What is P(B) given P(A|~B) = 1/2 and P(B?AUB) = 2/5?

[Yankel]

Hello all

I have a little problem with this short question, would appreciate your help.

Let A and B be two events such that:

P(A|~B) = 1/2 and P(B?AUB) = 2/5

(~B means not B)

Find P(B)

The final answer should be 1/4, I can't get there. I did some work with the conditional probability formula, got P(B)=2/5 * P(AUB) and P(B)=1-2*P(A and ~B). But where do I go from here ?

Thank you in advance !

 

[Jameson]

Hi Yankel!

I'm not sure what you mean by this: P(B?AUB) = 2/5. Could you explain? It's the question mark that's confusing me.

Anytime I see the phrase "conditional probability" I always write out the formula: \(\displaystyle P[A|B]=\frac{P[A \cap B]}{P}\).

In this case we're given: \(\displaystyle P[A|B']=\frac{P[A \cap B']}{P[B']}=\frac{1}{2}\).

So if you can explain my above question I'll try to help more. :)