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Difference between revisions of "2019 AIME I Problems/Problem 2"

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==Solution 1==
 
==Solution 1==
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The probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry.
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<math>B-J \ne 0</math> because <math>B \ne J</math>, so the probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry.
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<math>B-J \ne 0</math> because <math>B \ne J</math>.
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The probability that <math>B-J= 1</math> is <math>\frac{19}{20 \times 19} = \frac{1}{20}</math> because there are 19 pairs: <math>(B,J) = (2,1),.., (20,19)</math>.
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The probability that <math>B-J = 1</math> is <math>\frac{19}{20 \times 19} = \frac{1}{20}</math> because there are 19 pairs: <math>(B,J) = (2,1), \ldots, (20,19)</math>.
  
 
The probability that <math>B-J \ge 2</math> is <math>1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}</math>
 
The probability that <math>B-J \ge 2</math> is <math>1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}</math>
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-josephwidjaja
 
-josephwidjaja
 
==Solution 7 (Official MAA)==
 
==Solution 7 (Official MAA)==
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There are <math>\tbinom{20}{2}=190</math> equally likely pairs <math>\{J,B\}</math>. In <math>19</math> of these pairs <math>(\{1,2\},\{2,3\},\{3,4\}\dots,\{19,20\})</math>, the numbers differ by less than 2, so the probability that the numbers differ by at least 2 is <math>1-\tfrac{19}{190}=\tfrac9{10}</math>. Then <math>B-J\ge 2</math> holds in exactly half of these cases, so it has probability <math>\tfrac12\cdot\tfrac9{10}=\tfrac{9}{20}</math>. The requested sum is <math>9+20=29</math>.
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There are <math>\tbinom{20}{2}=190</math> equally likely pairs <math>\{J,B\}</math>. In <math>19</math> of these pairs <math>(\{1,2\},\{2,3\},\{3,4\}\dots,\{19,20\})</math>, the numbers differ by less than 2, so the probability that the numbers differ by at least 2 is <math>1-\tfrac{19}{190}=\tfrac9{10}</math>. Then <math>B-J\ge 2</math> holds in exactly half of these cases, so it has probability <math>\tfrac12\cdot\tfrac9{10}=\tfrac{9}{20}</math>. The requested sum is <math>9+20=\boxed{029}</math>.
  
 
==Video Solution #1(Easy Counting)==
 
==Video Solution #1(Easy Counting)==
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{{AIME box|year=2019|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2019|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Combinatorics Problems]]

Latest revision as of 18:47, 29 January 2025

Problem

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

$B-J \ne 0$ because $B \ne J$, so the probability that $B-J < 0$ is $\frac{1}{2}$ by symmetry.

The probability that $B-J = 1$ is $\frac{19}{20 \times 19} = \frac{1}{20}$ because there are 19 pairs: $(B,J) = (2,1), \ldots, (20,19)$.

The probability that $B-J \ge 2$ is $1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}$

Solution 2

By symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$, which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by $1$ (no zero, because the two numbers are distinct). There are $20 \cdot 19 = 380$ total possible combinations of $B$ and $J$, and $1 + 18 \cdot 2 + 1 = 38$ ones that form $P$, so $P = \frac{38}{380} = \frac{1}{10}$. Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$.

Solution 3

This problem is essentially asking how many ways there are to choose $2$ distinct elements from a $20$ element set such that no $2$ elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$, there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing $171$ by $380$, our desired probability is $\frac{171}{380} = \frac{9}{20}$. Thus, our answer is $9+20=\boxed{029}$. -Fidgetboss_4000

Solution 4

Create a grid using graph paper, with $20$ columns for the values of $J$ from $1$ to $20$ and $20$ rows for the values of $B$ from $1$ to $20$. Since $B$ cannot equal $J$, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$, we can mark the line where $B - J = 2$. Now we sum the number of squares that are on this line and below it. We get $171$. Then we find the number of total squares, which is $400 - 20 = 380$. Finally, we take the ratio $\frac{171}{380}$, which simplifies to $\frac{9}{20}$. Our answer is $9+20=\boxed{029}$.

Solution 5

We can see that if $B$ chooses $20$, $J$ can choose from $1$ through $18$ such that $B-J\geq 2$. If $B$ chooses $19$, $J$ has choices $1$~$17$. By continuing this pattern, $B$ will choose $3$ and $J$ will have $1$ option. Summing up the total, we get $18+17+\cdots+1$ as the total number of solutions. The total amount of choices is $20\times19$ (B and J must choose different numbers), so the probability is $\frac{18\cdot19\div2}{20\cdot19}=\frac{9}{20}$. Therefore, the answer is $9+20=\boxed{029}$

-eric2020

Solution 6

Similar to solution 4, we can go through the possible values of $J$ to find all the values of $B$ that makes $B-J\geq 2$. If $J$ chooses $1$, then $B$ can choose anything from $3$ to $20$. If $J$ chooses $2$, then $B$ can choose anything from $4$ to $20$. By continuing this pattern, we can see that there is $18+17+\cdots+1$ possible solutions. The amount of solutions is, therefore, $\frac{18\cdot19}{2}=171$. Now, because $B$ and $J$ must be different, we have $20\times19=380$ possible choices, so the probability is $\frac{171}{380}=\frac{9}{20}$. Therefore, the final answer is $9+20=\boxed{029}$

-josephwidjaja

Solution 7 (Official MAA)

There are $\tbinom{20}{2}=190$ equally likely pairs $\{J,B\}$. In $19$ of these pairs $(\{1,2\},\{2,3\},\{3,4\}\dots,\{19,20\})$, the numbers differ by less than 2, so the probability that the numbers differ by at least 2 is $1-\tfrac{19}{190}=\tfrac9{10}$. Then $B-J\ge 2$ holds in exactly half of these cases, so it has probability $\tfrac12\cdot\tfrac9{10}=\tfrac{9}{20}$. The requested sum is $9+20=\boxed{029}$.

Video Solution #1(Easy Counting)

https://youtu.be/JQdad7APQG8?t=245

Video Solution

https://www.youtube.com/watch?v=lh570eu8E0E

Video Solution 2

https://youtu.be/TSKcjht8Rfk?t=488

~IceMatrix

Video Solution 3

https://youtu.be/9X18wCiYw9M

~Shreyas S

See Also

2019 AIME I (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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