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2017 AMC 12A Problems/Problem 10

Revision as of 16:11, 9 February 2017 by Lphillips17 (talk | contribs) (→‎Alternate Solution: Geometric Probability)

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 and the area captured by $0 \leq x \leq 2017$, $0 \leq y \leq 4034$, and $y > x$ represents the possibilities of Laurent winning. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloé's, which is $\boxed {C=\frac{3}{4}}$.

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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