2017 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 15 Medium

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 Chloé's number? (Assume they cannot be equal). (2017 AMC 10A Problem, Question#15)

  • A.

    \dfrac{1}{2}

  • B.

    \dfrac{2}{3}

  • C.

    \dfrac{3}{4}

  • D.

    \dfrac{5}{6}

  • E.

    \dfrac{7}{8}

Answer:C

Denote "winning" to mean "picking a greater number". There is a \dfrac{1}{2} chance that Laurent chooses a number in the interval [2018, 4034] 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 \dfrac{1}{2} then the two people are symmetric, and each has a \dfrac{1}{2} chance of winning. Then the total probabilty is \dfrac{1}{2}*1+ \dfrac{1}{2}* \dfrac{1}{2}= (\rm C) \dfrac{3}{4}.

We can use geometric probability to solve this. Suppose a point (x,y) lies in the xy-plane. Let x be Chloe's number and y be Laurent's number. Then obviously we want y>x, which basically gives us a region above a line. We know that Chloe's number is in the interval [0, 2017] and Laurent's number is in the interval [0, 4034], so we can create a rectangle in the plane, whose length is 2017 and whose width is 4034. Drawing it out, we see that it is easier to find the probability that Chloe's number is greater than Laurent's number and subtract this probability from 1. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line y > x, which is \dfrac{2017 \cdot 2017}{2}. Instead of bashing this out we know that the rectangle has area 2017\cdot4034. So the probability that Laurent has a smaller number is \dfrac{2017\cdot2017}{2\cdot2017\cdot4034}. Simplifying the expression yields \dfrac{1}{4} and so 1-\dfrac{1}{4}=(\rm C)\dfrac{3}{4}.

Scale down by 2017 to get that Chloe picks from [0, 1] and Laurent picks from [0, 2]. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of 0.5. Therefore, Laurent has a range of 0.5 to 2 to pick from, on average, which is a length of 2 -0.5 =1.5 out of a total length of 2-0 = 2. Therefore, the probability is 1.5/2=15/20=3/4(\rm C).