2001 AMC 10 Problem 23

P21-25~4 minPrint

Basic Probability Conditional Probability & States Map the objects to something easier to count Exploit symmetry to reduce work or pair up objects

A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?

Show solution without answering

Key insight

Imagine drawing all five chips. The whites run out first exactly when the last chip in the box is red, and each chip is equally likely to be last.

Solution

Stopping early makes the problem awkward, so change it: keep drawing until all five chips are out. The chips drawn before the stopping point are the same either way, so nothing about the question changes.
Try an example order: $R, W, R, W, R$ . The two whites are both out after the fourth chip, before the third red. So the drawing would have stopped on a white chip. Notice the chip left at the end is red.
Another example: $W, R, R, R, W$ . All three reds are out after the fourth chip, so the drawing stops on a red. The chip left at the end is white.
This always happens. The color that runs out first is the one that is not in the last spot, because the last chip's color still has a chip left until the very end.
So "the last chip drawn is white" means "the fifth chip in the full order is red."
Every one of the five chips is equally likely to end up in the fifth spot. Three of the five chips are red. So the chance is $\frac{3}{5}$ .

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

Why this works

A rule like "stop when one color runs out" hides a simpler question. Letting the drawing finish gives a random order of all five chips, and "which color runs out first" becomes "which color is at the end." Every chip has the same chance of being at the end, which makes the answer quick.

Alternative approach

List all the orders. Only the positions of the two whites matter, and there are $10$ ways to choose $2$ of the $5$ positions:

$12,\ 13,\ 14,\ 15,\ 23,\ 24,\ 25,\ 34,\ 35,\ 45$ .

The whites run out first when both are in the first four positions, meaning neither is in position $5$ . That leaves $12, 13, 14, 23, 24, 34$ : $6$ of the $10$ . The probability is $\frac{6}{10} = \frac{3}{5}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Intermediate Counting & Probability, Chapter 4: Constructive Counting and 1-1 Correspondences (§4.4, §4.5, §4.6)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)