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). However, I can't really see this situation occurring here, could someone please explain this to me?

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?