# How can I show that $P(A \cap B) \gt P(A) + P(B) - 1$

How can I show that $P(A \cap B) \gt P(A) + P(B) - 1$

I know that $P(A \cap B)= P(A)P(B)$

But I don't see how that can help me get to that inequality. Can someone give me a hint on how to start this?

• $P(A \cap B) = P(A)\cap P(B)$ doesn't make sense. $P(A)$ and $P(B)$ are numbers so there is no idea of their intersection in the probabilistic sense. – SZN May 10 '17 at 21:05
• Are these events independent??? – ncmathsadist May 10 '17 at 21:05
• @ncmathsadist that is all I am given. There is nothing more to the question.I am having a hard time trying to figure it out – Hidaw May 10 '17 at 21:06
• If you let $A=B$ be an event that is always true you get $1>1$. – copper.hat May 10 '17 at 21:07
• use $B=A'$ to get $0\gt0$ – JonMark Perry May 11 '17 at 7:01

The $>$ sign should change to $\ge$ to make sense. This is because $$P(A\cup B)\le 1$$ $$P(A)+P(B)-P(A\cap B)\le 1$$ $$P(A)+P(B)-1\le P(A\cap B)$$

It's not true as stated, but with $\ge$ instead of $>$ it would be true. Note that $$P(A) + P(B) - P(A \cap B) = P(A \cup B) \le 1$$

The inequality is not correct, for example, let $A$ be the event that you get a head and $B$ be the even that you get a tail.

$$P(A \cap B)=0$$

$$P(A)+P(B)-1=0$$

$$0> 0$$ which is a contradiction.

• Oh I see so the inequality should be equal or greater not just greater. Is that correct? – Hidaw May 10 '17 at 21:09
• Yes, as proven by msm. – Siong Thye Goh May 10 '17 at 21:10
• Also, $P(A \cap B)=P(A)P(B)$ only holds when you have independence of events. – Siong Thye Goh May 10 '17 at 21:12

$$P(A\cap B)-P(A)-P(B)+1=P(A)P(B)-P(A)-P(B)+1=$$ $$=\left(1-P(A)\right)\left(1-P(B)\right)\geq0$$

$$P(A\cup B) =P(A) +P(B) - P(A\cap B)$$

$$P(S)=1 \ge P(A \cup B)$$

$$P(A∩B) \le P(A) + P(B) - 1$$