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

Latest revision as of 11:21, 11 June 2022

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?

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

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

~A genius ofc

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 $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\textbf{(D) }\dfrac35}$

Solution 3

The amount of ways to end with a white chip is by having $RRWW, RWW,$ and $WW$ . The amount of arrangements for $RRWW$ with $W$ at the end is $3$ , the number of arrangements of $RWW$ with $W$ at the end is $2$ , and the number of arrangements with $WW$ is just $1$ . This gives us $6$ total ways to end with white. Next, the cases to end with a red are $RWRR$ , and $RRR$ . $RWRR$ gives us $3$ ways and $RRR$ gives us $1$ way. So the number of ways to end with a red is $4$ . Thus, our answer is simply $\frac{6}{4+6}$ = $\boxed{\textbf{(D) }\dfrac35}$

Video Solution

https://youtu.be/zEvu7BTHdXg

https://youtu.be/QIqdhX9Ou-k

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 <math>
-\text{(A) }\frac {3}{10}
+\textbf{(A) }\frac {3}{10}
 \qquad
-\text{(B) }\frac {2}{5}
+\textbf{(B) }\frac {2}{5}
 \qquad
-\text{(C) }\frac {1}{2}
+\textbf{(C) }\frac {1}{2}
 \qquad
-\text{(D) }\frac {3}{5}
+\textbf{(D) }\frac {3}{5}
 \qquad
-\text{(E) }\frac {7}{10}
+\textbf{(E) }\frac {7}{10}
 </math>
-== Solution ==
+== 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 in white) is <math>\boxed{(\text{D})\frac {3}{5}}</math>.
+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 <math>3</math> 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 <math>\boxed{\textbf{(D) } \frac {3}{5}}</math>.
+~A genius ofc
+==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: <math>W, W, R, R, R</math> such that both <math>W</math>'s appear in the first <math>4</math>. We find the number of ways to arrange the white chips in the first <math>4</math> 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{\textbf{(D) }\dfrac35}</math>
+==Solution 3==
+The amount of ways to end with a white chip is by having <math>RRWW, RWW,</math> and <math>WW</math>. The amount of arrangements for <math>RRWW</math> with <math>W</math> at the end is <math>3</math>, the number of arrangements of <math>RWW</math> with <math>W</math> at the end is <math>2</math>, and the number of arrangements with <math>WW</math> is just <math>1</math>. This gives us <math>6</math> total ways to end with white. Next, the cases to end with a red are <math>RWRR</math>, and <math>RRR</math>. <math>RWRR</math> gives us <math>3</math> ways and <math>RRR</math> gives us <math>1</math> way. So the number of ways to end with a red is <math>4</math>. Thus, our answer is simply <math>\frac{6}{4+6}</math> = <math>\boxed{\textbf{(D) }\dfrac35}</math>
+==Video Solution==
+https://youtu.be/zEvu7BTHdXg
+https://youtu.be/QIqdhX9Ou-k
+~savannahsolver
 == See Also ==
+{{AMC10 box|year=2001|num-b=22|num-a=24}}
 {{AMC12 box|year=2001|num-b=10|num-a=12}}
-{{AMC10 box|year=2001|num-b=22|num-a=24}}
 {{MAA Notice}}

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