Difference between revisions of "2007 AMC 8 Problems/Problem 21"
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<math> \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} </math> | <math> \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} </math> | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/OOdK-nOzaII?t=1712 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/OOdK-nOzaII?t=1698 | https://youtu.be/OOdK-nOzaII?t=1698 | ||
| − | ==Solution== | + | ==Video Solution by WhyMath== |
| − | + | https://youtu.be/HZ0lxSAujrI | |
| + | |||
| + | ==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 <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}} | ||
Latest revision as of 13:16, 29 October 2024
Contents
Problem
Two cards are dealt from a deck of four red cards labeled 
, 
, 
, 
and four green cards labeled , , , . A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
Video Solution by OmegaLearn
https://youtu.be/OOdK-nOzaII?t=1712
~ pi_is_3.14
Video Solution
https://youtu.be/OOdK-nOzaII?t=1698
Video Solution by WhyMath
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 
. theepiccarrot7
Solution 3
We can use casework to solve this.
Case 
: 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 
.
Case 
: Same color
After choosing any letter, there are seven cards left. Three of them will make a winning pair, so the probability is 
.
Now that we have the probability for both cases, we can add them: 
.
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. 
