2020 AMC 10B Problem 18

1. What must happen. The urn starts with $1$ red and $1$ blue, and each draw adds one ball of the drawn color. To finish with $3$ of each, George must draw red exactly twice and blue exactly twice, in some order.
2. One order: red, red, blue, blue. Follow the urn step by step.

The chance of this order is
$$\frac12 \times \frac23 \times \frac14 \times \frac25 = \frac{4}{120} = \frac{1}{30}.$$
3. Another order: red, blue, blue, red. The chances are $\frac12$, then $\frac13$, then $\frac24$, then $\frac25$. The product is again $\frac{1 \times 1 \times 2 \times 2}{2 \times 3 \times 4 \times 5} = \frac{1}{30}$.
4. Why every order gives $\frac{1}{30}$. The denominators (bottoms) are always $2, 3, 4, 5$, because the urn gains one ball per draw. The numerators (tops) are the number of balls of the color drawn: the first red draw has $1$ red to pick, the second has $2$, and the same for blue. So the tops are always $1, 2, 1, 2$ in some order.
5. Count the orders. Choose which $2$ of the $4$ draws are red: $\binom42 = 6$ orders (RRBB, RBRB, RBBR, BRRB, BRBR, BBRR).
6. Total.
$$6 \times \frac{1}{30} = \frac{1}{5}.$$

The answer is $\boxed{\textbf{(B)}\ \frac15}$.

In this kind of urn, the chance of a sequence depends only on how many of each color were drawn, not on the order. So the chance of an ending mix equals (number of orders) times (chance of one order). Whenever draws change what is in the urn, compare two orders first. If they give the same product, you can skip drawing the whole tree.

Draw a probability tree and add up the paths that end at $3$ red, $3$ blue. By symmetry, the paths that start with red contribute the same as those that start with blue. Starting with red, the paths RRBB, RBRB and RBBR each have chance $\frac{1}{30}$, for a total of $\frac{3}{30} = \frac{1}{10}$. Doubling gives $\frac15$.

Ending with three of each color means exactly two red draws and two blue draws in some order. Compute the probability of one specific order, say red, red, blue, blue:

• Draw 1: $1$ red of $2$ balls, probability $\tfrac12$; urn becomes $2R, 1B$.
• Draw 2: $\tfrac23$; urn becomes $3R, 1B$.
• Draw 3: blue, $\tfrac14$; urn becomes $3R, 2B$.
• Draw 4: blue, $\tfrac25$.

Product: $\tfrac12 \cdot \tfrac23 \cdot \tfrac14 \cdot \tfrac25 = \tfrac{4}{120} = \tfrac{1}{30}$.

Any other order gives the same product: the denominators are always $2, 3, 4, 5$ (the urn grows by one each time), and the numerators are the current counts of the drawn color, which are $1$ then $2$ for red and $1$ then $2$ for blue regardless of interleaving. So each of the $\binom42 = 6$ orders has probability $\tfrac1{30}$.

Total probability: $6 \cdot \tfrac{1}{30} = \tfrac15$.

The answer is $\boxed{\textbf{(B)}\ \frac15}$.