Given the events $A, B$ the conditional probability of $A$ supposing that $B$ happened is:
$$P(A | B)=\frac{P(A\cap B )}{P(B)}$$
Can we write that for the Events $A,B,C$, the following is true?
$$P(A | B\cap C)=\frac{P(A\cap B\cap C )}{P(B\cap C)}$$
I have couple of problems with the equation above; it doesn't always fit my logical solutions.
If it's not true, I'll be happy to hear why.
Thank you.
Yes you can. I see no fault. Because if you put $K = B \cap C$ you obtain the original result
Yes and this is also known as multiple conditioning. In general, $$P(A_1 \cap \cdots \cap A_n) = P(A_1) \prod_{i=2}^{n} P(A_{i}|A_{1} \cap \cdots \cap A_{i-1})$$
Yes, $$ P(A|B \cap C) = \frac{{P(A \cap (B \cap C))}}{{P(B \cap C)}} = \frac{{P(A \cap B \cap C)}}{{P(B \cap C)}}. $$
Yes, your conclusion is correct. Note that the way you have defined conditional probability, you must insist that the denominator be non-zero.
