Quick probability proof

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

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

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

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

Originally Posted by Zennie
Sorry for the lack of clarity.

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

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