Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AMC 8 Problems/Problem 21"

(→‎Video Solution)
(→‎Solution 2)
Line 18: Line 18:
  
 
==Solution 2==
 
==Solution 2==
−
Notice that, no matter which card you choose, there are exactly 4 cards that either have the same color or letter as it. Since there are 7 cards left to choose from, the probability is <math>\frac{4}{7}</math>. theepiccarrot7
+
Notice that, no matter which card you choose, there are exactly 4 cards that either has the same color or letter as it. Since there are 7 cards left to choose from, the probability is <math>\frac{4}{7}</math>. theepiccarrot7
 +
 
 +
==Solution 3==
 +
We can use casework to solve this.
 +
 
 +
Case <math>1</math>: Same letter
 +
After choosing any letter, there are seven cards left, and only one of them will produce a winning pair. Therefore, the probability is <math>\frac17</math>.
 +
 
 +
Case <math>2</math>: Same color
 +
After choosing any letter, there are seven cards left. Three of them will make a winning pair, so the probability is <math>\frac37</math>.
 +
 
 +
Now that we have the probability for both cases, we can add them: <math>\frac17+\frac37=\boxed{\textbf{(D)} \frac47}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=20|num-a=22}}
 
{{AMC8 box|year=2007|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:30, 16 January 2023

Problem

Two cards are dealt from a deck of four red cards labeled $A$, $B$, $C$, $D$ and four green cards labeled $A$, $B$, $C$, $D$. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

$\textbf{(A)}\ \frac{2}{7}\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{4}{7}\qquad\textbf{(E)}\ \frac{5}{8}$

Video Solution by OmegaLearn

https://youtu.be/OOdK-nOzaII?t=1712

~ pi_is_3.14

Video Solution

https://youtu.be/OOdK-nOzaII?t=1698

Solution 1

There are 4 ways of choosing a winning pair of the same letter, and $2 \left( \dbinom{4}{2} \right) = 12$ ways to choose a pair of the same color.

There's a total of $\dbinom{8}{2} = 28$ ways to choose a pair, so the probability is $\dfrac{4+12}{28} = \boxed{\textbf{(D)}\ \frac{4}{7}}$.

Solution 2

Notice that, no matter which card you choose, there are exactly 4 cards that either has the same color or letter as it. Since there are 7 cards left to choose from, the probability is $\frac{4}{7}$. theepiccarrot7

Solution 3

We can use casework to solve this.

Case $1$: Same letter After choosing any letter, there are seven cards left, and only one of them will produce a winning pair. Therefore, the probability is $\frac17$.

Case $2$: Same color After choosing any letter, there are seven cards left. Three of them will make a winning pair, so the probability is $\frac37$.

Now that we have the probability for both cases, we can add them: $\frac17+\frac37=\boxed{\textbf{(D)} \frac47}$.

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 AJHSME/AMC 8 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_21&oldid=186854"