# Considering that the two selected balls are white, what is the probability that urn III was chosen?

Consider three numbered urns. Urn I contains 6 white balls and 2 red balls, urn II contains 2 white balls and 6 red balls, and urn III contains 4 white balls and 4 red balls. To choose an urn, roll a balanced die and consider: If faces 1, 2 appear, choose urn I, if faces 3, 4, 5 appear, choose urn II and if face 6 appear, choose urn III. Then 2 balls are chosen at random, one after the other, without replacement, from the selected urn. Considering that the two selected balls are white, what is the probability that urn III was chosen?

To solve this question I have used the Bayes Theorem for Conditional probability:

$$P(C_j|A)$$ = $${P(C_j)*P(A|C_j)} \over {\sum P(C_i)*P(A|C_i)}$$

So, since the first ball is white, let's calculate the probability of urn III.

$$P(UrnIII|White)$$ = $${P(UrnIII)*P(White|UrnIII)}\over{P(UrnI)*P(White|UrnI)+P(UrnII)*P(White|UrnII)+P(UrnIII)*P(White|UrnIII)}$$

P(UrnIII) = $$1\over 6$$
P(White|UrnIII) =$$4 \over 8$$
P(UrnI) = $$2\over 6$$
P(White|UrnI) = $$6\over 8$$
P(UrnII) = $$3 \over 6$$ P(White|UrnIII) = $$2 \over 8$$

Replacing those values in the equation we get:

$$P(UrnIII|White)$$ = $$2 \over 11$$

Now, I am supposed to to the same thing for the second ball. However, I am not quite sure of how I am going to evaluate considering that the first ball was not put back in the urn.

I thought about dividing the second part in cases. First, assume that the white ball was selected from the urnI. And recalculate all the probability in bayes theorem for the second ball. Then, I would consider the first white ball was selected from urnII. And lastly, consider the first white ball was selected from urnIII and calculate another probability.

However, this thought doesn't seem right to me, and I thought I could get some help on thoughts of how to solve this problem.

You can find $$P(\text{2 whites}|U_3)$$ combinatorially and directly as $$\frac{\binom42\binom40}{\binom82}=\frac6{28}$$ These probabilities are values of the hypergeometric distribution. Similarly the probabilities for $$U_1$$ and $$U_2$$ are $$\frac{15}{28}$$ and $$\frac1{28}$$ respectively. Then use total probability to get the answer as $$\frac{1/6\cdot6/28}{2/6\cdot15/28+3/6\cdot1/28+1/6\cdot6/28}=\frac2{13}$$