# Basic Probability / conditional probability

I have a question:

If $P(A)=0.5, P(B)=0.3, P(A\text{ and }B)=0.1$

Find $P(A\text{ and }B | A \cup B)$

My answer: $$P(A\text{ and }B | A \cup B)=\frac{P[(A\text{ and }B)\text{ and }(A \cup B)]}{P(A \cup B)}$$

The denominator is 0.5 + 0.3 - 0.1 = 0.7

By drawing the Venn diagram I know that $P[(A\text{ and }B) \text{ and } (A \cup B)]$ is actually $P(A \cup B) = 0.1$. But is there any way to do it mathematically ?

I cannot simply multiply $P(A\text{ and }B)$ and $P(A \cup B)$ because A and B are not independent right? So how can I do it?

• Hint regarding the events (A and B) and (A or B): one of those is a subset of the other. – Adrian Aug 20 '16 at 18:03
• So I can just say since (A and B) is a subset of (A or B), therefore $P[(A and B) and (A or B)] = P(A and B) ?$ – BigData Aug 20 '16 at 18:17
• Please add the self-study tag, and read its tag-wiki. – Glen_b Aug 20 '16 at 18:39

First note that "and" refers to intersection "$\cap$" and "or" refers to union "$\cup$".
Intersection distributions over union (similar to how multiplication distributes over addition). i.e. $$A \cap (B \cup C) = (A\cap B) \cup (A \cap C)$$
So in your scenario we have \begin{align*} (A\cap B) \cap (A \cup B) &= ((A\cap B) \cap A ) \cup ((A\cap B) \cap B)\\ &= (A\cap B) \cup (A\cap B)\\ &= (A\cap B) \end{align*}
$P[(A\text{ and }B) \text{ and } (A \cup B)]$ is actually $P(A \cup B) = 0.1$.
It should be $P(A \cap B) = 0.1$ (you had the number right).