Simplifying the Conditional probability

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To simplify the conditional probability P(S1 ∩ S2 ∩ S3 | r) when S2 and S3 are independent of r, one can utilize the property that allows conditioning on independent events. The equation P((X|Y)|Z) = P(X | (Y ∩ Z)) is suggested as a valid approach, though its acceptance may vary based on the interpretation of probability, whether through measure theory or simpler frameworks. Applying standard probability laws with the condition "| Z" can also aid in simplification, such as transforming P(A ∩ B | Z) into P((A|B)|Z) P(B | Z). The discussion encourages exploring these methods to make progress in simplifying the initial equation. Understanding the independence of events is crucial in this context.
bhathi123
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P(S1 \cap S2 \cap S3 | r)
How do I simplfy the above equation if S2 and S3 are independent of r ?
 
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I've don't recall seeing a book that gives the probability law
P((X|Y)|Z) = P(X | (Y \cap Z) )
but I think its true. How to argue it depends on whether you are approaching probability as measure theory or something simpler.

You can also apply the usual probability laws with a condition lilke "| Z" tagged onto every term. For example,
P(A \cap B) = P(A|B) P(B)
so
P(A \cap B | Z) = P( (A|B)|Z) P(B | Z)

See if you can make progress by applying those ideas.
 
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