# Finding P(AnB') (Homework)

#### NikW

##### New Member
Hi there, this is my first post on this forum.

I am trying to find the probability of P(BnA')

the data given is:
P(A) = 1/3
P(B) = 1/3
P(AnB) = 1/10

My attempt:
P(A') = 1-(1/3) = 2/3
P(BnA')= P(B) * P(A') = 1/3 * 2/3 = 2/9
The reason I think what I did was wrong is because I believe P(AnB) = P(A) * P(B) only applies to independent events. I don't think these are independent since 1/3 * 1/3 does not equal 1/10. I do not know where to go from here.

#### BGM

##### TS Contributor
There is a useful fact which you can use in these kinds of problems.

Let me give you an example first:

Note that

$$A$$ and $$A^c$$ are disjoint sets (in probability context we say they are mutually exclusive). Also they are collectively exhaustive - the union of them is the sample space. Therefore,

$$1 = P(S) = P(A \cup A^c) = P(A) + P(A^c)$$

which should be a well-known property taught in early probability lesson.

Now, consider $$B \cap A$$ and $$B \cap A^c$$.

Are they disjoint? What is the union of them?

Hopefully you can come up with a similar result here.

Once you understand this, you can try to extend the result to law of total probability which is extremely useful. If you are attempting to draw the Venn's Diagram, you should be drawing some puzzles inside the sample space.