# Quick probability proof

• September 15th 2010, 09:47 AM
Zennie
Quick probability proof
Let A and B be two events. Prove that $P(AB) \geq P(A) + P(B) - 1$
• September 15th 2010, 09:51 AM
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Quote:

Originally Posted by Zennie
Let A and B be two events. Prove that $P(AB) \geq P(A) + P(B) - 1$

Assuming $P(AB)$ means $P(A\cup B)$, use that $P(A\cup B) = P(A) + P(B) - P(A \cap B)$ and that $0\le P(X) \le1$ for any event $\,X$.
• September 15th 2010, 10:44 AM
Zennie
Sorry for the lack of clarity.

$P(AB)$ means $P(A \cap B)$. So prove that $P(A \cap B) \geq P(A) + P(B) - 1$
• September 15th 2010, 10:50 AM
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Quote:

Originally Posted by Zennie
Sorry for the lack of clarity.

$P(AB)$ means $P(A \cap B)$. So prove that $P(A \cap B) \geq P(A) + P(B) - 1$

If you solve for $P(A \cap B)$ you'll see it works out the same.