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Assume

$Pr("1"$is received$|$$"1"$is transmitted)=$Pr("0"$is received|$"0"$is transmitted)=$1-b$

$Pr("1"$is transmitted)=$p $ and $Pr("0"$is transmitted)=$1-p$

We transmit the symbol "1" a total of n times over the channel.At the output of the channel,we receive the symbol "1" three times in the n received bits,and that we receive a "1" at the n-th transmission.Given these observations,what is the probability that j-th received bit is "1" ?

I think according to the question,there are two $"0"$ bits in $n-1$ times,so the probability that j-th received bit is "1" is $C^{n-1}_2(1-b)^2(b)^{n-3}$

Is my ideal right?

1 Answers 1

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I think according to the question,there are two "0" bits in $n−1$ times, so the probability that j-th received bit is "1" is $\mathrm C^{n−1}_2 (1−b)^2 (b)^{n−3}$

No.   Well, you have the right idea, but are not quite there yet.

There are actually three okay bits (1 received given 1 sent), one of which is the last bit, and given this we want the conditional probability that the $j$-th bit is also one of other two the okay bits.

You have three events to consider.

$A$: we receive exactly 3 okay bits among the $n$.

$B$: the n-th bit is okay.

$C$: the j-th bit is okay; assuming $j

We want to find $\mathsf P(C\mid A\cap B)$ and so by the very definition of such:

$$\mathsf P(C\mid A\cap B) = \dfrac{\mathsf P(A\cap B\cap C)}{\mathsf P(A\cap B)}$$

Where we need to evaluate the probabilities for the events:

$A\cap B\cap C$ : only the j-th, n-th, and one other bit okay, and $n-3$ bits are bad.

$A\cap B$ : only the n-th bit and any two others are okay.

Hint: I assert that their will be a common factor of some polynomial of $b$, which will cancel out, and that the probability terms of the quotient will differ only by their binomial coefficient factors.

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    so is the answer equal to $\frac{C^n_3(1-b)^3b^{n-3}}{C^{n-1}_2(1-b)^2b^{n-3}}$ – 2017-02-03
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    No. Look at the events again, carefully. $A\cap B\cap C$ is the event where okay bits are in places $j$ and $n$ and one among the $n-2$ others, and bad bits are in the remaining $n-3$ places. $A\cap B$ is the event where okay bits are in place $n$ and two among the $n-1$ others, and bad bits are in the remaining $n-3$ places. – 2017-02-03