This week’s question, Don’t Lose Your Marbles, garnered a lot of responses, almost every single one of them correct. That’s awesome! What’s even more awesome is that at least five distinct approaches were taken across the various responses I got.

## Question

A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbles, and 5 magenta marbles. First, four marbles are removed at random, without replacement. Next, three more marbles are removed at random, without replacement. What is the probability that the remaining marbles are the same color?

(A) 1/7

(B) 1/8

(C) 1/120

(D) 7/120

(E) 11/120

## Answer and Explanation

Below, I’m going to present a very elegant—meaning the most straightforward—solution. After that, I’ll show another way that might even be more elegant (you be the judge!).

So after all the marbles are removed from the pouch (seven total), there are three remaining. How many different possibilities are there amongst those remaining? Well, there are ten total marbles and there are three “positions” so to speak. So we can use the combinations formula, because the other of the marbles doesn’t actually matter. That gives us 10C3 = 120.

The question is asking for all the possible combinations in which all three marbles are the same color. That means, three of the 5 magenta marbles or all 3 of the cerulean marbles would have to be in the pouch. There is only 1 possible way that occurs with the cerulean marbles. Yet, there are 5C3 ways in which the magenta marbles could end up there, since you could choose any three of the 5 magenta marbles to be the ones left in the pouch. 5C3 = 10. Adding that to the 1 possible set up of cerulean marbles, we get a total of 11, answer (E).

Now, another cool way of doing it is to realize that the question is just the reverse of asking what the probability of drawing three marbles of the same color from the pouch would be. It doesn’t matter, mathematically, whether the three marbles of the same color are the first out of the pouch or the last in the pouch. So what’s the probability that we pick all three magenta marbles: 5/10 x 4/9 x 3/8 = 1/12, or 1/120. Drawing all cerulean marbles give us the following: 3/10 x 2/9 x 1/8 = 1/120. Adding these together gives us 1/120 + 10/120 = 11/120. Therefore, there are eleven possible ways to choose three marbles of the same color, or end up with three marbles of the same color.

Hopefully, next week’s Brain Twister will warp the synapses a little more.

Since the marbles are being taken out in 2 steps, don’t we need to consider the different ways in which these are removed in those 2 steps.

For eg, when 3 magenta marbles are left in the end. We could have one where in first step we removed 2 AM and 2 CM and then in the second step 1 CM and 2 MM. OR in the first step 1 AM and 3 CM and in the second step 1 AM and 2 MM. There could be 40 such cases.

AM- alabaster marbles

CM- cerulean marbles

Hi Yash,

By focusing on the number of arrangements of the magenta marbles, i.e., the remaining marbles, it doesn’t matter what color any of the other marbles are. As long as there are n marbles that are not the three magenta marbles, that’s all that matters. That way, you can use the combinations formula. How many ways you can choose 3 marbles from a total of 10 is all that matters in finding the denominator. To find the numerator, you just have to find how many ways there are of choosing 3 magentas from a total of 5 magentas.

Hope that clears things up 🙂