Difference between revisions of "2015 AMC 12A Problems/Problem 9"
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== Solution== | == Solution== | ||
| + | If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is <math>\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}</math>. Since there are three different colors, our final probability is <math>3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2015|ab=A|num-b=8|num-a=10}} | ||
Revision as of 20:58, 4 February 2015
Problem
A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?
$\textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac16 \qquad\textbf{(C)}\ \frac15 \qquad\textbf{(D)}}\ \frac13 \qquad\textbf{(E)}\ \frac12$ (Error compiling LaTeX. Unknown error_msg)
Solution
If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is 
. Since there are three different colors, our final probability is 
.
See Also
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
