- #1
bitty
- 14
- 0
Homework Statement
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?
Homework Equations
Bayes law?
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?