Difference between revisions of "2017 AMC 10A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
Denote "winning" to mean "picking a greater number". | Denote "winning" to mean "picking a greater number". | ||
− | There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>[ | + | There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>[2018, 4034]</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{\textbf{(C)}\ \frac{3}{4}}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 17:18, 6 February 2018
Contents
[hide]Problem
Chloé chooses a real number uniformly at random from the interval . Independently, Laurent chooses a real number uniformly at random from the interval
. What is the probability that Laurent's number is greater than Chloé's number?
Solution 1
Denote "winning" to mean "picking a greater number".
There is a chance that Laurent chooses a number in the interval
. In this case, Chloé cannot possibly win, since the maximum number she can pick is
. Otherwise, if Laurent picks a number in the interval
, with probability
, then the two people are symmetric, and each has a
chance of winning. Then, the total probability is
Solution 2
We can use geometric probability to solve this.
Suppose a point lies in the
-plane. Let
be Chloe's number and
be Laurent's number. Then obviously we want
, which basically gives us a region above a line. We know that Chloe's number is in the interval
and Laurent's number is in the interval
, so we can create a rectangle in the plane, whose length is
and whose width is
. 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
. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line
, which is
. Instead of bashing this out we know that the rectangle has area
. So the probability that Laurent has a smaller number is
. Simplifying the expression yields
and so
.
Solution 3:
Scale down by to get that Chloe picks from
and Laurent picks from
. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of
. Therefore, Laurent has a range of
to
to pick from, on average, which is a length of
out of a total length of
. Therefore, the probability is
See Also
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 |
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