2017 AMC 12A Problems/Problem 10

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

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 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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