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

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Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: ''W, W, R, R, R'' such that both ''W's'' appear in the first 4. We find the number of ways to arrange the white chips in the first 4 and divide that by the total ways to choose all the chips. The probability of this occurring is <math>\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}</math>
 
Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: ''W, W, R, R, R'' such that both ''W's'' appear in the first 4. We find the number of ways to arrange the white chips in the first 4 and divide that by the total ways to choose all the chips. The probability of this occurring is <math>\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}</math>
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==Solution 3 ==
 
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If the last chip picked is white, then there are only three cases that could happen:
 
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Case 1: Only two red chips are picked.
 
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One arrangement can be <math>RRWW</math>, where the last <math>W</math> is set. There are only <math>3</math> ways to arrange this, and the probability of this happening is <math>\frac{3}{5}</math> <math>\cdot</math> <math>\frac{2}{4}</math> <math>\cdot</math> <math>\frac{2}{3}</math> <math>\cdot</math> <math>\frac{1}{2}</math> <math>=</math> <math>\frac{1}{10}</math>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 17:55, 14 May 2020

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 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 chips are picked. To satisfy this condition, we have to arrange the letters: W, W, R, R, R such that both W's appear in the first 4. We find the number of ways to arrange the white chips in the first 4 and divide that by the total ways to choose all the chips. The probability of this occurring is $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}$

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions
2001 AMC 12 (Problems • Answer Key • Resources)
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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