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This question already has an answer here:

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.

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

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

$P(a\cup b)\le 1\\\Rightarrow0\le 1-P(a\cup b)\\\Rightarrow 0\le1-[P(a)+P(b)-P(a \cap b)]\\\Rightarrow 0\le 1-[P(a)+P(b)]+P(a\cap b)\\\Rightarrow P(a)+P(b)-1\le P(a\cap b)$

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  • 2
    $\begingroup$ 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 $\endgroup$ – Dilip Sarwate Feb 21 '18 at 3:59

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