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

(Created page with "{{duplicate|2019 AMC 10A #22 and 2019 AMC 12A #20}} ==Problem== Real numbers between 0 and 1, inclusive, are chosen in th...")
 
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==Solution==
 
==Solution==
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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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The four cases are:
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#<math>x</math> is either <math>0</math> or <math>1</math> and <math>y</math> is either <math>0</math> or <math>1</math>
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#<math>x</math> is either <math>0</math> or <math>1</math> and <math>y</math> is chosen from the interval <math>[0,1]</math>
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#<math>x</math> is is chosen from the interval <math>[0,1]</math> and <math>y</math> is either <math>0</math> or <math>1</math>
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#<math>x</math> is is chosen from the interval <math>[0,1]</math> and <math>y</math> is chosen from the interval <math>[0,1]</math>
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Each case has a <math>\frac{1}{4}</math> chance of occurring.
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For case 1, we need <math>x</math> and <math>y</math> to be different. Therefore, the probability for success in case 1 is <math>\frac{1}{2}</math>.
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For case 2, if x is 0, we need y to be in the interval <math>(\frac{1}{2}, 1]</math>. If x is 1, we need y to be in the interval <math>[0, \frac{1}{2})</math>. Regardless of what x is, the probability for success for case 2 is <math>\frac{1}{2}</math>.
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By symmetry, case 3 has the same success rate as case 2.
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For case 4, we must use geometric probability because there are an infinite number of pairs (x, y) that can be selected. Graphing <math>|x-y| > \tfrac{1}{2}</math>, gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:
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<asy>
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filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1));
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filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1));
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draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
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</asy>
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The shaded area is <math>\frac{1}{4}</math>, which means the probability for success for case 4 is <math>\frac{1}{4}</math>.
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Adding up the success rates from each case, we get:
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 17:19, 9 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

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. 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:

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

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

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