Question: There are 3 urns: A, B and C.
Urn A has 2 white balls and 4 red balls
Urn B has 8 white balls and 4 red balls
Urn C has 1 white ball and 3 red balls
We pull one ball from each urn, and we get that 2 of them are white.
What is the probability that the ball we pulled from urn A is white?
This is my approach:
Given 2/3 of the balls are white, what is the probability we pulled a white ball from A:
$$\frac{P( \text{white from A})}{P(\text{white from A,B})+P(\text{white from A,C})+ P(\text{white from B,C})}$$
$=$
$$\frac{\frac{1}{3}}{(\frac{1}{3} \cdot \frac{8}{12} \cdot \frac{3}{4}) + (\frac{1}{3} \cdot \frac{4}{12} \cdot \frac{1}{4}) + (\frac{4}{6} \cdot \frac{8}{12} \cdot \frac{1}{4})} ~~~~~ (\geq 1)$$
However what I get is a number bigger than $1$ .. where am I wrong with this logic? If it is given $2/3$ balls are white then we need to calculate all the different ways we can get 2 white balls out of 3 urns.. that is: white from A and B OR white from A and C OR white from B and C..
I would appreciate your help!
Thank you!
Best Answer
Let $P(E)$ be the probability that the ball we pulled from urn A is white. There are $6 \cdot 12 \cdot 4 = 288$ possible combinations of balls drawn in our sample space.
If $P(F)$ is the probability that exactly two balls are chosen from the sample space then: $$P(F) = \frac{\text{Number of combinations of exactly 2 white balls}}{288}$$ $$= \frac{(2 \cdot 8 \cdot 3) + (2 \cdot 4 \cdot 1) + (4 \cdot 8 \cdot 1)}{288} = \frac{88}{288} = \frac{11}{36}$$ If $P(E \cap F)$ is the probability that the ball chosen from Urn A is white then: $$P(E \cap F) = \frac{\text{Number of combinations of F where the ball from Urn A is white}}{288}$$ $$= \frac{(2 \cdot 8 \cdot 3) + (2 \cdot 4 \cdot 1)}{288} = \frac{56}{288} = \frac{7}{36}$$
Then $P(E \mid F)$ probability that one of the 2 white balls we pulled is from urn A: $$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{7/36}{11/36} = \frac{7}{11}$$