# Difference between revisions of "2001 AMC 12 Problems/Problem 11"

The following problem is from both the 2001 AMC 12 #11 and 2001 AMC 10 #23, so both problems redirect to this page.

## Problem

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?

$\text{(A) }\frac {3}{10} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {1}{2} \qquad \text{(D) }\frac {3}{5} \qquad \text{(E) }\frac {7}{10}$

## Solution

Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is $\boxed{(\text{D}) \frac {3}{5}}$.

## Solution 2

We wish to arrange the letters: W,W, R, R, R such that R appears last. The probability of this occurring is simply $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}$

 2001 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions