Riddler Express: June 25, 2021

David Ding

June 27, 2021

Counting Marbles

Explanation:

Without loss of generality (WLOG), let us denote the three marble colors as red, green, and blue, and let \(r, g, b\) be the number of marbles of those colors, respectively. The first step in solving the problem is finding an expression for the probability of drawing three marbles, each of one distinct color, from the bag. Here we are assuming we are pulling out all three marbles at once, such that there will remain 97 marbles in the bag.

In combinatoric problems like these, it's often more straightforward to solve an easier problem, and then apply generalizations. Here, we will assume that the order of the colors matter. So, WLOG, let us compute the probability of drawing out three marbles in the following order: red-green-blue. In this case, since we are assuming the marbles are not placed back in the bag after each draw, the probability of drawing the red marble would be \(\frac{r}{100}\), then followed by green: \(\frac{g}{99}\), and finally blue: \(\frac{b}{98}\). Assuming the marbles are drawn independently of each other, the probability of the red-green-blue draw would be \(\frac{rgb}{(100)(99)(98)}\).

Now let's generalize. Clearly the order of the colors do not matter, as we are only interested in drawing three marbles each of a distinct color. Therefore, something like red-blue-green would also be valid, and if we compute that probability, we would also get \(\frac{rgb}{(100)(99)(98)}\). Clearly, there are \(3! = 6\) permutations of the drawing sequence that would lead us to the same desired result, and so the probability of drawing three marbles each of a distinct color without replacement is:

\begin{equation} 3! \frac{rgb}{(100)(99)(98)} \end{equation}

Now we just need to equate the above expression to the given value of 0.2:

\begin{align} 3! \frac{rgb}{(100)(99)(98)} &= 0.2 \\ rgb &= 32340 \\ &= 2^2 \times 3 \times 5 \times 7^2 \times 11 \end{align}

We will soon see how prime factoring 32340 will be helpful ;).

Now let us gather what we know:

\begin{align} rgb &= 32340 \\ r + g + b &= 100 \end{align}

At first glance, it seems that we cannot uniquely determine the values for \(r, g, b\), if they have solutions in the first place, because we have two equations and three unknowns. However, we must not overlook further conditions to those variables, namely they are all integers strictly between 1 and 98. They must be non-negative because they represent counts of marbles, and none of them can be zero otherwise the product \(rgb\) would equal to zero. This implies that we can assemble the prime factors of 32340 and keeping in mind the very restrictive integer value conditions for \(r, g, b\) and check if any of the trio of values add up to 100.

This is actually easier done than one would imagine, because the possibilities are very limited. For one, we cannot have 7 and 11 together as the product, 77, would take up too much "real estate" in the build-up to 100, as we still have two factors of 2, a 3, a 5, and another factor of 7 left. The trick is to even out the factors as much as possible. We can achieve this by pairing factors from the lower end of the real number line with those from the higher ends. For example, let us pair \(2^2\) with 11, then 3 with one copy of 7, and finally 5 with the other copy of 7, and get:

• Color 1: \(2^2 \times 11 = 44\)
• Color 2: \(3 \times 7 = 21\)
• Color 3: \(5 \times 7 = 35\)

Total: \(44 + 21 + 35 = 100\).

And we are done! We can check that \(3! \frac{(44)(21)(35)}{(100)(99)(98)} \) indeed equals to 0.2, as desired.