MHB Prove Theorem: Probability of A Subset B is Less than B

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To prove the theorem that if A is a subset of B, then the probability of A is less than or equal to the probability of B, the user begins by applying the properties of probability functions. They express P(B) as the sum of the probabilities of disjoint events, leading to the conclusion that P(B) is greater than or equal to P(A). The discussion also emphasizes the need to reference specific axioms of probability, including non-negativity, the certainty of events, and the additivity of mutually exclusive events. For the second part of the theorem, the user seeks clarification on how to express the difference between B and A in terms of unions to apply the axioms effectively. The conversation highlights the importance of foundational axioms in proving probability theorems.
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How do I prove the following theorem?

If $A \subset B$ then $P(A) \le P(B)$ and $P(B-A) = P(B)-P(A)$

$A$ and $B$ are events and $P$ is the probability function.

What I tried (but not sure if it's right or not):

$P(B) = P((B\setminus A) \cup (B \cap A)) = P(B\setminus A)+P(B \cap A) \ge 0+P(B \cap A) \ge P(A) $

Therefore $P(B) - P(A) \ge 0$. Hence $P(B) \ge P(A)$.
 
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Guest said:
How do I prove the following theorem?

If $A \subset B$ then $P(A) \le P(B)$ and $P(B-A) = P(B)-P(A)$

$A$ and $B$ are events and $P$ is the probability function.

What I tried (but not sure if it's right or not):

$P(B) = P((B\setminus A) \cup (B \cap A)) = P(B\setminus A)+P(B \cap A) \ge 0+P(B \cap A) \ge P(A) $

Therefore $P(B) - P(A) \ge 0$. Hence $P(B) \ge P(A)$.

Hi Guest! ;)

It looks right to me.
You may want to list the axioms you're using though.
There is some context missing, which is different in every textbook on the subject.
 
I like Serena said:
Hi Guest! ;)

It looks right to me.
You may want to list the axioms you're using though.
There is some context missing, which is different in every textbook on the subject.
Hi, I like Serena. :D Thanks for the reply.

I'm using the following three axioms:

Axiom 1: For every event $A$ in the class $C$,

$$P(A) \ge 0$$

Axiom 2: For the sure or certain event $S$ in the class $C$,

$$P(S) = 1$$

Axiom 3: For any number of mutually exclusive events $A_1, A_2,\cdots$, in the class C,

$$P(A_1 \cup A_2 \cup \cdots) = P(A_1)+P(A_2) +\cdots $$

How do I prove the second part of the theorem?

I know that $P(B-A) = P(B \cap A')$ but I don't know how to turn the intersection into union so that I can use axiom 3.
 
How about $B=(B-A)\cup A$?
 
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