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How to prove P(A ∩ B) ≤ P(A) using probability theory?

I understand this when drawn on a Venn Diagram but am unsure how it translates to a formal proof.

Ethan Bolker
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    As an instructor I would probably accept a Venn diagram as a proof. For more formality, write $A$ as a disjoint union one piece of which is $A \cap B$. – Ethan Bolker Jul 22 '19 at 22:30
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    $$A \cap B \subset A \implies P(A \cap B) \leq P(A),$$ by monotonicity. – Dzoooks Jul 22 '19 at 22:33
  • @Ethan Bolker I'm not sure a Venn diagram can be called a proof. Proofs must be formal, at least in mathematics. – Mark Jul 22 '19 at 22:44
  • @Mark True. But when I teach I tell my students (at least at the start) that I want to see that they have convinced themselves for good reason. One sentence along with a Venn diagram would do that here. – Ethan Bolker Jul 23 '19 at 00:32

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Here's one approach: Every measure, hence every probability measure $\mathbb{P}$ is monotone. This means $A \subset B$ implies $\mathbb{P}(A) \le \mathbb{P}(B)$. Can you finish the idea?

Another approach would be to notice that $A = (A \cap B) \cup (A \setminus B)$ is a disjoint union. Therefore, by additivity of the measure we have $$ \mathbb{P}(A) = \mathbb{P}(A \cap B) + \mathbb{P}(A \setminus B). $$ Can you finish the proof?

ViktorStein
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Since you are using a Venn Diagram, I presume that your sample space has equiprobable measure.

Let $U$ be the universal set.

Since $A \cap B \subseteq A$, then $\frac{n(A \cap B)}{n(U)} \leq \frac{n(A)}{n(U)}$; which inplies

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

DDS
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