# Homework Help: Probability P(A\B) = P(A) - P(B)

1. Sep 22, 2011

### magicarpet512

1. The problem statement, all variables and given/known data
Let $A \subseteq B \subseteq S$ where $S$ is a sample space.
Show that $P(A \setminus B) = P(A) - P(B)$

2. Relevant equations

$A \setminus B$ denotes set difference; these are probability functions.

3. The attempt at a solution
I have,
$P(A \setminus B) = P(A \cap B^{C}) = P(A) - P(A \cap B) = P(A) - [P(B) - P(A^{c} \cap B)] = P(A) - P(B) + P(A^{c} \cap B)$

It seems like i'm close, but i've spent a while trying to figure out how to get rid of the $P(A^{c} \cap B)$.

Any insight anyone?
Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 22, 2011

### spamiam

I think your inclusion is backwards: you mean $B \subseteq A \subseteq S$, right?

Try writing A as the union of two disjoint sets--this will give you P(A) in terms of something that can be rearranged into what you're trying to prove.

3. Sep 22, 2011

### magicarpet512

Yes, thank you.

I've tried that... unless i'm missing something?
A as the union of disjoint sets is $A = (A \cap B) \cup (A \cap B^{c})$.
So, $P(A) = P((A \cap B) + (A \cap B^{c})$.
When i plug this in and do some rearranging, i just get right back to where i ended up in the original post?

4. Sep 22, 2011

### spamiam

Ah I see, your calculations just went off in an unexpected direction after the third equality. Take a look at your third equality: since $B \subseteq A$, then what is $A \cap B$?