2020 AMC 10B Problem 18

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 $\boxed{\textbf{(B)}\ \frac15}$.

This is a Polya urn. Its hidden symmetry is that the probability of a sequence depends only on how many of each color were drawn, not on the order: the numerators for each color run $1, 2, 3, \dots$ in turn and the denominators run $2, 3, 4, \dots$. So probability of a final composition $=$ (number of orderings) $\times$ (probability of one ordering). Whenever sequential draws change the urn, check whether the products are order-independent before drawing a full tree.

Track states $(R, B)$ with a tree. From $(1,1)$: $(2,1)$ or $(1,2)$ each with $\tfrac12$. Continue to the four-draw level and add the probabilities of paths ending at $(3,3)$; by symmetry compute the paths through $(2,1)$ and double. Paths through $(2,1)$: RRBB, RBRB, RBBR each $\tfrac{1}{30}$, sum $\tfrac1{10}$; doubled gives $\tfrac15$.

• Multiplying probabilities as if the draws were independent with fixed probability 1/2, or forgetting to count all 6 orderings.
• Computing the probability of a single ordering ($\frac{1}{30}$) or of the three orderings starting with red ($\frac1{10}$) and stopping there.