Conditional probability and disjoint events

Question

If $\cup_{n=1}^\infty B_n=\Omega$ and $P(\Omega)=1$ then $\sum_{n=1}^\infty P(B_n)=1$, now $$P(A)=\sum_{n=1}^\infty P(A|B_n)P(B_n)=p\sum_{i=1}^\infty P(B_n)=p$$

If $X$ and $Y$ are independents $$P(X|Y=y)=\frac{P(X=x\cap Y=y)}{P(Y=y)}=\frac{P(X=x)P(Y=y)}{P(Y=y)}=P(X=x)$$ so $$a)\Leftrightarrow b)$$

Answer 1

1. No. The disjoint events $\{B_n\}$ are not independent. They are mutually exclusive.

Hint: Begin with the Law of Total Probability: $\mathsf P(A) = \sum\limits_{n=1}^\infty \mathsf P(A\mid B_n)\;\mathsf P(B_n)$

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2. Yes.

(But watch your shift key; you have a $P(X\mid y=y)$ typo. Additionally you need $X=x$ there not $X$ by itself).