Probability dependent on two independent factors If we know that $P(A|B) = R_1$ and $P(A|C) = R_2$ with $R_2 > R_1$ 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
 A: 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 $P(A|B) = P(A|C)= \frac 12$
A: @MaxFt - I love your simple answer, but I will fight for votes with this one.  :)
An example in words: 


*

*Among Bald people, a fraction $R_1$ are Americans.  

*Among Clueless people, a fraction $R_2$ are Americans.  

*But it turns out there is only one person who is both Bald AND Clueless, named Donald.

*Now is Donald American?  Who knows?
A: It is impossible to predict without more information on the sets A, B and C. To understand this, consider the different possible cases:


*

*$ B \cap C$ is a set with very small size. In such a case, if the area common between B and C is also common with A, probability will be almost 1. If it is not, probability will be almost zero

*$ A \subseteq C \subseteq B $ In this case, $ P \left[ A | B 
 \cap C \right] $ is simply $P \left[ A|C\right]$


Many more such possibilities are possible, all satisfying the above criterion. Therefore the probability cannot be estimated or even predicted to be within certain bounds without more information about the system / sets.
A: 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 $\dfrac 12 > \dfrac 13$ 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.
