2017 AMC 10A Problem 15

Question

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?

\dfrac{1}{2}

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{5}{6}

\dfrac{7}{8}

Break Laurent’s interval into the lower half and upper half

In the lower half, symmetry decides which chosen number is larger

If Laurent chooses a number in the interval

(2017, 4034],

then there is no way that Chloé can have the greater number.

This means that Laurent has a

\frac{1}{2}

chance of automatically winning.

Otherwise, Laurent chooses a number in the interval

[0, 2017].

The probability that he gets a greater number than Chloé is the same as Chloé getting a greater number than Laurent.

This means that Laurent has a

\frac{1}{2}

chance of getting a greater number (when working with real intervals, the probability of a tie is essentially

0

due to the infinite size of the intervals).

Laurent’s total chance of getting a greater number is

\dfrac{1}{2} \cdot 1 + \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{3}{4}.

Thus, C is the correct answer.

Problem 15 in other years