Probability union and intersections Is it sometimes true/sometimes false that $P(AB)\ge P(A \cup B)$ when $A$ and $B$ are two events? Using the example of rolling a die I am able to show that it is sometimes false, but am having a hard time thinking of a situation with rolling a die that makes it true. Any ideas?
 A: Sometimes $AB$ is used to denote the intersection of the sets $A$ and $B$, though the notation $A \cap B$ is more common. We first look at the problem under the assumption that by $AB$ you mean $A\cap B$.  
For any two sets $A$ and $B$, we  have $P(A\cap B)\le P(A\cup B)$. The easiest way to see this is geometrically, by using a Venn Diagram. Information about Venn Diagrams can be found here. 
Drawing a simple picture shows that $A\cap B \subseteq A \cup B$, and $P(A\cap B) \le P(A\cup B)$ follows. We can also prove that $A\cup B \subseteq A \cup B$ without a diagram. For if $x\in A\cap B$, then $x\in A$, and therefore $x\in A\cup B$.
The only remaining issue is whether we can have $P(A\cap B)=P(A\cup B)$.  Indeed we can, but it requires very special circumstances.  If $A=B$, then $A\cap B=A\cup B$, so the probabilities are equal. When else? 
We have 
$$A \cup B=(A\cap B) \cup ((A\cup B)\setminus(A\cap B)),$$
where in general if $E$ and $F$ are events then $E\setminus F$ consists of the points that are i $E$ but not in $F$.   Since the events $A\cap B$ and 
$(A\cup B)\setminus(A\cap B)$ are mutually exclusive,
$$P(A\cup B)=P(A\cap B) + P((A\cup B)\setminus(A\cap B)) .$$ 
Thus $P(A\cup B)=P(A\cap B)$ if and only if $P((A\cup B)\setminus(A\cap B))=0$. This can happen with $A \ne B$. For instance, let random variable $X$ have, say, standard normal distribution, let $A$ be the event $1/3 \le X <1/2$, and let $B$ be the event $1/3<X \le 1/2$. Then $P(A\cup B)=P(A\cap B)$ since $P(X=1/3)=P(X=1/2)=0$.
Comment: Things get more interesting if $P(AB)$ is a typo for $P(A)P(B)$.  It is easy to find examples for which $P(A)P(B)<P(A \cup B)$. For instance toss a fair coin once, and let $A$ be the event we get a head, and $B$ the event we get a tail.  We can also have $P(A)P(B)=P(A \cup B)$, for example if each of $P(A)$ and $P(B)$ is equal to $1$.  
It remains to ask whether we can have $P(A)P(B)>P(A \cup B)$. 
Let $P(A)=a$ and P(B)=b$. 
By looking at a Venn Diagram, or otherwise, we can see that 
$$P(A \cup B) \ge \max(P(A), P(B))=\max(a,b).$$
Since $0 \le a \le 1$ and $0 \le b \le 1$, we always have $ab \le \max(a,b)$. So we cannot have $P(A)P(B)>P(A \cup B)$.
Even $ab=\max(a,b)$ can only happen in very special circumstances. Equality certainly does happen if $a=b=0$. It cannot happen if one of $a$ or $b$ is $0$ and the other is not.  Now assume that neither $a$ nor $b$ is $0$. Then $ab=\max(a,b)$ if and only if $a=b=1$.
A: Your title is "Probability union and intersections".  
So in words, you are asking if it is sometimes false that the probability of $A$ and $B$ is greater than or equal to the probability of $A$ and $B$.  Yes this is sometimes false.
Either the probability of $A$ and $B$ is less than the probability of $A$ or $B$ or the probability of $A$ or $B$ is equal to the probability of $A$ or $B$; the latter is true when  the probability of $A$ and not $B$ and the probability of $B$ and not $A$ are each zero.  It is not possible that the probability of $A$ and $B$ is greater than the probability of $A$ or $B$. 
You may have originally intended to ask about $P(A\cap B)\le P(A\cup B)$.  This is always true.
A: $A \cup B$ is the union of three mutually exclusive events $A \cap B$, $A^c \cap B$, and $A \cap B^c$ where $A^c$ denotes the complementary event that occurs exactly when $A$ does not occur, and similarly for $B^c$.  Hence,
$$P(A \cup B) = P(A \cap B) + P(A^c \cap B) + P(A \cap B^c).$$
Since probabilities are nonnegative, $P(A \cap B) > P(A \cup B)$ can 
never be true (it would require  $P(A \cap B)$ or $P(A^c \cap B)$ or
both to be negative), and $P(A \cap B) = P(A \cup B)$ will hold exactly
when $P(A \cap B^c) = P(A^c \cap B) = 0$.  Note that this implies that
$$P(A) = P(A\cap B^c) + P(A \cap B) = P(A \cap B)$$ and similarly
$$P(B) = P(A^c\cap B) + P(A \cap B) = P(A \cap B)$$ also .  Thus,
$P(A \cap B) = P(A \cup B)$ holds exactly when $P(A) = P(B) = P(A\cap B)$.
In typical real-life applications to finite sample spaces,
$P(A \cap B) = P(A \cup B)$ holds exactly when $A = B = A\cap B = A\cup B$
because only the empty event has zero probability.    More generally, suppose that $X$ is a continuous random variable,
 uniformly distributed on $(0, 1)$.
Let $A = \{0.5 \leq X < 0.6\}, B = \{0.5 < X \leq 0.6\}$ so that $A \neq B$, but $P(A) = P(B) = 0.6 - 0.5 = 0.1$.  Then, $A \cup B = \{0.5 \leq X \leq 0.6\}$ and $A \cap B = \{0.5 < X < 0.6\}$ are different from $A$ and $B$ but
both have the same probability $0.6 - 0.5 = 0.1$ as $A$ and $B$.
