Probability P(A\B) = P(A) - P(B)

[magicarpet512]

Homework Statement

Let A \subseteq B \subseteq S where S is a sample space.
Show that P(A \setminus B) = P(A) - P(B)

Homework Equations

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

The Attempt at a Solution

I have,
P(A \setminus B) = P(A \cap B^{C}) <br /> = P(A) - P(A \cap B) <br /> = P(A) - [P(B) - P(A^{c} \cap B)] <br /> = 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!

 

[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.

 

[magicarpet512]

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

Yes, thank you.

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.

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?

 

[spamiam]

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?

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?