2001 AMC 12 Problems/Problem 11

Revision as of 23:49, 2 February 2019 by Chenr28 (talk | contribs) (Solution 1)
The following problem is from both the 2001 AMC 12 #11 and 2001 AMC 10 #23, so both problems redirect to this page.


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 1

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

Let's assume we don't stop picking until all of the balls are picked. To satisfy this condition, we have to arrange the letters: W, W, R, R, R such that both R's appear in the first 4. We find the number of ways to arrange the red balls in the first 4 and divide that by the total ways to chose all the balls. The probability of this occurring is $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}$

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 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

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