# probability of the union and intersection of sets A and B

If I have two sets A and B and take

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

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

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

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.

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.

• How do you know Commented Jan 31, 2020 at 16:44
• 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 $F$. Commented Jan 31, 2020 at 17:01

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.