A slip of paper is given to an individual A, who marks it with either a plus or a minus sign; the probability of his writing a plus sign is 0.35. A passes the slip to B, who may either leave it alone or change the sign before passing it to C. Next, C passes the slip to D (after perhaps changing the sign or keeping the sign unaltered). Finally, D passes the paper to an umpire. It is unknown if D altered the sign on the paper or kept it unchanged. The umpire observes a plus sign on the slip.
It is known that B, C and D are the only individuals who can alter a sign and the probability of a change is the same for each of them and independent with probability 0.65.
What is the probability that A originally wrote a plus sign?
Let p = .65 be the probability that a particular individual, B, C or D, will change the sign.
The a priori probability of zero alterations would take place is (1p)^3; ... of exactly one alteration, is 3p(1p)^2; ... of exactly two alterations, is 3p^2(1p); and of three alterations is p^3.
The probability that either zero or 2 alterations would take place is then 1  3p + 6p^2  4p^3. As p=.65, this comes out to .4865.
So the probability that A would write a + and that it would be either unchanged or ultimately restored is .35 * .4865. The probability that he would write a  and have it changed ultimately to + is .65 * (1  .4865).
Using Baye's theorem, the probability that it was originally a + is the probability that that would happen and a + would be observed divided by the probability that a + would be observed in either of the two ways of that happening (original + or original ), so:
.35 * .4865 / (.35 * .4865 + .65 * .5135) = .337813708957445
is the probability that A had originally chosen a plus sign. Worked out as a fraction (just leave out the decimal points and simplify, as that's just multiplying the numerator and the denominator each by a million), that comes out to 6811/20162.

Posted by Charlie
on 20061110 09:04:14 