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Question Within a communications network one of two likely messages are sent to a hub. The first is 000 and other is 111. While the message is being transmitted there a likely hood of receiving a bit in error is .3 ( i.e. 0 is actually a 1, and vice versa).

The message received is 010, what is the probability it is 111?

Attempt at a solution

Probability of receiving P({111}) and P({000}) is a 50/50 chance. So it is .5.

If we received 010, the chance of it being 111 means that the first and the third bit have been received in error.

That means there are two messages received in error meaning, the chance that it is 111 is:

P({111}) = (.3)(.5) + 0 + (.3)(.5) = .3

BUT THEN I THOUGHT ABOUT IT, WHY DOES IT MATTER IF THE BITS ARE WARPED??

We know we have a probability .5 of getting either 111 or 000. The fact that the bits are warped doesn't matter. Meaning P({111}) = .5. Is my solution correct???

Thanks!

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I think you want Bayes Theorem. Let $I$ be the message intended and $R$ be the message received: \begin{align*} P(I = 111 \, | \, R = 010 ) & = \frac{ P( R = 010 \, | \, I = 111 )P(I = 111)}{P(R = 010 \, | \, I = 111)P(I = 111) + P(R = 010 \, | \, I = 000)P(I = 000)} \\ & = \frac{(0.3 \times 0.7 \times 0.3) \times 0.5}{(0.3 \times 0.7 \times 0.3) \times 0.5 + (0.7 \times 0.3 \times 0.7) \times 0.5} \end{align*}

Whilst you are correct that, given no evidence to the contrary, $P(I = 111) = 0.5$, the fact that you received $010$ makes the intention of $111$ less likely. Imagine you walk in a room with three men, one of them dead. There's a 50% chance one of the remaining men is the murderer. But then you notice one of them is holding a smoking gun. Is it still 50% chance? $R=010$ is the smoking gun being held by $I = 000$

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  • $\begingroup$ I have a question, shouldn't the numerator be... (.3 x .3 X .3) (.5) as both 1 and 0 have a .3 chance? $\endgroup$ – user363657 Aug 27 '16 at 2:57
  • $\begingroup$ Ignore the comment above, I understand why it is .7 now. It's the conditional property. $\endgroup$ – user363657 Aug 27 '16 at 3:04
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Probability of starting with 111 is 0.5. Probability that the bits went out as 010 is 0.5 x (0.3 * 0.7 * 0.3) = 0.0315

Similarly, probability of starting with 000 is 0.5 and the they went out as 010 = 0.5 X (0.7 * 0.3 * 0.7) = 0.0735

P(010) = 0.0315 + 0.0735 = 0.105

P(111|010) = P(111 intersect 010)/P(010) = 0.0315/0.105 = 0.3

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  • $\begingroup$ Shai beat me to the answer and much better written too :) $\endgroup$ – Srini Aug 25 '16 at 2:48
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WHY DOES IT MATTER IF THE BITS ARE WARPED??

Because you received $010$.   Which bits were warped and how liable that is matters.

Thus the probability that $111$ was sent instead of the equally likely $000$ is the probability that you received two errors and an okay rather than one error and two okay.

$$\begin{align}\mathsf P(S=`111` \mid R=`010`, S\in\{`000`,`111`\}) ~=~& \dfrac{0.3^2 \cdot 07}{0.3^2 \cdot 0.7 + 0.7^2 \cdot 0.3} \\\;\\ =~& 0.3\end{align}$$

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