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

Latest revision as of 13:16, 29 October 2024

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

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 $\frac{4}{7}$ . 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: $\frac17+\frac37=\boxed{\textbf{(D)} \frac47}$ .

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.

@@ Line 1: / Line 1: @@
-There are 4 ways of choosing a winning pair of the same number, and <math>2 \left( \dbinom{4}{2} \right) = 12</math> ways to choose a pair of the same color.
+==Problem ==
+Two cards are dealt from a deck of four red cards labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and four green cards labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
-There's a total of <math>\dbinom{8}{2} = 28</math> ways to choose a pair, so the probability is <math>\dfrac{4+12}{28} = \boxed{\dfrac{4}{7}}</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==
+https://youtu.be/OOdK-nOzaII?t=1698
+==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==
+{{AMC8 box|year=2007|num-b=20|num-a=22}}
+{{MAA Notice}}

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