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




Re: A question on conditional probability!
Posted:
May 29, 2006 4:05 AM


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 = 1p1, 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=1A=a) = s(a) and P(B=0A=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=7B=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.



