Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10A Problems/Problem 22"

m (Slightly improved formatting)
m (→‎Problem)
Line 3: Line 3:
 
==Problem==
 
==Problem==
  
−
Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval <math>[0,1]</math>. Two random numbers <math>x</math> and <math>y</math> are chosen independently in this manner. What is the probability that <math>|x-y| > \tfrac{1}{2}</math>?
+
Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval <math>[0,1]</math>. Two random numbers <math>x</math> and <math>y</math> are chosen independently in this manner. What is the probability that <math>|x-y| > \tfrac{1}{2}</math>?
  
 
<math>\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}</math>
 
<math>\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}</math>

Revision as of 14:25, 18 January 2020

The following problem is from both the 2019 AMC 10A #22 and 2019 AMC 12A #20, so both problems redirect to this page.

Problem

Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$?

$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$

Solution

There are several cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$.

The four cases are:

Case 1: $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$.

Case 2: $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$.

Case 3: $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$.

Case 4: $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$.

Each case has a $\frac{1}{4}$ chance of occurring (as it requires two coin flips).

For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\frac{1}{2}$.

For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\frac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \frac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\frac{1}{2}$.

By symmetry, Case 3 has the same success rate as Case 2.

For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \tfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:

[asy] filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1)); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); [/asy]

The shaded area is $\frac{1}{4}$, which means the probability for success for case 4 is $\frac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$).

Adding up the success rates from each case, we get:

$\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\right) = \boxed{\textbf{(B) }\frac{7}{16}}$.

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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
2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_22&oldid=114936"