From: Lee Corbin (lcorbin@tsoft.com)
Date: Mon Jul 07 2003 - 21:57:03 MDT
Oops. I am so sorry. I honestly do not know how I can be so
sloppy.
> > For the red-majority barrel, the probability of
> > drawing a red each time is 3/4, blue 1/4; so
> > the odds of 3 blues, 7 reds is p1 = (1^3)*(3^7)/(4^10).
> > For the blue-majority barrel, it's p2 = (3^3)*(1^7)/(4^10).
> > Therefore the overall probability (conditioning
> > on the 50-50 prior) is the average of those two
> > probabilities. A posteriori we know that this is
> > what happened, so the odds that it was the
> > blue-majority barrel are
>
> > p2/(p1+p2) = 27/(27+2187) = 1/82
>
> Well, I get 27/(27+1054). Is that right?
I meant 27/(27 + 1024). And just in case I did another typo
just then, my calculation is
P(A) ~ (1/4)^7 * (3/4)^3
P(B) ~ (1/2)^7 * (1/2)^3 = 1/1024
and the answer is P(A) divided by P(A)+P(B). No?
Lee
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