Difference between revisions of "2017 AMC 12A Problems/Problem 10"
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==Alternate Solution: Geometric Probability== | ==Alternate Solution: Geometric Probability== | ||
− | Let <math>x</math> be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is interval <math> [ 0, 2017 ] </math>, <math> 0 \leq x \leq 2017</math>. Next, let <math>y</math> be the number chosen randomly by Laurent. Because it is given that the number Chloé choose is interval <math> [ 0, 2017 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is Chloé's we write the equation <math>y > x</math>. When these three inequalities are graphed the area captured by <math> 0 \leq x \leq 2017</math> and <math> 0 \leq y \leq 4034</math> represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area <math> 4034 * 2017</math>. The area captured by <math> 0 \leq x \leq 2017</math>, <math> 0 \leq y \leq 4034</math>, and <math>y > x</math> represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases | + | Let <math>x</math> be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is interval <math> [ 0, 2017 ] </math>, <math> 0 \leq x \leq 2017</math>. Next, let <math>y</math> be the number chosen randomly by Laurent. Because it is given that the number Chloé choose is interval <math> [ 0, 2017 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is Chloé's we write the equation <math>y > x</math>. When these three inequalities are graphed the area captured by <math> 0 \leq x \leq 2017</math> and <math> 0 \leq y \leq 4034</math> represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area <math> 4034 * 2017</math>. The area captured by <math> 0 \leq x \leq 2017</math>, <math> 0 \leq y \leq 4034</math>, and <math>y > x</math> represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus making an area <math>2017 *\frac{4034+2017}{2}</math>. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloé's, which is <math> \boxed {C=\frac{3}{4}}</math>. |
==See Also== | ==See Also== |
Revision as of 21:19, 9 February 2017
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 Chloe's number?
Solution
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 probability . Since each case is equally likely, the probability of Laurent's number being greater is , so the answer is C.
Alternate Solution: Geometric Probability
Let be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is interval , . Next, let be the number chosen randomly by Laurent. Because it is given that the number Chloé choose is interval , . Since we are looking for when Laurent's number is Chloé's we write the equation . When these three inequalities are graphed the area captured by and represents all the possibilities, forming a rectangle 2017 in width and 4034 in height. Thus making its area . The area captured by , , and represents the possibilities of Laurent winning, forming a trapezoid with a height 2017 in length and bases 4034 and 2017 length, thus making an area . The simplified quotient of these two areas is the probability Laurent's number is larger than Chloé'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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