P\left(R_1\right)$

My attempt: $$ P\left(R_1|T_1\right) =\frac{P\left(R_1\cap T_1\right)}{P\left(T_1\right)} $$

therefore $$ P\left(R_1\cap T_1\right) = P\left(T_1\right)P\left(R_1|T_1\right) $$

and that's as far as i can go, is there a way i can separate $P\left(R_1\cap T_1\right)$ to get just $P\left(R_1\right)$

-Thanks!

edit:

The following is also given:

For simpilicity we assume that the transmitterd signal satisfies

$P\left(T_o\right)$=$0.6$, and $P\left(T_1\right)$ = $0.4$

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  • $\begingroup$ Hi there, and welcome! I have edited your post this time. But please take a look at this mathjax tutorial. $\endgroup$
    – Chinny84
    Aug 2, 2014 at 21:57
  • $\begingroup$ As for your question can you fill out the details of the probabilities and indeed the original question? i.e. what is $R_1$? As it stands (my knowledge of Bayes is not complete by any means) but I think some further information is required. $\endgroup$
    – Chinny84
    Aug 2, 2014 at 21:59
  • $\begingroup$ Thanks for cleaning up my question :). i have edited the initial question and copied it out as it is in my text book. $\endgroup$
    – HappyFeet
    Aug 2, 2014 at 22:07

1 Answer 1

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$\textbf{hint:}$ $$ P\left(T_1|R_1\right) = 1 $$ I am assuming (due to the level of the problem) that the only events that can occur from a given transmission is either the correct bit or no bit i.e. the $R_e$ event. therefore the probability that the transmitting 1 given that the system received a one is unity.

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  • $\begingroup$ $$ P\left(R_1|T_1\right) =\frac{P\left(T_1|R1\right)P\left(R_1\right)}{P\left(T_1\right)} $$ is this what you have in mind?.. $$P\left(R_1\right) =\frac{P\left(R_1|T_1\right)P\left(T_1\right)}{P\left(T_1|R_1\right)}$$ $\endgroup$
    – HappyFeet
    Aug 3, 2014 at 7:36
  • $\begingroup$ Exactly :). To be honest it doesn't make use of the entire set of information you provided, but i assume this is only one part of a question? $\endgroup$
    – Chinny84
    Aug 3, 2014 at 8:17
  • $\begingroup$ yes it doesn't, i figured if i could do the first ones, then i''ll manage the rest of the questions! -Thanks! $\endgroup$
    – HappyFeet
    Aug 3, 2014 at 8:21

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