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

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(4) This one is a little bit tricky. You could use geometric probability shown above, but if you don't understand it or happened to not think of it, there is another solution that involves using the multiple choice answers.  
 
(4) This one is a little bit tricky. You could use geometric probability shown above, but if you don't understand it or happened to not think of it, there is another solution that involves using the multiple choice answers.  
  
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First, the probability we have so far is <math>\frac{3}{8}</math>.This is greater than <math>\frac{1}{3}</math> so <math>A</math> is the not the answer. It is either <math>A, B, C,</math> or <math>D</math>. Let's draw out the probability with two parallel lines on a paper with represent a number line from <math>0-1</math>. The concept is to compare one point and find the fraction of the other line that contains points that are at least <math>\frac{1}{2}</math> away from it. If the point we choose is on the endpoints, then the fraction of the other line that works is exactly <math>\frac{1}{2}</math>. But as we move this point closer to the middle, the deadzone (area where the other point cannot be) grows, diminishing the probability. Finally, when it is directly in the middle, there are no points that pass the requirements except at <math>1</math> and <math>0</math>.
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First, the probability we have so far is <math>\frac{3}{8}</math>.This is greater than <math>\frac{1}{3}</math> so <math>A</math> is the not the answer. It is either <math>B, C, D,</math> or <math>E</math>. Let's draw out the probability with two parallel lines on a paper with represent a number line from <math>0-1</math>. The concept is to compare one point and find the fraction of the other line that contains points that are at least <math>\frac{1}{2}</math> away from it. If the point we choose is on the endpoints, then the fraction of the other line that works is exactly <math>\frac{1}{2}</math>. But as we move this point closer to the middle, the deadzone (area where the other point cannot be) grows, diminishing the probability. Finally, when it is directly in the middle, there are no points that pass the requirements except at <math>1</math> and <math>0</math>.
  
 
So, looking at the choices again, we have <math>\frac{7}{16}</math>, <math>\frac{1}{2}</math>, <math>\frac{9}{16}</math>, and <math>\frac{2}{3}</math>.  
 
So, looking at the choices again, we have <math>\frac{7}{16}</math>, <math>\frac{1}{2}</math>, <math>\frac{9}{16}</math>, and <math>\frac{2}{3}</math>.  
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-jackshi2006
 
-jackshi2006
 +
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==Solution 3 (with Table)==
 +
First, calculate the conditional probabilities:
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<math>P(x\in[0,\frac{1}{2}) \mid x\in[0,1]) = \frac{1}{2}</math>,  <math>P(|x-y|>\frac{1}{2} \mid x\in[0,1], y=1) = \frac{1}{2}</math>
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<math>P(x\in(\frac{1}{2},1] \mid x\in[0,1]) = \frac{1}{2}</math>,  <math>P(|x-y|>\frac{1}{2} \mid x\in[0,1], y=0) = \frac{1}{2}</math>
 +
 +
<math>P(y\in[0,\frac{1}{2}) \mid y\in[0,1]) = \frac{1}{2}</math>,  <math>P(|x-y|>\frac{1}{2} \mid x=1, y\in[0,1]) = \frac{1}{2}</math>
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<math>P(y\in(\frac{1}{2},1] \mid y\in[0,1]) = \frac{1}{2}</math>,  <math>P(|x-y|>\frac{1}{2} \mid x=0, y\in[0,1]) = \frac{1}{2}</math>
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As stated in Solution 1, <math>P(|x-y|>\frac{1}{2} \mid x\in[0,1], y\in[0,1]) = \frac{1}{4}</math>
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Next, calculate the dependent event probabilities by <math>P(A \cap B)= P(A \mid B) \cdot P(B)</math>, draw a table to calculate <math>P(|x-y|>\frac{1}{2})</math> by the dependent events:
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<cmath>\[ \begin{array}[b]{cccc}
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& P(x=0)=\frac{1}{4} & P(x=1)=\frac{1}{4} & P(x \in [0,1])=\frac{1}{2} \\
 +
&  &  &  \\
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P(y=0)=\frac{1}{4} & 0 & \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} \\
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&  &  &  \\
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P(y=1)=\frac{1}{4} & \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} & 0 & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} \\
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&  &  &  \\
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P(y \in [0,1])=\frac{1}{2} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16}
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\end{array} \]</cmath>
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The sum of all the cell values is <math>\boxed{\textbf{(B)}\frac{7}{16}}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 4==
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Notice the probability that <math>|x-y| > \tfrac{1}{2}</math> is equal to <math>2</math> times the probability that <math>x-y > \tfrac{1}{2}</math>.
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For <math>x-y > \tfrac{1}{2}</math> to be true; there are <math>4</math> cases.
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Case <math>1</math>: <math>x</math> flips <math>T</math> and <math>y</math> flips <math>T</math>
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When <math>x</math> flips <math>T</math> and <math>y</math> flips <math>T</math>, both <math>x</math> and <math>y</math> are randomly chosen from the interval <math>[0,1]</math>. The probability that <math>x-y > \tfrac{1}{2}</math> when both <math>x</math> and <math>y</math> are randomly chosen from the interval <math>[0,1]</math> is <math>\frac18</math>, as illustrated in the geometric probability diagram below.
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<asy>
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size(3cm,0);
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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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Therefore, the probability for Case <math>1</math> is <math>\frac12 \cdot \frac12 \cdot \frac18 = \frac{1}{32}</math>
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Case <math>2</math>: <math>x</math> flips <math>HT</math> and <math>y</math> flips <math>T</math>
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When <math>x</math> flips <math>HT</math> and <math>y</math> flips <math>T</math>, <math>x=1</math>, and <math>y</math> is randomly chosen from the interval <math>[0,1]</math>. The probability that <math>x-y > \tfrac{1}{2}</math> when <math>x=1</math>, and <math>y</math> is randomly chosen from the interval <math>[0,1]</math> is <math>\frac12</math> (<math>y</math> has to be between <math>[0,\frac12]</math>).
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Therefore, the probability for Case <math>2</math> is <math>\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}</math>
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Case <math>3</math>: <math>x</math> flips <math>T</math> and <math>y</math> flips <math>HH</math>
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When <math>x</math> flips <math>T</math> and <math>y</math> flips <math>HH</math>, <math>x</math> is randomly chosen from the interval <math>[0,1]</math>, and <math>y = 0</math>. The probability that <math>x-y > \tfrac{1}{2}</math> when <math>x</math> is randomly chosen from the interval <math>[0,1]</math>, and <math>y = 0 </math>is <math>\frac12</math> (<math>x</math> has to be between <math>[\frac12,1]</math>).
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Therefore, the probability for Case <math>3</math> is <math>\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}</math>
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Case <math>4</math>: <math>x</math> flips <math>HT</math> and <math>y</math> flips <math>HH</math>
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When <math>x</math> flips <math>HT</math> and <math>y</math> flips <math>HH</math>, <math>x=1</math>, and <math>y = 0</math>. The probability that <math>x-y > \frac{1}{2}</math> when <math>x=1</math>, and <math>y = 0</math> is <math>1</math>.
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Therefore, the probability for Case <math>4</math> is <math>\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 \cdot 1= \frac{1}{16}</math>
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Hence, the answer is <math>2(\frac{1}{32} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}) = \boxed{\textbf{(B)}\frac{7}{16}}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==Video Solution 1==
 
==Video Solution 1==
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Education, the Study of Everything
 
Education, the Study of Everything
  
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==Video Solution 2==
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== Video Solution 2 ==
−
https://www.youtube.com/watch?v=fe5bdwszsos
 
−
 
 
−
== Video Solution 3 ==
 
 
https://youtu.be/ZhAZ1oPe5Ds?t=263
 
https://youtu.be/ZhAZ1oPe5Ds?t=263
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
== Video Solution by Richard Rusczyk  ==
 +
https://www.youtube.com/watch?v=Fe5bDwSzsOs
  
 
==See Also==
 
==See Also==

Latest revision as of 10:46, 8 October 2023

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:

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:

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


Solution 2

Taking into account that there are two options for the result of the first coin flip, there are four possible combinations with equal possibility of initial coins flips.

(1) x: heads, y: heads

(2) x: heads, y: tails

(3) x: tails, y: heads

(4) x: tails, y: tails

(1) Because x and y both must be integers, so it's simple to see that the probability that $|x-y| > \tfrac{1}{2}$ is $\frac{1}{2}$.

(2) Since x has to be an integer it is also easy to see that the probability is again $\frac{1}{2}$.

(3) This is the same as case 2 so the probability is $\frac{1}{2}$.

(4) This one is a little bit tricky. You could use geometric probability shown above, but if you don't understand it or happened to not think of it, there is another solution that involves using the multiple choice answers.

First, the probability we have so far is $\frac{3}{8}$.This is greater than $\frac{1}{3}$ so $A$ is the not the answer. It is either $B, C, D,$ or $E$. Let's draw out the probability with two parallel lines on a paper with represent a number line from $0-1$. The concept is to compare one point and find the fraction of the other line that contains points that are at least $\frac{1}{2}$ away from it. If the point we choose is on the endpoints, then the fraction of the other line that works is exactly $\frac{1}{2}$. But as we move this point closer to the middle, the deadzone (area where the other point cannot be) grows, diminishing the probability. Finally, when it is directly in the middle, there are no points that pass the requirements except at $1$ and $0$.

So, looking at the choices again, we have $\frac{7}{16}$, $\frac{1}{2}$, $\frac{9}{16}$, and $\frac{2}{3}$.

$\frac{1}{2}$ is exactly $\frac{1}{8}$ more than the probability we had before. Notice that this is impossible because we proved that the average probability of the fourth case is lower than $\frac{1}{2}$, so the answer is $\boxed{B}$.

-jackshi2006

Solution 3 (with Table)

First, calculate the conditional probabilities:

$P(x\in[0,\frac{1}{2}) \mid x\in[0,1]) = \frac{1}{2}$, $P(|x-y|>\frac{1}{2} \mid x\in[0,1], y=1) = \frac{1}{2}$

$P(x\in(\frac{1}{2},1] \mid x\in[0,1]) = \frac{1}{2}$, $P(|x-y|>\frac{1}{2} \mid x\in[0,1], y=0) = \frac{1}{2}$

$P(y\in[0,\frac{1}{2}) \mid y\in[0,1]) = \frac{1}{2}$, $P(|x-y|>\frac{1}{2} \mid x=1, y\in[0,1]) = \frac{1}{2}$

$P(y\in(\frac{1}{2},1] \mid y\in[0,1]) = \frac{1}{2}$, $P(|x-y|>\frac{1}{2} \mid x=0, y\in[0,1]) = \frac{1}{2}$

As stated in Solution 1, $P(|x-y|>\frac{1}{2} \mid x\in[0,1], y\in[0,1]) = \frac{1}{4}$

Next, calculate the dependent event probabilities by $P(A \cap B)= P(A \mid B) \cdot P(B)$, draw a table to calculate $P(|x-y|>\frac{1}{2})$ by the dependent events:

\[ \begin{array}[b]{cccc} & P(x=0)=\frac{1}{4} & P(x=1)=\frac{1}{4} & P(x \in [0,1])=\frac{1}{2} \\ & & & \\ P(y=0)=\frac{1}{4} & 0 & \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} \\ & & & \\ P(y=1)=\frac{1}{4} & \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} & 0 & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} \\ & & & \\ P(y \in [0,1])=\frac{1}{2} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} & \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{16} \end{array} \]

The sum of all the cell values is $\boxed{\textbf{(B)}\frac{7}{16}}$

~isabelchen

Solution 4

Notice the probability that $|x-y| > \tfrac{1}{2}$ is equal to $2$ times the probability that $x-y > \tfrac{1}{2}$.

For $x-y > \tfrac{1}{2}$ to be true; there are $4$ cases.

Case $1$: $x$ flips $T$ and $y$ flips $T$

When $x$ flips $T$ and $y$ flips $T$, both $x$ and $y$ are randomly chosen from the interval $[0,1]$. The probability that $x-y > \tfrac{1}{2}$ when both $x$ and $y$ are randomly chosen from the interval $[0,1]$ is $\frac18$, as illustrated in the geometric probability diagram below.

[asy] size(3cm,0); 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]

Therefore, the probability for Case $1$ is $\frac12 \cdot \frac12 \cdot \frac18 = \frac{1}{32}$

Case $2$: $x$ flips $HT$ and $y$ flips $T$

When $x$ flips $HT$ and $y$ flips $T$, $x=1$, and $y$ is randomly chosen from the interval $[0,1]$. The probability that $x-y > \tfrac{1}{2}$ when $x=1$, and $y$ is randomly chosen from the interval $[0,1]$ is $\frac12$ ($y$ has to be between $[0,\frac12]$).

Therefore, the probability for Case $2$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}$

Case $3$: $x$ flips $T$ and $y$ flips $HH$

When $x$ flips $T$ and $y$ flips $HH$, $x$ is randomly chosen from the interval $[0,1]$, and $y = 0$. The probability that $x-y > \tfrac{1}{2}$ when $x$ is randomly chosen from the interval $[0,1]$, and $y = 0$is $\frac12$ ($x$ has to be between $[\frac12,1]$).

Therefore, the probability for Case $3$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}$

Case $4$: $x$ flips $HT$ and $y$ flips $HH$

When $x$ flips $HT$ and $y$ flips $HH$, $x=1$, and $y = 0$. The probability that $x-y > \frac{1}{2}$ when $x=1$, and $y = 0$ is $1$.

Therefore, the probability for Case $4$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 \cdot 1= \frac{1}{16}$

Hence, the answer is $2(\frac{1}{32} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}) = \boxed{\textbf{(B)}\frac{7}{16}}$

~isabelchen

Video Solution 1

https://youtu.be/wimynTAVIx8

Education, the Study of Everything

Video Solution 2

https://youtu.be/ZhAZ1oPe5Ds?t=263

~ pi_is_3.14

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=Fe5bDwSzsOs

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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