How do determine the probability of an event occurring given that two other events occurred

Question

I know that $P(E\,|\,A) = .3$ and that $P(E\,|\,B) = .2$.

I have two questions:

1. How do I go about computing $P(E\,|\,A \cap B)$? (If the information above is not sufficient, what other information would be needed in order to compute it?)
2. Given the information above, what is the best estimate I can make for $P(E\,|\,A \cap B)$?

My research:

Looking at Bayes rule, we have $$P(E\,|\,A \cap B) = \frac{P(E\ \cap A \cap B)}{P(A \cap B)}.$$

That formula, unfortunately, doesn't seem to have any relation to the specified equations above. That is, I don't think I can use $P(E\,|\,A)$ or $P(E\,|\,B)$ in order to determine $P(E\,|\,A \cap B)$. So, my guess for question (1) is that you would need to know $P(E\ \cap A \cap B)$ and $P(A \cap B)$, and that the equations are useless.

For question (2) however, my instinct tells me that given the information above, we should at least be able to come up with some estimate for $E$. Since we know that at least $A$ occurs, I feel like a good estimate would have to be at least greater than or equal to $.3\,$. My instinct also says that since we know at least A occurs, the higher $P(E\,|\,B)$ is, the higher our estimate should be, but I'm not sure of any exact formula for that.

So, I'm in conflict. One part of me says that nothing can be calculated, and another part of me says that we can calculate an estimate, but I'm not sure how.

Any clarification on those two questions would be appreciated.

Answer 1

We cannot make estimates for $P(E\,|\,A \cap B)$ except trivial: $$ 0\leq P(E\mid A \cap B) \leq 1. $$ Let us consider a simple example. Define $\Omega=\{1,2,\ldots,20\}$ with equally probable elementary events: $P(\{i\})=\frac1{20}$.

If we take $$ A=\{1,\ldots,10\},\quad B=\{10,11,\ldots,19\},\quad E=\{1,2,3,18,19\}, $$ then $$P(A)=P(B)=.5, \quad P(A\cap E)=\frac{3}{20}=.15, \quad P(B\cap E)=\frac{2}{20}=.1 $$ and $$ P(E\mid A) = .3, \quad P(E\mid B) = .2. $$ But $$P(E\mid A\cap B) = 0.$$ If we take $E=\{8,9,10,11\}$ then $P(E\mid A) = .3$, $P(E\mid B) = .2$ still holds but $$P(E\mid A\cap B) = 1.$$