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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?

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What does AB mean? The probability that both A and B happen? –  mixedmath Oct 14 '11 at 4:57
    
By writing P(AB) I am asking about the intersection of two events A and B. –  mathstudent Oct 14 '11 at 13:39

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$A \cup B$ 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 $B$.

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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 $a=b=1$.

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The only other case of equality being when $B\subseteq A$. –  Did Oct 14 '11 at 5:47
    
Are there typographical errors that need fixing in the statement "The only remaining issue is whether we can have $P(A\cap B)=P(A\cap B)$. and indeed we can. For example, if $A \subseteq B$, then $A \cap B = A \cup B = B$, and in particular $P(A \cap B) = P(A \cup B)$? $P(A\cap B)=P(A\cap B)$ is a tautology that we always have, and if $A \subseteq B$, then $A \cap B = A$, not $B$, and it is not always true that $A \cap B = A \cup B$. I think the short answer is that $P(A \cap B) > P(A \cup B)$ can never hold, and $P(A \cap B) = P(A \cup B)$ only when $P(A\cap B^c) = P(A^c \cap B) = 0$. –  Dilip Sarwate Oct 14 '11 at 11:33
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@Dilip Sarwate: Thanks. Calling the errors typographical was an understatement. –  André Nicolas Oct 14 '11 at 14:50

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.

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Regarding your last point that the probability of A or B is equal to the probability of A and B: I see that this happens when the probability of A and not B and the probability of B and not A are each zero, but I cannot seem to think of an example when this could occur when rolling a die. For example, when rolling a six sided die numbered 1-6, P(getting a 7)=0 and P(getting an 8)=0 so P(AB)=0 and P(A U B)=0 like you said. However, aren't those not possible events since they are not in the sample space where S={1,2,3,4,5,6}? Can you think of an example that would show that P(A∩B)=P(A∪B)? –  mathstudent Oct 14 '11 at 13:41
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@mathstudent As I commented in my answer, "In typical real-life applications to finite sample spaces, $P(A \cap B) = P(A \cup B)$ holds exactly when $A = B$." In your die rolling example, you will not get any examples. If $X$ is uniformly distributed on $(0, 1)$, take $A = \{0.5 \leq X < 0.6\}, B = \{0.5 < X \leq 0.6\}$ so that $A \neq B$, $P(A) = P(B) = 0.6 - 0.5 = 0.1$. Then, $A \cup B = \{0.5 \leq X \leq 0.6\}$ and $A \cap B = \{0.5 < X < 0.6\}$ both have the same probability $0.6 - 0.5 = 0.1$ –  Dilip Sarwate Oct 14 '11 at 14:35

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