# Three urns with two types of balls problem

I've got the following homework question:

Consider 3 urns. Urn A contains 2 white and 4 black balls; urn B contains 8 white and 4 black balls; and urn C contains 1 white and 3 black balls.

1. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white, given that exactly 2 white balls were selected?

2. Let X be the set of all balls which have been selected. Now we randomly select one ball from each urn (without returning the balls in X). What is the probability that Y contains exactly 2 white balls, given that X contains exactly one white ball?

I managed to solve 1 using $P(A | B)=\frac{P(A\cap B )}{P(B)}$ , but got stuck when I was trying to solve 2 in a similar way - I couldn't figure out how to calculate the probability of the intersection of the two events.

Any help will be greatly appreciated.

• I've interpreted the second question to be a new experiment... Nov 15, 2015 at 17:06

Let $F_{i,j,k}$ denote that the result of the first experiment is $i$ white balls from the first box, $j$ white balls from the second box, and $k$ white balls from the third box. Let $S_{i,j,k}$ denote the same in the case of the second experiment. ($i,j,k\in\{0,1\}$)

The task is to calculate the following conditional probability

$$P(S_{0,1,1}\lor S_{1,0,1}\lor S_{1,1,0}\mid F_{0,0,1}\lor F_{0,1,0} \lor F_{1,0,0}).$$

W need to calculate the following probabilities:

$$P((S_{0,1,1}\lor S_{1,0,1}\lor S_{1,1,0})\land (F_{0,0,1}\lor F_{0,1,0} \lor F_{1,0,0})) \tag 1$$

and $$P(F_{0,0,1}\lor F_{0,1,0} \lor F_{1,0,0}).\tag 2$$

For $(1)$:

$$(S_{0,1,1}\lor S_{1,0,1}\lor S_{1,1,0})\land (F_{0,0,1}\lor F_{0,1,0} \lor F_{1,0,0})=$$

$$=(S_{0,1,1}\land F_{0,0,1}) \lor (S_{0,1,1} \land F_{0,1,0}) \lor (S_{0,1,1} \land F_{1,0,0})\lor$$ $$\lor(S_{1,0,1}\land F_{0,0,1})\lor (S_{1,0,1}\land F_{0,1,0})\lor (S_{1,0,1}\land F_{1,0,0})\lor$$ $$\lor (S_{1,1,0}\land F_{0,0,1})\lor(S_{1,1,0}\land F_{0,1,0})\lor(S_{1,1,0}\land F_{1,0,0}).$$

Since the events in the form of $S_{i,k,l}\land F_{u,v,z}$ are mutually excluding, we can add the corresponding probabilities. Then for example

$$P(S_{1,0,1}\land F_{1,0,0})=P(S_{1,0,1}\mid F_{1,0,0})P(F_{1,0,0}).$$

Here the probability that we select one white ball from the first box and one black ball from the second box and one white ball from the third box given that we already removed one white ball from the first box and one black ball from the second box and one black ball from the third box is easy to calculate:

$$P(F_{1,0,0})=\frac26\frac4{12}\frac34$$ and $$P(S_{1,0,1}\mid F_{1,0,0})=\frac15\frac3{11}\frac13.$$

So,

$$P(S_{1,0,1}\land F_{1,0,0})=\frac26\frac4{12}\frac34\frac15\frac3{11}\frac13.$$ For $(2)$ it is enough to note again that the events in question are mutually excluding. So,

$$P(F_{0,0,1}\lor F_{0,1,0} \lor F_{1,0,0})=P(F_{0,0,1})+P(F_{0,1,0})+P(F_{1,0,0})=$$ $$=\frac46\frac4{12}\frac14+\frac46\frac8{12}\frac34+\frac26\frac4{12}\frac34.$$ I hope that it is easy to go on.