Difference between revisions of "2015 AMC 12A Problems/Problem 9"
Burunduchok (talk | contribs) m (→Problem) |
m (Fixed some LaTeX) |
||
| (5 intermediate revisions by 4 users not shown) | |||
| Line 3: | Line 3: | ||
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? | 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? | ||
| − | <math> \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D) | + | <math> \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2} </math> |
== Solution 1== | == Solution 1== | ||
| − | 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 | + | 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 certain color is <math>\frac{4}{6} \cdot \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{15}</math>. Since there are three different colors, our final probability is <math>3 \cdot \frac{1}{15} = \textbf{ (C)} \frac{1}{5} </math>. |
==Solution 2== | ==Solution 2== | ||
The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is <math>\frac{1}{5} \textbf{ (C)}</math>. | The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is <math>\frac{1}{5} \textbf{ (C)}</math>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | The total number of ways they can draw is <math>{6 \choose 2}</math> <math>{4 \choose 2}</math> <math>{2 \choose 2}</math>. Let Cheryl draw first and since there are three colors, there are <math>{3 \choose 1}</math> ways she can get 2 marbles of the same color. The other two pick two each, which leads to <math>{4 \choose 2}</math> and <math>{2 \choose 2}</math>, respectively. | ||
| + | <math>\frac{{3 \choose 1}{4 \choose 2}{2 \choose 2}}{{6 \choose 2}{4 \choose 2}{2 \choose 2}} = \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 12:30, 1 August 2021
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?
Solution 1
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 certain color is 
. Since there are three different colors, our final probability is 
.
Solution 2
The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is 
.
Solution 3
The total number of ways they can draw is 
. Let Cheryl draw first and since there are three colors, there are 
ways she can get 2 marbles of the same color. The other two pick two each, which leads to and , respectively.
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 | |
