1. ## Probability Proof

Problem:

Let A and B be two events, show that:

P(A) + P(B) -1 $\leq$ P(A U B) $\leq$ P(A)+P(B)

Questions/Attempts:

Am I supposed to assume that A and B are the only events in the sample space, and that therefore P(A)+P(B)=1?

If I assume that, and begin the proof with the first axiom of probability, I get this far:

1. 0 $\leq$ P(A) $\leq$ 1

2. P(A)+P(B)=1 (Since P(S)=1, where P(S) is the sample space)

so

3. 0 $\leq$ P(A) $\leq$ P(A) + P(B)

There's a couple things I can do from here but it seems to make things more complicated than they need to be.

Thanks for any help in advance!

2. Use $P(A)+P(B)-P(A\cap B)=P(A\cup B)\le 1$ and $0\le P(A\cap B)\le 1$

3. Questions:

1. If I reduce the problem statement to an identity, is that considered "proving" it?

2. In this problem, can I assume P(A)+P(B)=1?

Attempt:

1.P(A) + P(B) -1 $\leq$ P(A U B) $\leq$ P(A)+P(B)

2. P(A) + P(B) -1 $\leq$ P(A)+P(B)- P(A $\cap$ B ) $\leq$ P(A)+P(B)

but P(A)+P(B)=1 so,

3. 0 $\leq$ 1 - P(A $\cap$ B ) $\leq$ 1

and P(A $\cap$ B ) is between 0 and 1.

It almost seems like the proof went backwards here, from a statement to an identity. Is this correct? Thanks for your help Plato I appreciate it.

4. Originally Posted by divinelogos
1. If I reduce the problem statement to an identity, is that considered "proving" it?

2. In this problem, can I assume P(A)+P(B)=1?
No to both of those.
From the axioms we get:
$P(A)+P(B)=P(A\cup B)+P(A\cap B)\le P(A\cup B)+1$

OR $P(A)+P(B)-1\le P(A\cup B).$

5. How do you start the proof? What axiom do you start with?