# Why is $P(A\cup B)\leq P(A\cap B)$, then $P(A)=P(B)$?

$P(A\cup B)\leq P(A\cap B)$, then $P(A)=P(B)$.

I'm confused by this question. I'm trying to imagine this as a Venn diagram and it seems that if the probabilities of two events are equal, they are essentially the same circle. In that case =, would those events be considered independent? Then it makes sense that the union of those events would just be 2 times one of the events?

Just want to know why this statement is true? It sort of makes sense but when I dig too deep, I'm failing to grasp it.

Edit: Can someone illustrate this with Venn diagrams, if that is possible?

• immediate that $A\cup B\subset A\cap B$ so if $P$ is a probability then the only possible situation is that $P(A\cup B)=P(A\cap B)$. Hence the only part of A or B with nonzero measure is $A\cap B$, hence $P(A)=P(B)$ – enthdegree May 23 '18 at 22:06
• Isn't it intuitively obvious that the likelihood of one event or another happening is MORE than the likelihood of both happening that that is impossible? And that if the likelihood is equal then one can never occur with out the other and the probabilities must be equal? – fleablood May 23 '18 at 22:18
• A little late, I know, but notice that you when you were trying to imagine the picture, you were imagining the "then" clause first, when you should've imagined the "if" part first. When you do this, the answers below are the result. – WaveX May 24 '18 at 1:46

All of the colored regions (red and blue), taken together, represent the event $A \cup B.$ The event $A \cap B$ is represented by the red region alone.

This shows that $A \cap B \subseteq A \cup B$; everything in the intersection of the sets is also in their union. You can't be in $A\cap B$ without also being in $A \cup B.$

It follows that $P(A\cap B) \leq P(A \cup B).$ This is always true.

Now what can we conclude if it is also true that $P(A\cup B) \leq P(A \cap B)$?

If both inequalities are true, we can put them together like this: $$P(A\cap B) \leq P(A \cup B)\leq P(A \cap B).$$ Notice how we have $P(A \cap B)$ on both ends. That is, we have a chain of inequalities like this: $x \leq y \leq x.$ The only way for this to be true is if $x = y.$

Conclusion: $P(A\cap B) = P(A \cup B).$ The probability of anything occurring in the blue regions is zero. The only way for any of the events $A$, $B$, or $A\cup B$ to occur is when $A\cap B$ occurs; and if $A\cap B$ occurs then the other three events have also occurred. Since all four events either occur together or none of them occurs, we have $P(A\cap B) = P(A \cup B) = P(A) = P(B).$

• God bless...you explain like I'm five, and I love you for that. – Asker123 May 24 '18 at 2:46

Note that $$P(B)\leq P(A\cup B)\leq P(A\cap B)\leq P(A)$$ since $$B\subseteq A\cup B\quad \text{and}\quad A\cap B\subseteq A.$$

Moreover $$P(A)\leq P(A\cup B)\leq P(A\cap B)\leq P(B)$$ since $$A\subseteq A\cup B\quad \text{and}\quad A\cap B\subseteq B.$$ In both cases we use the fact that $$A\subseteq B\implies P(A)\leq P(B)$$

Just basic common sense if $A\cap B \implies A$ and $A\cap B \implies B$ and $A \implies A\cup B$ and $B\implies A\cup B$ so $A\cap B\implies A\cup B$.

(Two events occurring imply one or the other occurring.)

So $P(A\cap B)\le P(A) \le P(A\cup B)$ and $P(A\cap B) \le P(B) \le P(A\cup B)$.

So if $P(A\cup B) \le P(A\cap B)$ then $P(A\cap B)\le P(A) \le P(A\cup B)\le P(A\cap B)$ and $P(A\cap B) \le P(B) \le P(A\cup B)\le P(A\cap B)$. So $P(A\cap B) = P(A)=P( B) = P(A\cup B)$.

(This can be interpreted as if two events are as equally or more likely to occur than one or the other than the two events then the two events can never occur separately and they are equally likely to occur. That is because since every time two events occur one or the other occur, so if one or the other event could occur without the other then one or the other event would occur more frequently and be more likely than bot events. As that's a contradiction neither event can ever occur alone. Either both events occur or neither and their probabilities are equal.)

====

I can't/won't draw the venn diagram but $A\cup B = (A \cap B^c) \cup (A^c \cap B) \cup (A\cap B)$ where $(A \cap B^c)$ and $(A^c \cap B)$ and $(A\cap B)$ are independent.

So we have $P(A\cup B) = P(A \cap B^c) + P(A^c \cap B) + P(A\cap B) \le P(A\cap B)$ so $P(A \cap B^c) + P(A^c \cap B) \le 0$. As probabilities can't be $< 0$ we have $P(A \cap B^c)= P(A^c \cap B) = 0$. Which means $A\subset B$ and $B\subset A$.