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Topic: A question on conditional probability!
Replies: 1   Last Post: May 29, 2006 4:05 AM

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Ray Koopman

Posts: 3,382
Registered: 12/7/04
Re: A question on conditional probability!
Posted: May 29, 2006 4:05 AM
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Sticker wrote:
> Then any idea of how to calculate this?

The hardest part of all this is getting a notation.
The following isn't perfect, but...

Let P(A1=1) = p1 and P(A1=0) = q1 = 1-p1,
and similarly for A2 and A3.

Since A1,A2,A3 are independent, we have

a1 a2 a3 P(A1=a1 & A2=a2 & A3=a3)
0 0 0 q1*q2*q3
0 0 1 q1*q2*p3
0 1 0 q1*p2*q3
0 1 1 q1*p2*p3
1 0 0 p1*q2*q3
1 0 1 p1*q2*p3
1 1 0 p1*p2*q3
1 1 1 p1*p2*p3,

where p1,p2,p3 are free to be specified.

Treat the binary variables A1,A2,A3
as a single variable A that takes on 8 values:
read the combination a1,a2,a3 as a binary number a,
and relabel the rightmost column above as r(a) = P(A=a), a = 0...7.

Let P(B=1|A=a) = s(a)
and P(B=0|A=a) = t(a) = 1 - s(a), a = 0...7,

where s(0)...s(7) are free to be specified.

Then P(A=a & B=1) = r(a)*s(a)
and P(A=a & B=0) = r(a)*t(a), a = 0...7.

You asked about P(A=7|B=1), which is
P(A=7 & B=1) / sum_a P(A=a & B=1)
= r(7)*s(7) / sum_a r(a)*s(a).

In general, that expression does not simplify.




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