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

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==Problem==
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==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?
 
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?
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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>
  
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==Solution==
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== Video Solution by OmegaLearn ==
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There are 4 ways of choosing a winning pair of the same letter, and <math>2 \left( \dbinom{4}{2} \right) = 12</math> ways to choose a pair of the same color.
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https://youtu.be/OOdK-nOzaII?t=1712
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/OOdK-nOzaII?t=1698
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==Video Solution by WhyMath==
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https://youtu.be/HZ0lxSAujrI
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==Solution 2==
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Notice that, no matter which card you choose, there are exactly <math>4</math> cards that either have the same color or letter as it. Since there are <math>7</math> cards left to choose from, the probability is <math>\boxed{\textbf{(D)}\frac47}</math>.
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-theepiccarrot7
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==Solution 3==
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We can use casework to solve this.
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Case <math>1</math>: Same letter
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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>.
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Case <math>2</math>: Same color
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After choosing any letter, there are seven cards left. Three of them will make a winning pair, so the probability is <math>\frac37</math>.
  
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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{\textbf{(D)}\ \frac{4}{7}}</math>.
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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 12:05, 24 December 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 have the same color or letter as it. Since there are $7$ cards left to choose from, the probability is $\boxed{\textbf{(D)}\frac47}$. -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

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