2017 AMC 10A Problems/Problem 15

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Problem

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

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

Solution

Denote "winning" to mean "picking a greater number". There is a $\frac{1}{2}$ chance that Laurent chooses a number in the interval $(2017, 4032]$. In this case, Chloé cannot possibly win, since the maximum number she can pick is $2017$. Otherwise, if Laurent picks a number in the interval $[0, 2017]$, with probability $\frac{1}{2}$, then the two people are symmetric, and each has a $\frac{1}{2}$ chance of winning. Then, the total probability is $\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\frac{3}{4} (C)}$

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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