Probability: 3 Urns, pulling out 3 balls

Question

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!

Answer 1

You got it somewhat right. But the desired probability you are looking for is:

$\frac{P(\text{white from A,B})+P(\text{white from A,C})}{P(\text{white from A,B})+P(\text{white from A,C})+ P(\text{white from B,C})}$

This is a conditional probability question. You need to find out the Probability of "ball from Urn A being white" GIVEN "you get two white balls". Do you know the formula for conditional probability?

So define two events E and F such that E is the event that the ball chosen from Urn A is white and F is the event that 2 out 3 balls are white.

What you need to do is calculate $P(E|F)$ which is probability of E given F and is equal to Probability of both E and F happening at the same time divided by Probability of F, i.e. $P(E|F)=\frac{P(E\cap F)}{P(F)}$

It's your numerator that you need to work on. Remember, for the numerator it can't just be the probability of getting a white ball from Urn A. It needs to be the probability of the INTERSECTION of the given event F and the event E. So the numerator becomes the probability of getting a white ball from urn A AND getting a 2nd white ball from any other urn.

In conclusion:

$\frac{P(\text{white from A,B})+P(\text{white from A,C})}{P(\text{white from A,B})+P(\text{white from A,C})+ P(\text{white from B,C})} = \frac{\frac{7}{36}}{\frac{11}{36}} = \frac{7}{11}$

Answer 2

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: $$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{7/36}{11/36} = \frac{7}{11}$$