# Show $P(A\cap B)\geq P(A)+P(B)-1$

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

How to show that $P(A\cap B)\geq P(A)+P(B)-1$? Can anyone show me to solve this kind of question? Thanks in advance.

Rearranging the equation we can get $$P(A\cap B)=P(A)+P(B) - P(A\cup B)$$ Now it is clear that $P(A\cup B) \leq 1$ which allows you to conclude.

• Yea I did this. But I didn't know that I could conclude after that. Thanks in advance! – Mathxx Jan 18 '17 at 9:36

Note that $\displaystyle\left(\bigcap_{i=1}^nA_i\right)^c=\bigcup_{i=1}^nA_i^c\quad$, by De Morgan's law.

Then, taking the probability function on both sides we have, $$P\left(\bigcap_{i=1}^nA_i\right)=1-P\left(\bigcup_{i=1}^nA_i^c\right)$$

Now, using Boole's inequality we have, $\displaystyle P\left(\bigcup_{i=1}^nA_i^c\right)\le\sum_{i=1}^nP\left(A_i^c\right)$

$\displaystyle\implies 1-P\left(\bigcup_{i=1}^nA_i^c\right)\ge1-\sum_{i=1}^nP\left(A_i^c\right)$

$\displaystyle\implies P\left(\bigcap_{i=1}^nA_i\right)\ge1-\sum_{i=1}^n\left(1-P\left(A_i\right)\right)=1-n+\sum_{i=1}^nP\left(A_i\right)$

i.e., $\displaystyle P\left(\bigcap_{i=1}^nA_i\right)\ge \sum_{i=1}^nP\left(A_i\right)-(n-1)$, which is Bonferroni's inequality.

For $n=2$, you have your required result.

Because $P(A\cup B) \leq 1$ (and $-P(A\cup B) \geq -1$ ), we get:

$P(A\cap B) = P (A)+P(B) - P(A\cup B) \geq P (A)+P(B) - 1$