# Help me find P(A) and P(B)

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• Aug 22nd 2011, 11:01 AM
Hanga
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!
• Aug 22nd 2011, 11:21 AM
Siron
Re: Help me find P(A) and P(B)
Use the definition of a conditional probability.
• Aug 22nd 2011, 11:32 AM
Hanga
Re: Help me find P(A) and P(B)
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? :)
• Aug 22nd 2011, 11:45 AM
Plato
Re: Help me find P(A) and P(B)
Quote:

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 $\displaystyle P(A)~\&~P(B)~?$
Could you have been asked to find bounds on those values?

Why not double check the wording?
• Aug 22nd 2011, 11:52 AM
Siron
Re: Help me find P(A) and P(B)
modified, sorry I made a mistake in it, my apologies.
I agree with Plato.
• Aug 22nd 2011, 12:03 PM
Hanga
Re: Help me find P(A) and P(B)
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 :(
• Aug 22nd 2011, 12:39 PM
mr fantastic
Re: Help me find P(A) and P(B)
Quote:

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.
• Aug 22nd 2011, 12:50 PM
Siron
Re: Help me find P(A) and P(B)
My attempt (sorry for my previous post):
$\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$
$\displaystyle P(B|A)=\frac{P(B\cap A)}{P(A)}$
And $\displaystyle P(A|B)\neq P(B|A)$ whereas $\displaystyle P(A\cap B)=P(B\cap A)$ and therefore:
$\displaystyle P(A|B)\cdot P(B)=P(B|A)\cdot P(A)$

If we enter the given information we get:
$\displaystyle 0,8P(B)=0,5P(A) \Leftrightarrow P(A)=\frac{8}{5}P(B) \Leftrightarrow P(B)=\frac{5}{8}P(A)$

We know also this formula:
$\displaystyle P(A\cap B)=P(A)+P(B)-P(A\cup B)$
If we enter the given information we get:
$\displaystyle P(A|B)\cdot P(B)= P(A)+P(B) - P(A\cup B)$
$\displaystyle 0,8\cdot \frac{5}{8}P(A)=P(A)+\frac{5}{8}P(A)-0,9$
$\displaystyle P(A)=\frac{4}{5}$
and so $\displaystyle P(B)=\frac{1}{2}$
• Aug 22nd 2011, 01:56 PM
CaptainBlack
Re: Help me find P(A) and P(B)
Quote:

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
• Aug 22nd 2011, 03:57 PM
Plato
Re: Help me find P(A) and P(B)
I just could not clear my brain enough to see through this problem.
But now I think it is real interesting.

Given $\displaystyle P(A|B),~P(B|A)~\&~P(A\cup B)$ then

\displaystyle \begin{align*}P(A)&=\frac{P(A|B)\cdot P(A\cup B)}{P(A|B)+P(B|A)-P(A|B)\cdot P(B|A)}\\ P(B) &=\frac{P(B|A)\cdot P(A\cup B)}{P(A|B)+P(B|A)-P(A|B)\cdot P(B|A)} \end{align*}
• Aug 22nd 2011, 10:05 PM
Hanga
Re: Help me find P(A) and P(B)
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!
• Aug 23rd 2011, 01:05 AM
mr fantastic
Re: Help me find P(A) and P(B)
Quote:

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.
• Aug 24th 2011, 08:45 PM
MAX09
Re: Help me find P(A) and P(B)
i too got the same answer....