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Difference between revisions of "2007 AMC 8 Problems/Problem 24"

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The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4). The number of ways to choose three digits out of four is 4. So, the probability is <math>\frac{1}{2}</math>.
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==Problem==
  
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Answer : C
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A bag contains four pieces of paper, each labeled with one of the digits <math>1</math>, <math>2</math>, <math>3</math> or <math>4</math>, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of <math>3</math>?
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<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math>
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==Solution 1==
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The number of ways to form a 3-digit number is <math>4 \cdot 3 \cdot 2 = 24</math>. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is <math>3! + 3! = 12</math>.
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Therefore, the probability is <math>\frac{12}{24} = \boxed{\frac{1}{2}}</math>.
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~abc2142
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==Solution 2==
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Only (1,2,3) and (2,3,4) are possible combinations that sum to 3. The total ways to choose three numbers from the four numbers is 4 choose 3 which equals 4. The answer is <math>\frac{2}{4} = \boxed{\frac{1}{2}}</math>.
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==Video Solution by WhyMath==
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https://youtu.be/MnnxFJBV87A
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==See Also==
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{{AMC8 box|year=2007|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 13:17, 29 October 2024

Problem

A bag contains four pieces of paper, each labeled with one of the digits $1$, $2$, $3$ or $4$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$? $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

Solution 1

The number of ways to form a 3-digit number is $4 \cdot 3 \cdot 2 = 24$. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is $3! + 3! = 12$. Therefore, the probability is $\frac{12}{24} = \boxed{\frac{1}{2}}$.

~abc2142

Solution 2

Only (1,2,3) and (2,3,4) are possible combinations that sum to 3. The total ways to choose three numbers from the four numbers is 4 choose 3 which equals 4. The answer is $\frac{2}{4} = \boxed{\frac{1}{2}}$.

Video Solution by WhyMath

https://youtu.be/MnnxFJBV87A

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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