Probability dependent on two independent factors If we know that $P(A|B) = R_1$ and $P(A|C) = R_2$ with $R_2 &gt; R_1$ what is the probability of $A$ given both $B$ and $C$ are true (perhaps denoted $P(A|B \wedge C)$)

Intuitively it would seem like it is greater than $R_1$ and less than $R_2$ but I can't seem to figure out what it is exactly.

I apologize if this is a duplicate, I might just not know the terminology to search for this correctly
Both are false $P(A|B\cap C)$ could take any value from 0 to 1:

Consider that we choose an uniformly randomly number $X$ from $\{1,2,3\}$

Let $A = \{ X= 1 \ \mathrm{or} \ 2\}$, $B=\{X=2 \ \mathrm{or} \ 3 \}$, $C=\{X=1 \ \mathrm{or} \ 3 \}$

Well $P(A|B\cap C)=0$ and $P(A|B) = P(A|C)= \frac 12$

Now let's say $A=\{X=3  \}$ 

Well $P(A|B\cap C)=1$ and $P(A|B) = P(A|C)= \frac 12$
I think it must be larger than both $R_1$ and $R_2$.

For example, suppose we roll a fair die. Let $A$ be the event that we get a $6$, $B$ the event that we get a multiple of $2$, $C$ the event we get a multiple of $3$.

Then $\Bbb P (A|B)=\dfrac 13$ and $\Bbb P (A|B)=\dfrac 12$ where $\dfrac 12 &gt; \dfrac 13$ as required.

However, $\Bbb P (A|B \cap C)=\Bbb P (\text{roll a }6|\text{multiple of 6})=1$



EDIT:

Note that we need the condition that $B \cap C \neq \emptyset$ since  $\Bbb P(A| \emptyset )$ is undefined.