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My question is as follows:

In a nonsymmetric binary channel, $0$ and $1$ are transmitted independently in the proportion of $1:4$. If a $0$ was transmitted, we will receive $0$ with a probability of $0.9$. If a $1$ transmitted, we will receive $1$ with a probability of $0.95$.

a) What is the probability that a received symbol is "$1$"?

b) A "$1$" has been received what is the probability that a "$1$" actually been transmitted?

c) A "$0$" has been received what is the probability that a "$0$" actually been transmitted?

I found below answers but I am not sure:

a)

$P(R_1)=\frac{1}{5}*P(T_0|R_1)+ \frac{4}{5}*P(T_1|R_1)$=$\frac{1}{5}*0.1+ \frac{4}{5}*0.95=0.78$

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  • $\begingroup$ For a) I found 0.78. I suppose bayes will be sufficient but I am not sure $\endgroup$
    – sajjad
    Commented Mar 3, 2018 at 20:02
  • $\begingroup$ @sajjad Show your work, please! Thanks. $\endgroup$ Commented Mar 3, 2018 at 20:03
  • $\begingroup$ Please put it in your question $\endgroup$
    – Remy
    Commented Mar 3, 2018 at 20:05
  • $\begingroup$ Please press the "edit" link and add that to the actual post. This is part of what we want from good question posts here. How far you've gotten on (b) and (b) would also be nice if you said something about. $\endgroup$
    – Arthur
    Commented Mar 3, 2018 at 20:05
  • $\begingroup$ I am editing... for part B and C (as they are similar) I have no answer $\endgroup$
    – sajjad
    Commented Mar 3, 2018 at 20:13

1 Answer 1

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$(a)$ looks good

Hint for $(b)$

Use Bayes' Theroem:

$$\begin{align*} P(\text{1 transmitted | 1 received}) &=\frac{P(\text{1 transmitted} \cap 1\text{ received})}{P(\text{1 received})}\\\\ \end{align*}$$

where $P(\text{1 received})$ is what you solved correctly in part $(a)$

Hint for $(c)$

This calculation is similar noting that $$P(\text{0 received})=1-P(\text{1 received})$$

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  • $\begingroup$ Thank you. I am trying right now $\endgroup$
    – sajjad
    Commented Mar 3, 2018 at 20:17
  • $\begingroup$ For $B$ if found this: $P(T_1|R_1)= \frac{P(T_1 \cap R_1)}{P(R_1)}= \frac{P(R_1|T_1)*P(T_1)}{P(R_1)}= \frac{0.95*4/5}{0.78}=0.97$ $\endgroup$
    – sajjad
    Commented Mar 3, 2018 at 20:27
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    $\begingroup$ That's correct! $\endgroup$
    – Remy
    Commented Mar 3, 2018 at 20:29
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    $\begingroup$ with same noting I found : $P(T_0|R_0)=0.81$ Thank you for hint. $\endgroup$
    – sajjad
    Commented Mar 3, 2018 at 20:37
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    $\begingroup$ Yes, except it rounds to $0.82$ $\endgroup$
    – Remy
    Commented Mar 3, 2018 at 20:39

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