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Difference between revisions of "2017 AMC 12A Problems/Problem 10"
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<math> \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} </math>
 
<math> \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} </math>
  
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==Solution==
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==Solution 1==
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Suppose Laurent's number is in the interval <math> [ 0, 2017 ] </math>. Then, by symmetry, the probability of Laurent's number being greater is <math>\dfrac{1}{2}</math>. Next, suppose Laurent's number is in the interval <math> [ 2017, 4034 ] </math>. Then Laurent's number will be greater with probability <math>1</math>. Since each case is equally likely, the probability of Laurent's number being greater is <math>\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}</math>, so the answer is C.
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Suppose Laurent's number is in the interval <math> [ 0, 2017 ] </math>. Then, by symmetry, the probability of Laurent's number being greater is <math>\dfrac{1}{2}</math>. Next, suppose Laurent's number is in the interval <math> [ 2017, 4034 ] </math>. Then Laurent's number will be greater with a probability of <math>1</math>. Since each case is equally likely, the probability of Laurent's number being greater is <math>\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}</math>, so the answer is <math> \boxed{C}</math>.
  
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==Alternate Solution: Geometric Probability==
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==Solution 2 (Geometric Probability)==
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Let <math>x</math> be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval <math> [ 0, 2017 ] </math>, <math> 0 \leq x \leq 2017</math>.  Next, let <math>y</math>  be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval <math> [ 0, 4034 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is greater than Chloe's we write the equation <math>y > x</math>. When these three inequalities are graphed the area captured by <math> 0 \leq x \leq 2017</math> and <math> 0 \leq y \leq 4034</math> represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area <math> 4034 * 2017</math>. The area captured by <math> 0 \leq x \leq 2017</math>, <math> 0 \leq y \leq 4034</math>, and <math>y > x</math> represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus making an area <math>2017 *\frac{4034+2017}{2}</math>.  The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which is <math> \boxed {C=\frac{3}{4}}</math>.
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Let <math>x</math> be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval <math> [ 0, 2017 ] </math>, <math> 0 \leq x \leq 2017</math>.  Next, let <math>y</math>  be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval <math> [ 0, 4034 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is greater than Chloe's we write the equation <math>y > x</math>. When these three inequalities are graphed the area captured by <math> 0 \leq x \leq 2017</math> and <math> 0 \leq y \leq 4034</math> represents all the possibilities, forming a rectangle <math>2017</math> in width and <math>4034</math> in height. Thus making its area <math>4034 * 2017</math>. The area captured by <math> 0 \leq x \leq 2017</math>, <math> 0 \leq y \leq 4034</math>, and <math>y > x</math> represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases <math>4034</math> and <math>2017</math> length, thus making an area <math>2017 *\frac{4034+2017}{2}</math>.  The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which is <math> \boxed {C=\frac{3}{4}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 14:15, 19 December 2020

Problem

Chloe chooses a real number uniformly at random from the interval $[ 0,2017 ]$. Independently, Laurent chooses a real number uniformly at random from the interval $[ 0 , 4034 ]$. What is the probability that Laurent's number is greater than Chloe's number?

$\textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8}$

Solution 1

Suppose Laurent's number is in the interval $[ 0, 2017 ]$. Then, by symmetry, the probability of Laurent's number being greater is $\dfrac{1}{2}$. Next, suppose Laurent's number is in the interval $[ 2017, 4034 ]$. Then Laurent's number will be greater with a probability of $1$. Since each case is equally likely, the probability of Laurent's number being greater is $\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}$, so the answer is $\boxed{C}$.

Solution 2 (Geometric Probability)

Let $x$ be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval $[ 0, 2017 ]$, $0 \leq x \leq 2017$. Next, let $y$ be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval $[ 0, 4034 ]$, $0 \leq y \leq 4034$. Since we are looking for when Laurent's number is greater than Chloe's we write the equation $y > x$. When these three inequalities are graphed the area captured by $0 \leq x \leq 2017$ and $0 \leq y \leq 4034$ represents all the possibilities, forming a rectangle $2017$ in width and $4034$ in height. Thus making its area $4034 * 2017$. The area captured by $0 \leq x \leq 2017$, $0 \leq y \leq 4034$, and $y > x$ represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases $4034$ and $2017$ length, thus making an area $2017 *\frac{4034+2017}{2}$. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which is $\boxed {C=\frac{3}{4}}$.

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AMC 10 Problems and Solutions
2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 AMC 12 Problems and Solutions

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