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If I have two sets A and B and take

$$ P((A\cap B) \cap (A\cup B)),$$

is this the same as $P(A\cup B)$?

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  • $\begingroup$ Not quite: $(A \cap B) \cap(A\cup B) = (A \cap B) $ while $(A \cap B) \cup(A\cup B) = (A \cup B) $ $\endgroup$
    – Henry
    Commented Jan 31, 2020 at 8:58
  • $\begingroup$ Draw a Venn Diagram of the two events and compare. $\endgroup$
    – whuber
    Commented Jan 31, 2020 at 16:28

3 Answers 3

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Using Venn diagrams, you can easily see, $(A \cap B) \cap(A\cup B) = (A \cap B) $ $\\ $

But $(A \cap B) = (A \cup B)$, only when $A$ and $B$ are subsets of each other, i.e they both have the same elements.$\\$

Hence, $P(A \cap B) = P(A \cup B)$, only when A and B have the same elements.

Or $P(A \cap B) \cap(A\cup B) = P(A \cup B) $, only when A and B have the same elements.

I hope this helps.

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Guide:

  • Notice that if $C \subset D$, then $C \cap D = C$.

  • For $A \cap B$ and $A \cup B$, one of them is a subset of the other. Using the first pointer, you should be able to simplify your expression.

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  • $\begingroup$ How do you know $\endgroup$
    – caston1414
    Commented Jan 31, 2020 at 16:44
  • $\begingroup$ give it and attempt to prove it,, to prove two sets are equal, you show that they are subset of each other. To show that a set $E$ is a subset of $F$, you take an element of $E$ and show that it is an element of $F$. $\endgroup$ Commented Jan 31, 2020 at 17:01
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No, it is not the same because the union contains all the elements in the intersection, but not the vice versa; therefore their intersection is actually the intersection of the original sets.

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