Re: Help: Coding theory and Probability




Proginoskes wrote:
> blazer320@xxxxxxxx wrote:
> > I have a code C={000, 111}. Suppose v in C is sent and 110 is
> > received. What is the probability 111 was sent? Here p=.9
> > where p is the probability that a digit sent is the digit
> > received.
> >
> > Obviously an error occured in the channel with 110 as the only
> > 2 possible words that could have been sent are 000 and 111.
> > I don't really have much of a probability background, so I'm
> > at a loss as to how to compute this. I would really appreciate
> > someone showing me how to do this problem.
> >
> > I tried (.9*.9*.1)/(.9*.9*.1+.1*.1*.9) but that just equalled .9
so
> I
> > must have tried the wrong thing.
>
> I thought it was Bayes's Formula, which you almost have above.
> Bayes's Formula is:
>
> P(S111 : R110) = P(S111) * P (R110 : S111) /
> [P(S111) * P (R110 : S111) + P(S000) * P (R110 : S000)],
>
> where S111 is the event that 111 was sent, R110 is the event that 110
> was received, etc. and I've used : instead of the usual vertical bar
> "|".
>
> You've calculated P (R110 : S111) (.081) and P (R110: S000) (.009).
If
> we let q = P(S111), then the conditional formula reads:
>
> P(S111 : R110) = .081 q / (.081 q + .009 (1 - q)).
>
> I strongly suspect that there may be insufficient information here.

Yes, as you indicate with your unknown q, you need to know
what the probability was that 000 or 111 was sent. If the
sender never transmits 111, then the probability that the
received 110 indicates a sent 111 is obviously 0.

In the absence of other information to the contrary, it's
reasonable to assume the two codes are transmitted
equally often.

- Randy

.



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