# Another proof using the axioms of probability

1. Mar 14, 2013

### phosgene

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

Axiom 1: $P(n) \geq 0$

Axiom 2: $P(S)=1$

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

3. 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?

Last edited: Mar 14, 2013
2. Mar 14, 2013

### vela

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

3. Mar 14, 2013

### phosgene

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

4. Mar 14, 2013

### vela

Staff Emeritus
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.

5. Mar 14, 2013

### phosgene

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