Thread: Help me find P(A) and P(B)

Hey all!

GIVEN:
P(A|B) = 0.8
P(B|A) = 0.5
P(AuB) = 0.9

The question:
Find P(A) and P(B).

My attempt:
I've done many questions like this, however they have some substantial data in them and it's easy to get the general feel on how to solve them. This question however I cannot solve at all.

My biggest problem is that I do not understand what to do with P(AuB) if I dont know weather P(A) and P(B) are independent events; P(AuB) = P(A)*P(B). How do I solve this without that assumption? I have no laws about it in my book and the three I do have just gets me in a circle and back to nothing. Please help!

Use the definition of a conditional probability.

P(A|B) = P(AnB)/P(B) ? Then I dont have P(AnB).. and P(AnB) = P(A) + P(B) - P(AuB). Trying many ways I still have three unknowns and two equations. Is there something i've missed?

Originally Posted by Hanga

P(A|B) = P(AnB)/P(B) ? Then I dont have P(AnB).. and P(AnB) = P(A) + P(B) - P(AuB). Trying many ways I still have three unknowns and two equations. Is there something i've missed?

I agree that something is missing here.
Are you sure that you were asked to find
Could you have been asked to find bounds on those values?

Why not double check the wording?

modified, sorry I made a mistake in it, my apologies.
I agree with Plato.

The exact wordings are:

For two events A and B we have P(A|B) = 0.8, P(B|A) = 0.5, P(AuB)=0.9
Decide P(A) and P(B).

I'm still trying to figure this out but no dice

Originally Posted by Hanga

The exact wordings are:

For two events A and B we have P(A|B) = 0.8, P(B|A) = 0.5, P(AuB)=0.9
Decide P(A) and P(B).

I'm still trying to figure this out but no dice

Dice are irrelevant here.

Pr(A) = 0.8 Pr(B) .... (1)

Pr(B) = 0.5 Pr(A) .... (2)

Solve equations (1) and (2) simultaneously for Pr(A) and Pr(B) etc.

My attempt (sorry for my previous post):


And whereas and therefore:


If we enter the given information we get:


We know also this formula:

If we enter the given information we get:



and so

Originally Posted by mr fantastic

Dice are irrelevant here.

Pr(A) = 0.8 Pr(B) .... (1)

Pr(B) = 0.5 Pr(A) .... (2)

Solve equations (1) and (2) simultaneously for Pr(A) and Pr(B) etc.

Errr...hemm..

Pr(AnB)=Pr(A|B)Pr(B)=Pr(B|A)Pr(A)

which means you have enough information to determine the ratio of Pr(A) and Pr(B). So Pr(B)=(0.5/0.8)Pr(A)

0.5=Pr(AuB)=Pr(A)+Pr(B)-Pr(AnB)=Pr(A)(1+5/8)-Pr(B|A)Pr(A)

etc

I just could not clear my brain enough to see through this problem.
But now I think it is real interesting.

Given then

Thank you very much guys! I'm going to study the method today and make sure I understand it perfectly, I belive the key was P(A|B)P(B)=P(B|A)P(A) and some smart moves on your part!
Have a nice day!

Originally Posted by CaptainBlack

Errr...hemm..

Pr(AnB)=Pr(A|B)Pr(B)=Pr(B|A)Pr(A)

which means you have enough information to determine the ratio of Pr(A) and Pr(B). So Pr(B)=(0.5/0.8)Pr(A)

0.5=Pr(AuB)=Pr(A)+Pr(B)-Pr(AnB)=Pr(A)(1+5/8)-Pr(B|A)Pr(A)

etc

Yes, I'll have to learn to post only when I'm awake.

i too got the same answer....