2017 AMC 12A Problems/Problem 10: Difference between revisions

Revision as of 12:54, 9 February 2017

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.

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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 ==See Also==
+{{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}
 {{AMC12 box|year=2017|ab=A|num-b=9|num-a=11}}
 {{MAA Notice}}

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2017 AMC 12A Problems/Problem 10: Difference between revisions

Revision as of 12:54, 9 February 2017

Problem

Solution

See Also