How to find $p(a \mid b \cap c)$ given $p(a \mid b)$ and $p(a \mid c)$?

Question

How to find $p(a \mid b \cap c)$ given $p(a \mid b)$ and $p(a \mid c)$?

Can this be done? I'm trying to write code for a simulation where an event a might be dependent two or more other known events.

For example, if $A$ happens .5 of the time when $B$ happens, and .8 of the time when C happens, what are the odds of $A$ if both $B$ and C already happened? Can I figure that out?

Answer 1

(Standard notation for probability theory tends to use uppercase letters for events so I will write A,B,C instead of a,b,c in this answer.)

It's going to depend on the specific events $A,B,C$ and how they depend on each other. Based on your comment on my deleted post it seems you know lots of conditional probabilities, such as $P(B|C)$ and $P(C|B)$, so maybe you are interested in an approach like this.

By the definition of conditional probability, $$ P(A | B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)} .$$

You can calculate $P(B \cap C)$ from $P(B|C)$ and $P(C)$, for example:

$$ P(B \cap C) = P(B|C) P(C) .$$

You can calculate $P(A \cap B \cap C)$ also in lots of different ways. For example

$$ P(A \cap B \cap C) = P(C) P(A \cap B | C) .$$

Plugging these in gives $$ P(A | B \cap C) = \frac{P(A \cap B | C)}{P(B|C)} .$$

Answer 2

I obtain answer in some particular case, if it helps in someway.

Suppose $b,c$ are independent i.e.$P(bc)=P(b)P(c)$. Then $$P(a|bc)=\frac{P(abc)}{P(ac)P(ab)}\cdot P(a|b)P(a|c)$$ Is "coefficient" $\frac{P(abc)}{P(ac)P(ab)}$ convenient?

In general case, when we know nothing more, then $P(a|bc)$ can be any:

1. if $a \subset bc \Rightarrow P(a|bc) =1$
2. if $a \cap bc=\emptyset \Rightarrow P(a|bc) =0$