# Conditional Probability and Marbles in different bags

There is a white marble in one bag and a red marble in another bag. You have a white marble in your hand and put it in one of the bags. You draw a white marble from one of the bags. What is the probability there is a white marble in the other bag?

This isn't a simple tree diagram example because we are dealing with separate events: Two different marble bags. I'm stuck.

It should not be hard to construct a probability tree. At each step of the process you have either one or two things that can happen. Just draw the branches of the tree methodically starting at the first step. You could get something like this: I leave it as an exercise to assign the probabilities to each node of the tree. Once you have done so, you merely need to find which nodes represent events where you draw a white ball, and of those events, in which cases was there a white ball in the other bag.

• Yes, Thanks David. This works well, I just couldn't visualize it to draw it. I got a probability of 1/2 as the answer following the tree. – Jessi Mar 15 at 16:44
• Better check the probabilities at each node, because it should not be possible to select a subset of "W" outcomes from the bottom row whose probabilities add up to $1/2$ times the probability of all "W" outcomes added together. – David K Mar 15 at 19:09

To avoid triviality, I assume we "forget" where we put the white marble, as well as which bags contained which marbles to begin with. In any case, we can use Bayes's Theorem:

$$P(H\,|\,E)=\frac{P(E\,|\,H)P(H)}{P(E)}$$

Our evidence, $$E$$, is that we drew a white marble from our bag; the hypothesis, $$H$$, is that a white marble is in the other bag.

Exactly one of the following is true of our bag before our draw: it has one white marble; it has one red marble; it has two white marbles; it has one white and one red marble. We favor none of these options. The hypothesis rules out only the third case, so we ought to assign it a prior probability of $$P(H)=\frac{3}{4}$$. The prior probability for our evidence should be

$$P(E) = \frac{1}{4} (1) + \frac{1}{4} (0) + \frac{1}{4} (1) + \frac{1}{4} \left ( \frac{1}{2} \right) = \frac{5}{8},$$

and the likelihood of our evidence given the hypothesis should be

$$P(E \, | \, H) = \frac{1}{3} (1) + \frac{1}{3} (0) + \frac{1}{3} \left ( \frac{1}{2} \right) = \frac{1}{2}$$,

where we remember that the hypothesis rules out only the third case. So

$$P(H\,|\,E)=\frac{3}{5}$$.

To draw a white marble either:

• $$A:$$ Both white marbles were placed in the same bag and you drew from that.
• $$B:$$ The white marbles were placed in different bags, and you drew from the bag with one marble.
• $$C:$$ The white marbles were placed in different bags, you drew from the bag with two marbles, and drew the white one.

There will not be a white marble in the other bag if event $$A$$ happened, only in the event of $$B$$ or $$C$$.

Therefore your task is to determine: $$\mathsf P(B\cup C\mid A\cup B\cup C)$$. Clearly by definition this is:

$$\mathsf P(B\cup C\mid A\cup B\cup C)=\dfrac{\mathsf P(B)+\mathsf P(C)}{\mathsf P(A)+\mathsf P(B)+\mathsf P(C)}$$

So just evaluate the prior probabilities for these events and calculate.

Alternative method: In essence you have four marbles, two white, one red, and one chameleon which mimics the colour of the marble its placed with. You place a white marble in one bag, and the red in the other, then randomly put a second marble in each bag.

So you now have one bag with two white marbles and, with equal probability, the other bag either contains two red marbles, or one marble of each colour.

That is $$\require{cancel}\sf WW:R\cancelto RC$$ or $$\sf W\cancelto WC:RW$$.

Thus there are so many equally probable ways to draw a white marble, and of these some have a white marble in the other bag.