Difference between revisions of "2015 AMC 12A Problems/Problem 9"

(Solution 1)
 
(2 intermediate revisions by 2 users not shown)
Line 6: Line 6:
  
 
== 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 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>.
+
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} * \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} = \textbf{ (C)} \frac{1}{5} </math>.
  
 
==Solution 2==
 
==Solution 2==
Line 13: Line 13:
 
==Solution 3==
 
==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.
 
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}</math>
+
<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}}

Latest revision as of 01:55, 9 January 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?

$\textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2}$

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 $\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}$. Since there are three different colors, our final probability is $3 * \frac{1}{15} = \textbf{ (C)} \frac{1}{5}$.

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 $\frac{1}{5} \textbf{ (C)}$.

Solution 3

The total number of ways they can draw is ${6 \choose 2}$ ${4 \choose 2}$ ${2 \choose 2}$. Let Cheryl draw first and since there are three colors, there are ${3 \choose 1}$ ways she can get 2 marbles of the same color. The other two pick two each, which leads to ${4 \choose 2}$ and ${2 \choose 2}$, respectively. $\frac{{3 \choose 1}{4 \choose 2}{2 \choose 2}}{{6 \choose 2}{4 \choose 2}{2 \choose 2}} = \frac{1}{5} \textbf{ (C)}$

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS