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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 :).

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10  
$P(A\cap B)=P(A)P(B)$ holds only when $A$ and $B$ are independent; in which case the result should not be surprising. –  David Mitra Dec 17 '12 at 21:52
    
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? –  Fidilip Dec 17 '12 at 21:57
    
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. –  Johannes Kloos Dec 17 '12 at 22:02
    
I think it might be instructive to go through a bunch of examples. See: ams.sunysb.edu/~jsbm/courses/311/conditioning.pdf –  Amzoti Dec 17 '12 at 22:03
2  
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. –  André Nicolas Dec 17 '12 at 22:31

1 Answer 1

The easiest way to picture conditional probability is by viewing a Venn diagram. (Sorry, I am doing this quickly and have not learned to generate one in MathJAX yet.) In a simple diagram with two sets A and B, think of conditional probability as the ratio of the area of the intersection of the two circles to the area of the "given" circle (for A given B, that circle is B).

Imagine redefining the universe to be the given set (B in the previous example); the conditional probability is then the probability of A in that universe.

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