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I have been struggling with understanding the solution to one of exercises in Blitzstein's book Introduction to Probability. My question is: in part b), why can we just drop $X=15$ in the term $P(Y=k|X=15,B)$? As I understand it, we could do this if $B$ happened which would imply that $X=15$ must have happened too in which case %X=15% gives us no new information and we could drop $X=15$ (or vice versa). This is, I think, equivalent to saying that $X=15 \subset B$ so that $P(X=15,B)=P(X=15)$ (or, again, vice versa). Please correct me if I'm wrong as I want to be crystal clear on that. However, I can't really see this situation occurring here, could someone please explain this to me?

scary exercise

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It is not correct to say that "if B happened", the X=15 happened too." Instead, if B "happened", that is the new treatment is better/works 60% of the time, then what happened previously does not tell us more about the distribution of future outcomes X. Another way of saying that is: conditional on B being true, future events Y do not depend on X.

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  • $\begingroup$ I agree, but for me, intuitively, it feels like if B and X=15 happened, then X=15 does give us a hint about the effectiveness of the new treatment, since it worked for 15/20 people in the pilot study, so it is more probable that it will work, let's say, Y=15 again. I am still not convinced that X=15 does not give us any new information abut the distribution of Y. $\endgroup$
    – slazien
    Commented Aug 23, 2016 at 6:30

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