# Conditional Probability and Urns in balls

1. Feb 2, 2012

### bitty

1. The problem statement, all variables and given/known data
You have 3 urns: Urn 1 has 3 red balls, Urn 2 has 2 red balls, 1 blue ball. Urn 3 has 2 red balls 2 blue balls. You pick a ball from each urn and place it into Urn 4.
You draw 2 balls from Urn 4 and they are red. What is the conditional probability that the 3rd ball is also red?

2. Relevant equations
Bayes law?

3. The attempt at a solution
The third ball can be from urn 1, 2, or 3. So I initially tried
P(3rd red given first 2 are red)=P(3rd ball is red and from Urn 1)+P(3rd ball is red and from Urn 2)+P(3rd ball is red and from Urn 3)
=1/3(1+2/3+1/2)=13/18.

But this doesn't seem right, I feel like I'm neglecting the conditions of the first 2.
what if I try P(3rd red given first 2 are red)=P(1st two balls being red from urn 1 and urn 2)*P(3rd ball red from urn 3)+P(1st two balls being red from urn 1 and urn 3)*P(3rd ball red from urn 2)+ P(1st two balls being red from urn 2 and urn 3)*P(3rd ball red from urn 1)

Is that formulation correct?

2. Feb 2, 2012

### Ray Vickson

By far the easiest (and most error-free) way is to get the probabilities of the various numbers of red in Urn 4, and to forget about which of Urns 1-3 they came from. Letting R = number red in Urn 4, we have R = 1, 2 or 3: we always have at least 1 red in U4 because all balls in U1 are red. {R=1} = {blue from U2 & blue from U3}, {R=2} = {red from U1 and blue from U2} or {blue from U1 & red from U2}, and {R=3} = {red from U2 & red from U3}. Once you know the values of P{R=1}, P{R=2} and P{R=3}, you can find P{R=3 | R >= 2}, which is what you want.

RGV

3. Feb 2, 2012

### bitty

Let's see:
P(R=1)=1*1/3*1/2=1/6
P(R=2)=1*2/3*1/2+1*1/3*1/2=1/2
P(R=3)=1*2/3*1/2=1/3
which add to 1 as desired.

Then, by Bayes' Rule: P(R=3|R>=2)= P(R=3)/(P(R=2)+P(R=3))=2/5

---My first question is: did I use Bayes rule correctly? It seems, correctly, that P(R=1) is irrelevant given the condition.

I checked this answer a different way: we have 3*3*4=36 possible outcomes of 3 balls.
Of these,3*2*2=12 have 3 red balls, 3*2*2+3*1*2=18 have 2red, 1 white, and 6 have 1 red 2 white.

Restricting ourselves to >=2 reds, we have 12/(12+18)=2/5, which agrees.

---My 2nd question is: we are findingP{R=3 | R >= 2} the probability that 3 balls are red, given at least 2 are. Is this the same as finding the probability that the 3rd ball, given 1st are 2 red?

Thanks

4. Feb 2, 2012

### Ray Vickson

OK, I see what you mean: the events {R >= 2} and {1st two red} could be different, because one of the possibilities in {R >= 2} is RBR. So *my second suggestion was incorrect*. Instead, compute P{1st red and 2nd red} as P{1st red and 2nd red|R=2}*P{R=2} + P{1st red & 2nd red | R=3}*P{R=3}. We know P{R=2} and P{R=3}. We have P{1st red & 2nd red | R=3} = 1 of course, and getting P{1st red & 2nd red | R=2} is not very hard. Note that here, 1st and 2nd refer to drawings from U4, not from U1--U3.

RGV