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Hi,
This assertion is true
Franck
Matt Dobrin a écrit :
Does P(A*B|C)=P(A|C)*P(B|A*C)? If not, what does it equal? Thanks in
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Hi,
This assertion is true
Franck
Matt Dobrin a écrit :
Does P(A*B|C)=P(A|C)*P(B|A*C)? If not, what does it equal? Thanks in
It's true. If you are concerned with proof, following the this belove
according to conditional probability p(a|b)=p(a,b)/p(b)
(1) P(A,B|C)=P(A,B,C)/P(C)
(2) P(A,B,C)=P(A,C)*P(B|A,C)
(3) P(A,C)=P(C)*P(A|C)
WITH (2) AND (3) WE GET
(4) P(A,B,C)=P(C)*P(A|C)*P(B|A,C)
TAKING (1) AND (4) WE GET
Michael Cohen ([EMAIL PROTECTED]) wrote:
: : On Wed, 1 Dec 1999, Yonah Russ wrote:
: :
: : how do you solve a problem like this one?
: : thanks in advance
: : ---
: : X is a chance variable such that X(omega)={1,2,3...,n}
: : and for every i in {1,2,3...n}, 4P(X=i+2)=5P(X=i+1)-P(X=i)
:
Donald F. Burrill ([EMAIL PROTECTED]) wrote:
: Haven't seen a public response to this question that I find credible,
: and am curious. Is the problem as described solvable?
: If n is finite, what meaning attaches to "P(X=n+2)" and "P(X=n+1)"?
: If n is infinite, shouldn't the
