Difference between revisions of "2017 AMC 12A Problems/Problem 10"
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<math> \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} </math> | <math> \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Suppose Laurent's number is in the interval <math> [ 0, 2017 ] </math>. Then, by symmetry, the probability of Laurent's number being greater is <math>\dfrac{1}{2}</math>. Next, suppose Laurent's number is in the interval <math> [ 2017, 4034 ] </math>. Then Laurent's number will be greater with probability <math>1</math>. Since each case is equally likely, the probability of Laurent's number being greater is <math>\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}</math>, so the answer is C. | + | Suppose Laurent's number is in the interval <math> [ 0, 2017 ] </math>. Then, by symmetry, the probability of Laurent's number being greater is <math>\dfrac{1}{2}</math>. Next, suppose Laurent's number is in the interval <math> [ 2017, 4034 ] </math>. Then Laurent's number will be greater with a probability of <math>1</math>. Since each case is equally likely, the probability of Laurent's number being greater is <math>\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}</math>, so the answer is <math> \boxed{C}</math>. |
− | == | + | ==Solution 2 (Geometric Probability)== |
− | Let <math>x</math> be the number chosen randomly by | + | Let <math>x</math> be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the 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 Laurent chooses is in the interval <math> [ 0, 4034 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is greater than Chloe'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 <math>2017</math> in width and <math>4034</math> 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 <math>2017</math> in length and bases <math>4034</math> and <math>2017</math> 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 Chloe's, which is <math> \boxed {C=\frac{3}{4}}</math>. |
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/1f2JaybCZCY | ||
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+ | ~Education, the Study of Everything | ||
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+ | == Video Solution== | ||
+ | https://youtu.be/LwtoLiBwO-E?t=79 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}} | ||
{{AMC12 box|year=2017|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2017|ab=A|num-b=9|num-a=11}} | ||
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+ | [[Category:Introductory Probability Problems]] | ||
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{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:58, 10 June 2023
Contents
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
.
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/LwtoLiBwO-E?t=79
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.