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Difference between revisions of "2017 AMC 10A Problems/Problem 15"

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

Revision as of 15:10, 8 February 2017

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 $\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\frac{3}{4} (C)}$

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