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Difference between revisions of "2019 AMC 10A Problems/Problem 22"
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<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>
  
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==Solution==
+
==Solution 1==
  
 
There are several cases depending on what the first coin flip is when determining <math>x</math> and what the first coin flip is when determining <math>y</math>.
 
There are several cases depending on what the first coin flip is when determining <math>x</math> and what the first coin flip is when determining <math>y</math>.
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<math>(\frac{1}{4}) \cdot (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}) = \boxed{\textbf{(B) }\frac{7}{16}}</math>.
 
<math>(\frac{1}{4}) \cdot (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}) = \boxed{\textbf{(B) }\frac{7}{16}}</math>.
 +
 +
==Solution 2==
 +
We can have either <math>|x-y| > \frac{1}{2}</math> , <math>|x-y| < \frac{1}{2}</math> , or <math>|x-y| = \frac{1}{2}</math>.
 +
 +
The probability that <math>|x-y| = \frac{1}{2}</math> is <math>\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}</math> , so the probability of this NOT happening is <math>1 - \frac{1}{8} = \frac{7}{8}</math>.
 +
 +
Now, by symmetry, the probability that <math>|x-y| > \frac{1}{2}</math> and the probability that <math>|x-y| < \frac{1}{2}</math> is equal, so each must be <math>\frac{\frac{7}{8}}{2} = \boxed{\textbf{(B) }\frac{7}{16}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:29, 11 February 2019

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 1

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:

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

Each case has a $\frac{1}{4}$ chance of occurring.

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 $(\frac{1}{2}, 1]$. If $x$ is 1, we need $y$ to be in the interval $[0, \frac{1}{2})$. 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 $\frac{1}{4}$.

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

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

Solution 2

We can have either $|x-y| > \frac{1}{2}$ , $|x-y| < \frac{1}{2}$ , or $|x-y| = \frac{1}{2}$.

The probability that $|x-y| = \frac{1}{2}$ is $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$ , so the probability of this NOT happening is $1 - \frac{1}{8} = \frac{7}{8}$.

Now, by symmetry, the probability that $|x-y| > \frac{1}{2}$ and the probability that $|x-y| < \frac{1}{2}$ is equal, so each must be $\frac{\frac{7}{8}}{2} = \boxed{\textbf{(B) }\frac{7}{16}}$.

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

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