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Disjoint events A and B with positive probability must not be independent as $P(A)$, $P(B) \gt 0$, $P(A\cap B) = P(\emptyset) = 0 \Rightarrow P(A\cap B) \lt P(A)P(B)$. What are some real world examples that can help better understanding this?

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Consider the experiment of rolling a fair dice and define $A=\{\text{roll a }1,\text{ or a } 2, \text{ or a } 3\}$, $B=\{\text{roll a } 4,\text{ or a } 5, \text{ or a } 6\}$. Clearly, $A$ and $B$ are disjoint and $P(A)=P(B)=1/2 \gt 0$, which satisfy the conditions. – user10525 Dec 29 '12 at 23:37
    
Consider any attribute which is normally regarded as categorical (eye color, sex, educational level). Any of them work - e.g. if I have blue eyes, even though brown eyes are very common, I know I don't have brown eyes - not-having-brown-eyes is utterly dependent on having-blue-eyes. – Glen_b Dec 30 '12 at 0:09
    
@Procrastinator Excuse me, but it is not obvious to me that $A$ and $B$ are dependent. – qazwsx Dec 30 '12 at 3:32
    
@Glen_b, $A$ and $B$ being "having blue eye" and "having brown eye", shouldn't you be making a statement on those? Why you said something about $A$ and $B^c$? In other words, it is still not obvious to me that "having blue eye" and "having brown eye" are dependent. – qazwsx Dec 30 '12 at 3:34
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By definition of independent events. You can see that this is not satisfied. Intuitively, if $A$ occurs, then $B$ cannot occur. This reflects the dependency. – user10525 Dec 30 '12 at 4:05
up vote 1 down vote accepted

Consider any attribute which is normally regarded as categorical (eye color, sex, educational level). Any of them work - e.g. if I have blue eyes, even though brown eyes are very common, I know I don't have brown eyes - not-having-brown-eyes is utterly dependent on having-blue-eyes.

[If your eyes are brown, they aren't blue.* If your eyes are blue, they aren't brown. If one occurs, the other doesn't.]

* Note that "one blue and one brown eye" is neither "having brown eyes" nor "having blue eyes".

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