If an event $A$ is independent from $B$ and $C$, is A also independent from $B \cap C$? $A$ and $B$ are independent. $A$ and $C$ are independent. Are $A$ and $B \cap C$ also independent?
 A: 
Let $A$ and $B$ be independent events, and let $A$ and $C$ be independent events. How do I show that $A$ and $B\cap C$ are independent events as well?

You cannot show this result because it does not hold for all $A, B, C$
enjoying these properties. Consider the following counter-example.
Consider two independent tosses of a fair coin. Let $B=\{HT,HH\}$ and $C=\{HT,TT\}$ be the events that the first and second tosses resulted in Heads and Tails respectively. Let $A=\{HT,TH\}$ be the event that exactly one toss resulted in Heads.
Then, $P(A)=P(B)=P(C) = \frac 12$ while $P(A\cap B) = P(A\cap C) =
\frac 14$ and so $A$ and $B$ are independent events as are $A$ and
$C$ independent events. Indeed, $B$ and $C$ are also independent
events (that is, $A$, $B$, and $C$ are pairwise independent events).
However, 
$$P(A) = \frac 12 ~ \text{and}~ P(B\cap C)=\frac 14 ~ \text{while}~ P(A\cap(B\cap C)) =\frac 14 \neq P(A)P(B\cap C)$$
and so $A$ and $B\cap C$ are dependent events.

Notice that whether $B$ and $C$ are independent 
or not is not relevant to the
issue at hand: in the counter-example above, $B$ and $C$ were
independent events and yet $A = \{HT, TH\}$ and $B\cap C = \{HT\}$ were
not independent events.If $A$, $B$, and $C$ are mutually independent events (which
requires not just independence of $B$ and $C$ but also for
$P(A\cap B \cap C) = P(A)P(B)P(C)$ to hold), then $A$ and $B\cap C$
are indeed independent events. Mutual independence 
of $A$, $B$ and $C$ is a sufficient condition.
A: No. Pick two numbers $x,y$ from $\{0,1\}$ independently and with prob 1/2 each. Let $B$ be the event that $x$ is even, $C$ be the event that $y$ is even and $A$ be the event that $x+y$ is even.
A: $\implies P(A|B)=P(A),P(B|A)=P(B)$
$\implies P(A|C)=P(A),P(C|A)=P(C)$
$P(A|(B∩C))=P(A∩B∩C)/P(B∩C)$
Which after simplification using,
$P(X|Y)=P(X∩Y)/P(Y)$
$P(A|(B∩C)) ≠ P(A)$
