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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.

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The formula for conditional probability is correct only when $P(B)\neq 0$, not when $B$ can occur. There are plenty of events that can occur that have zero probability. Like picking $1/2$ out of a uniform distribution on $[0,1]$ or flipping a coin and having every even flip come up heads. – Aaron May 24 '11 at 19:14
up vote 7 down vote accepted

Yes you can. I see no fault. Because if you put $K = B \cap C$ you obtain the original result

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Yes and this is also known as multiple conditioning. In general, $$P(A_1 \cap \cdots \cap A_n) = P(A_1) \prod_{i=2}^{n} P(A_{i}|A_{1} \cap \cdots \cap A_{i-1})$$

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The way you've written that is (very) prone to misinterpretation. – cardinal May 24 '11 at 19:16

Yes, $$ P(A|B \cap C) = \frac{{P(A \cap (B \cap C))}}{{P(B \cap C)}} = \frac{{P(A \cap B \cap C)}}{{P(B \cap C)}}. $$

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Note that the symbol $ \cap $ may be replaced by "and". – Shai Covo May 24 '11 at 18:57
In addition, see Section 1.5 in – Shai Covo May 24 '11 at 20:23

Yes, your conclusion is correct. Note that the way you have defined conditional probability, you must insist that the denominator be non-zero.

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