2017 AMC 12A Problems/Problem 10
Problem
Chloe 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 Chloe's number?
Solution 1
Suppose Laurent's number is in the interval . Then, by symmetry, the probability of Laurent's number being greater is . Next, suppose Laurent's number is in the interval . Then Laurent's number will be greater with a probability of . Since each case is equally likely, the probability of Laurent's number being greater is , so the answer is .
Solution 2 (Geometric Probability)
Let be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval , . Next, let be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval , . Since we are looking for when Laurent's number is greater than Chloe's we write the equation . When these three inequalities are graphed the area captured by and represents all the possibilities, forming a rectangle in width and in height. Thus making its area . The area captured by , , and represents the possibilities of Laurent winning, forming a trapezoid with a height in length and bases and length, thus making an area . The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which 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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 12 Problems and Solutions |
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