2017 AMC 12A Problems/Problem 10

Problem

Chloé 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

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 probability $1$ . 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 C.

Alternate Solution: Geometric Probability

Let $x$ be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is 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 Chloé choose is interval $[ 0, 2017 ]$ , $0 \leq y \leq 4034$ . Since we are looking for when Laurent's number is Chloé'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 Chloé's, which is $\boxed {C=\frac{3}{4}}$ .

2017 AMC 10A (Problems • Answer Key • Resources)
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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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2017 AMC 12A Problems/Problem 10

Contents

Problem

Solution

Alternate Solution: Geometric Probability

See Also