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

m (Fixed formatting of answer choices)
m (Slightly improved formatting)
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The four cases are:
 
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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'''Case 1''': <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>.
 +
 
 +
'''Case 2''': <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>.
 +
 
 +
'''Case 3''': <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>.
 +
 
 +
'''Case 4''': <math>x</math> is is chosen from the interval <math>[0,1]</math>, and <math>y</math> is also chosen from the interval <math>[0,1]</math>.
 +
 
 +
Each case has a <math>\frac{1}{4}</math> chance of occurring (as it requires two coin flips).
  
 
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>.
 
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 <math>x</math> is 0, we need <math>y</math> to be in the interval <math>(\frac{1}{2}, 1]</math>. If <math>x</math> is 1, we need <math>y</math> to be in the interval <math>[0, \frac{1}{2})</math>. Regardless of what <math>x</math> is, the probability for success for Case 2 is <math>\frac{1}{2}</math>.
+
For Case 2, if <math>x</math> is 0, we need <math>y</math> to be in the interval <math>\left(\frac{1}{2}, 1\right]</math>. If <math>x</math> is 1, we need <math>y</math> to be in the interval <math>\left[0, \frac{1}{2}\right)</math>. Regardless of what <math>x</math> is, the probability for success for Case 2 is <math>\frac{1}{2}</math>.
  
 
By symmetry, Case 3 has the same success rate as Case 2.
 
By symmetry, Case 3 has the same success rate as Case 2.
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</asy>
 
</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>.
+
The shaded area is <math>\frac{1}{4}</math>, which means the probability for success for case 4 is <math>\frac{1}{4}</math> (since the total area of the bounding square, containing all possible pairs, is <math>1</math>).
  
 
Adding up the success rates from each case, we get:
 
Adding up the success rates from each case, we get:
  
 
<math>\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}}</math>.
 
<math>\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}}</math>.
−
 
−
<!--==Solution 2==
 
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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>.
 
−
 
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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>.
 
−
 
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Now, by symmetry, the probability that <math>|x-y| > \frac{1}{2}</math> and the probability that <math>|x-y| < \frac{1}{2}</math> are 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 00:50, 27 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:

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

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