Another proof using the axioms of probability

[phosgene]

Homework Statement

If A and B are events, use the axioms of probability to show that:

if B \subset A, then P(B) \leq P(A)

Homework Equations

Axiom 1: P(n) \geq 0

Axiom 2: P(S)=1

Axiom 3: If A₁,A₂,... are disjoint sets, then P(\bigcup _{i} A_{i}) = \sum_{i} P(A_{i})

The Attempt at a Solution

I start with using the law of total probability to define the set A:

A= (A \cap B) \cup (A \cap B^{C})

Then I use axiom 3 to get turn it into a probability:

P(A) = P(A \cap B) + P(A \cap B^{C})

Since B \subset A, P(A \cap B) = P(B)

So

P(A) = P(B) + P(A \cap B^{C})

P(B)=P(A) - P(A \cap B^{C})

And as axiom 1 states that a probability must be greater than or equal to 0,

P(B) \leq P(A)

As for proving the equality case, this means that P(A \cap B^{C}) = 0, but then doesn't that just mean that A=B. Since the question states that B is a *proper* subset of A, am I incorrect in thinking that it might be a typo?

 

[vela]

And as axiom 1 states that a probability must be greater than 0

It says the probability must be greater than or equal to 0.

 

[phosgene]

Ok, fixed. But suppose that P(A \cap B^{C}) = 0, doesn't this mean that A=B?

 

[vela]

Not necessarily. For example, say you have a continuous random variable X that's uniformly distributed on [0,1]. Let A=[0,1] and B=(0,1). Both P(A)=P(B)=1, but A≠B.

 

[phosgene]

I see, interesting..so basically, using axiom 1, I will get P(B) \leq P(A), which completes my proof. Thanks :)