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Suppose $A,B,C$ are events. My tutor wrote that Bayes theorem is

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

I've just seen that $$ P(A|B)P(B) = P(B|A)P(A) $$ How does the former follow from the latter?

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    [This question was asked on stats.SE](http://stats.stackexchange.com/q/258379/6633) just last week, and it might be worth your while to read the answers given there. – 2017-02-05

2 Answers 2

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We have that $\Pr (A \vert E) \Pr (E) = \Pr (E \vert A) \Pr (A) = \Pr (A \cap E)$.

Let $E = B \cap C$, so that $$\Pr (A \vert B\cap C) \Pr (B\cap C) = \Pr (B\cap C \vert A) \Pr (A) = \Pr (A \cap B \cap C)$$

The left and right expressions give us that

$$ \Pr(A \vert B \cap C) = \frac{\Pr (A \cap B \cap C)}{ \Pr(B \cap C)} = \frac{\Pr(A\cap B \vert C) \Pr (C)}{\Pr (B \vert C) \Pr (C)} = \frac{\Pr(A \cap B \vert C)}{\Pr(B \vert C)} $$

where the second equality follows from the definition of conditional probability, similar to the second displayed equation above.

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Assume that for any events $X$ and $Y$, we have that:

$$\Pr[X \mid Y] = \frac{\Pr[X \cap Y]}{\Pr[Y]}$$

Then observe that: $$ \frac{\Pr[A \cap B \mid C]}{\Pr[B \mid C]} = \frac{\dfrac{\Pr[A \cap B \cap C]}{\Pr[C]}}{\dfrac{\Pr[B \cap C]}{\Pr[C]}} = \frac{\Pr[A \cap B \cap C]}{\Pr[B \cap C]} = \Pr[A \mid B \cap C] $$