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conditional probability is defined us:

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

But if (and there is the part where i might be missing something) $P(A \cap B) = P(A) \cdot P(B)$ then:

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

which doesn't make any sense. Why would the formula exist in the first place. I'm confused.

Thanks for helping me :).

  • 11
    $P(A\cap B)=P(A)P(B)$ holds only when $A$ and $B$ are [independent](http://en.wikipedia.org/wiki/Independence_%28probability_theory%29); in which case the result should not be surprising.2012-12-17
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    So how do i count $P(A|B)$ if $A$ and $B$ are dependent? What is formula for $P(A \cap B)$, $A$ and $B$ beeing dependent?2012-12-17
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    If $A$ and $B$ are dependent, there is no formula for $P(A \cap B)$ in terms of $P(A)$ and $P(B)$. In fact, depending on the events in question, $P(A \cap B)$ can be more-or-less anything.2012-12-17
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    I think it might be instructive to go through a bunch of examples. See: http://www.ams.sunysb.edu/~jsbm/courses/311/conditioning.pdf2012-12-17
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    Thanx you all. I think i have it clear now. I'm little sorry there is no abstract formula for $P(A \cap B)$ that would be common for all kinds of probabilities.2012-12-17
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    The formula $\Pr(A\cap B)=\Pr(B|A)\Pr(A)$ that you quoted is precisely such a formula. It is fairly often the case that we can find $\Pr(A)$ and $\Pr(B|A)$ quite easily.2012-12-17

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