2001 AMC 10 Problem 23

Suppose the drawing did not stop but continued until the box was empty. The chips already drawn before the stopping point are unaffected, so the color of the "last chip drawn" in the original process is determined by the full random order of all five chips.

In the full order, the process stops on a white chip exactly when the second white appears before the third red, that is, when both whites are among the first four positions. Equivalently, the fifth (final) position holds a red chip.

By symmetry, each of the five chips is equally likely to sit in the final position, and three of them are red. So the probability is $\frac{3}{5}$.

The answer is $\boxed{\textbf{(D)}\ \frac{3}{5}}$.

A stopping rule does not change the underlying random permutation, so extend the experiment to the full ordering, where positional symmetry is available. "Which color is used up first" is the same event as "which color occupies the last slot," and the last slot is uniform over the chips.

Count orderings of $RRRWW$: there are $\binom{5}{2} = 10$ placements of the two whites, and $\binom{4}{2} = 6$ of them put both whites in the first four slots. Probability $\frac{6}{10} = \frac{3}{5}$.

• Misreading which color runs out and answering 2/5, the probability that the final chip of the full sequence is white.
• Doing casework on when the second white appears (draw $2$, $3$, or $4$) and forgetting to multiply by the number of arrangements of the earlier reds.