P(A n B) \leq P(A)

**dwsmith** · January 25th 2011, 11:28 AM

Prove:

$P(A\cap B)\leq P(A)$

using the axioms of probability.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\geq 0$

$P(A\cap B)\leq P(A)+P(B)$

Now, I am stuck with a P(B).

**FernandoRevilla** · January 25th 2011, 11:36 AM

Hint :

$A\cap B\subset A$

Fernando Revilla

**Plato** · January 25th 2011, 11:41 AM

Suppose that $C\subseteq D$ .
Then $\mathcal{P}(D)=\mathcal{P}(C)+\mathcal{P}(D\cap C^c)\ge \mathcal{P}(C)$ .
Probability is monotone.

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