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I have to prove that $A\subseteq B\implies\Bbb P(A) \le\Bbb P(B)$, where $\Bbb P$ is probability.

So far I have learned about direct proof, mathematical induction and proof by contraposition.

This exact proposition is solved in my literature (*Elementare Einführung in die Wahrscheinlichkeitsrechnung* by Karl Bosch), but it isn’t explained at all and I don’t want to copy sth. I don’t understand.

- Expected area of the intersection of of triangles made up random points inside a circle, all the triangles must contain the origin
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- Probability of two integers' square sum divisible by $10$
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- Understanding Borel sets
- A game of guessing a chosen number

Which method should I choose and how is it done?

Thanks!

(c means “is subset of”)

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Since $A\subseteq B\rightarrow N(A)\leq N(B)\rightarrow \frac{N(A)}{N(\Omega)}\leq \frac{N(B)}{N(\Omega)}\rightarrow P(A)\leq P(B)$.

Probability is a kind of measure measuring the size of sets, which are called events. And basically, measure requires two properties. One, non-negative; Two, measure of disjoint union of sets equals to sum of measures of each set. Write in language of probability, for any disjoint events $A,B$, we have

$$P(A\cup B)=P(A)+P(B)\geq0$$

So, if $A\subseteq B$, then

$$P(B)=P(A\cup(B/A))=P(A)+P(B/A)\geq P(A)$$

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