An example of three probability events with certain property I am trying to find three events $A,B,C$ with the following properties
\begin{gather}
P(A|B)>P(A),\\
P(A|C)>P(A),\\
P(A|B\cup C)<P(A).
\end{gather}
I have not been able to come up with events satisfying all three, and would appreciate some help.
 A: Consider drawing one number from $1$ to $5$ and define following events


*

*$A:$ Number drawn is $1,2$

*$B:$ Number drawn is $1,3$

*$C:$ Number drawn is $1,4$
A: In fact, I think I've just spotted an answer.
Consider picking a number at random from $\{1,2,\dots,7\}$.
Let $A$ be the event $1$ or $2$ is picked, let $B$ be the event that one of $1$, $3$, or $4$ is picked, and $C$ be the event that one of $1$, $5$ or $6$ is picked. 
A: Here's an example we can reason through intuitively.
Suppose you're buying two lottery tickets with a $\frac1{1000}$ chance of winning. (This chance can be made as small as you like, as long as it's less than $\frac12$.) Let $A$ be the event that neither or both of the tickets win. (They tickets are for different drawings, so they're independent.)
Now,


*

*Define $B$ to be the event that the first ticket fails to win. Then $\Pr[A \mid B] > \Pr[A]$ because the likeliest way by far for $A$ to happen is if neither ticket wins, and conditioning on $B$ makes that likelier.

*Define $C$ to be the event that the second ticket fails to win. We have $\Pr[A \mid C] > \Pr[A]$ for exactly the same reason as above.


However, $B \cup C$ is just the event that at most one ticket wins: the negation of "both tickets win". So $\Pr[A \mid B\cup C] < \Pr[A]$ because conditioning on $B \cup C$ removes one (very unlikely) way for $A$ to happen, and does nothing else.
