# Need help with proof [duplicate]

I am trouble proving $P(a\cap b)\ge P(a)+P(b)-1$

$P(a \cup b)=P(a)+P(b)-P(a \cap b)$

so

$P( a \cap b)= P(a)+P(b)-P(a \cup b)$

I am not a student.

## marked as duplicate by Michael Chernick, Community♦Feb 21 '18 at 19:04

• use $P(a\cup b)\le 1$ – Deep North Feb 21 '18 at 2:10
• No probability can be greater than 1 and therefore subtracting 1 make the quantity smaller then subtracting P(a U b). – Michael Chernick Feb 21 '18 at 2:44
• If P( a u b)<=1 then P(a u b) +P(a u b)’ <=1. ? – larry mintz Feb 21 '18 at 4:32
• I thought of using the p(a)+p(a)’=1 so P(A’)=1-P(A) Prior was a typo – larry mintz Feb 21 '18 at 4:43
• @larrymintz Please add the self-study tag. Also, this question gets asked periodically, see: stats.stackexchange.com/questions/126901/… – Jim Feb 21 '18 at 16:59

$P(a\cup b)\le 1\\\Rightarrow0\le 1-P(a\cup b)\\\Rightarrow 0\le1-[P(a)+P(b)-P(a \cap b)]\\\Rightarrow 0\le 1-[P(a)+P(b)]+P(a\cap b)\\\Rightarrow P(a)+P(b)-1\le P(a\cap b)$
• More simply, $$P(A)+P(B) - P(A\cap B) = P(A\cap B) \leq 1 \implies P(A) + P(B)-1 \leq P(A\cap B)$$ merely by moving $P(A\cap B)$ and $1$ to the other side of the inequality – Dilip Sarwate Feb 21 '18 at 3:59