# Homework Help: Marble Probability

1. Jan 30, 2016

### Michele Nunes

1. The problem statement, all variables and given/known data
A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?
a) 1/10
b) 1/6
c) 1/5
d) 1/3
e) 1/2
2. Relevant equations

3. The attempt at a solution
I really just don't understand probability. Here is the solution to the problem: "If Cheryl gets 2 marbles of the same color, then Claudia and Carol must take all 4 marbles of the 2 other colors. The probability of this happening, given that Cheryl has 2 marbles of a certian color is . Since there are 3 different colors, our final probability is ." I don't understand how multiplying all of those fractions gives the probability of Claudia and Carol taking 4 marbles, I just really don't understand where all those fractions came from. If Cheryl gets 2 marbles of the same color, then wouldn't that only leave 3 other possible combinations for Claudia and Carol to choose from? This just makes no sense to me.

2. Jan 30, 2016

### Samy_A

Let's for the moment just say that Cheryl will in the end get the two yellow marbles.
Carol picks the first marble. We don't want her to pick a yellow marble, but she can pick one of the two red marbles, or one of the two green marbles.
What are the odds that the first marble picked by Carol is not yellow?

3. Jan 30, 2016

### Michele Nunes

There are 6 equally likely possibilities and Carol can pick any 1 out of the 4 non-yellow marbles, so 4/6

4. Jan 30, 2016

### Samy_A

Exactly. That's where the 4/6 in the expression 4/6*3/5*2/4*1/3 comes from.
Now Carol has to pick a second marble. We still don't want her to pick a yellow one. What are the odds that the next marble picked by Carol is not yellow?

5. Jan 30, 2016

### Michele Nunes

Oh okay I understand where the fractions came from now. But why exactly do you multiply them?

6. Jan 30, 2016

### geoffrey159

Assume that your marbles are artificially distinguishable and numbered : 1,2 for the two red marbles, 3,4 for the two green marbles, and 5,6 for the two yellow marbles.

Then the set of all possible output for your random experiment can be modeled by $\Omega = \{ \omega = (\omega_1,...,\omega_6) \text{ is a permutation of } [1..6] \}$ where each $\omega\in \Omega$ is a sequence representing the 2 draws of Carol, then the 2 draws of Claudia, and the 2 draws of Cheryl.

Each of these output are equally likely to happen.

Now you are interested in the event $A = \{ \omega\in \Omega :\ (w_5,w_6) \in \{ (1,2), (2,1), (3,4), (4,3), (5,6), (6,5) \} \}$

The probability of such event is $P(A)= |A| / |\Omega|$

Now what is $|A|$ and $|\Omega|$ ?

7. Jan 30, 2016

### Samy_A

The general rule is: to find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities.
In Carol's case, the odds are 4/6 that the first marble will not be yellow, and 3/5 that the second marble will not be yellow.
Said differently, 4 times in 6 we will be glad with Carol's choice. Of these 4 times, what are the odds that the second marble will not be yellow either? Well: 3/5.
So in total, for Carol, the odds for a good outcome are first 4/6, and then 3/5 of these .That's precisely 4/6*3/5.

8. Jan 30, 2016

### Ray Vickson

IF Cheryl went first, her first ball would be some color, then for a match her second ball would need to be the same color as the first. After the first draw there are 5 marbles left, and only one has the correct color, so her match probability would be 1/5.

OK, that is what would happen if Cheryl went first. However, would you believe, the same probability applies if she goes second or goes third? In other words, as long as Cheryl does not know anything about the colors drawn by Claudia and Carol, her probability of drawing a match is still 1/5, even there are only 2 balls left. It is true!

This is hard to swallow, I know, so let's look at it from a sample-space point of view. Numbering the balls from 1 to 6, every permutation of the numbers from 1 to 6 is equally likely. In the actual experiment (where Cheryl is last) we are looking at the last two places in a permutation of the numbers 1,2,3,4,5,6, and the probability her colors match is the number of permutations where the colors match in the last two places of the permutation, divided by 6! (the total number of permutations altogether). However, the exact same argument applies if we are looking at Carol's draws (first two places in the permutation) or Claudia's draws (positions 3 and 4 in the permutation).

9. Jan 31, 2016

### HallsofIvy

Since we do not know what marbles Carol and Claudia get, they have no affect on the "a-priori" probability. (That is what Ray Vickson is saying in his last paragraph.) This is exactly the same as "There are 6 marbles, 2 red, 2 green, and 2 yellow. If Cheryl takes two marbles, what is the probability she gets two of the same color? What ever color Cheryl gets first, there remain 5 marbles, one of the same color she got the first time, 4 of other colors. The probability she gets the same color the second time is 1/5 so the probability of getting "two of the same color" is 1/5.

10. Jan 31, 2016

### geoffrey159

The trick is to see that there is a surjection that sends exactly six permutations of $\Omega$ to exactly one output $\omega' \in \Omega'$ of the real random experiment where marbles are undistinguishable inside the same color.
The same remark applies between $A$ and $A'$ (event where Cheryl picks two identical colors and the marbles are undistinguishable inside the same color).

This is to say $|\Omega'| = |\Omega|/6$ and $|A'| = |A|/6$

Since $(\Omega', {\cal P}(\Omega') )$ is fitted with the uniform probability (each output being equally likely), the probability of event $A'$ is $|A'|/|\Omega'| = |A|/|\Omega|$, which is also the probability of event $A$ in $(\Omega, {\cal P}(\Omega))$ fitted with the uniform probability.

As a conclusion, you would find exactly the same result by artificially allowing a distinction between marbles.

11. Jan 31, 2016

Staff Emeritus
Read this again. And again and again and again. This is an incredibly useful fact.

12. Jan 31, 2016

### geoffrey159

Is this message for me ?