Difference between revisions of "2019 AIME I Problems/Problem 2"
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− | The | + | ==Problem== |
+ | Jenn randomly chooses a number <math>J</math> from <math>1, 2, 3,\ldots, 19, 20</math>. Bela then randomly chooses a number <math>B</math> from <math>1, 2, 3,\ldots, 19, 20</math> distinct from <math>J</math>. The value of <math>B - J</math> is at least <math>2</math> with a probability that can be expressed in the form <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | == | + | ==Solution 1== |
− | + | <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. | |
− | ==Solution== | + | 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> |
+ | |||
+ | ==Solution 2== | ||
+ | By symmetry, the desired probability is equal to the probability that <math>J - B</math> is at most <math>-2</math>, which is <math>\frac{1-P}{2}</math> where <math>P</math> is the probability that <math>B</math> and <math>J</math> differ by <math>1</math> (no zero, because the two numbers are distinct). There are <math>20 \cdot 19 = 380</math> total possible combinations of <math>B</math> and <math>J</math>, and <math>1 + 18 \cdot 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{38}{380} = \frac{1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \rightarrow \boxed{029}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | This problem is essentially asking how many ways there are to choose <math>2</math> distinct elements from a <math>20</math> element set such that no <math>2</math> elements are adjacent. Using the well-known formula <math>\dbinom{n-k+1}{k}</math>, there are <math>\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171</math> ways. Dividing <math>171</math> by <math>380</math>, our desired probability is <math>\frac{171}{380} = \frac{9}{20}</math>. Thus, our answer is <math>9+20=\boxed{029}</math>. | ||
+ | -Fidgetboss_4000 | ||
+ | |||
+ | ==Solution 4== | ||
+ | Create a grid using graph paper, with <math>20</math> columns for the values of <math>J</math> from <math>1</math> to <math>20</math> and <math>20</math> rows for the values of <math>B</math> from <math>1</math> to <math>20</math>. Since <math>B</math> cannot equal <math>J</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=\boxed{029}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | We can see that if <math>B</math> chooses <math>20</math>, <math>J</math> can choose from <math>1</math> through <math>18</math> such that <math>B-J\geq 2</math>. If <math>B</math> chooses <math>19</math>, <math>J</math> has choices <math>1</math>~<math>17</math>. By continuing this pattern, <math>B</math> will choose <math>3</math> and <math>J</math> will have <math>1</math> option. Summing up the total, we get <math>18+17+\cdots+1</math> as the total number of solutions. The total amount of choices is <math>20\times19</math> (B and J must choose different numbers), so the probability is <math>\frac{18\cdot19\div2}{20\cdot19}=\frac{9}{20}</math>. Therefore, the answer is <math>9+20=\boxed{029}</math> | ||
+ | |||
+ | -eric2020 | ||
+ | |||
+ | ==Solution 6== | ||
+ | Similar to solution 4, we can go through the possible values of <math>J</math> to find all the values of <math>B</math> that makes <math>B-J\geq 2</math>. If <math>J</math> chooses <math>1</math>, then <math>B</math> can choose anything from <math>3</math> to <math>20</math>. If <math>J</math> chooses <math>2</math>, then <math>B</math> can choose anything from <math>4</math> to <math>20</math>. By continuing this pattern, we can see that there is <math>18+17+\cdots+1</math> possible solutions. The amount of solutions is, therefore, <math>\frac{18\cdot19}{2}=171</math>. Now, because <math>B</math> and <math>J</math> must be different, we have <math>20\times19=380</math> possible choices, so the probability is <math>\frac{171}{380}=\frac{9}{20}</math>. Therefore, the final answer is <math>9+20=\boxed{029}</math> | ||
+ | |||
+ | -josephwidjaja | ||
+ | ==Solution 7 (Official MAA)== | ||
+ | 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>. | ||
+ | |||
+ | ==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== | ==See Also== | ||
{{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}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 22:15, 28 December 2023
Contents
Problem
Jenn randomly chooses a number from
. Bela then randomly chooses a number
from
distinct from
. The value of
is at least
with a probability that can be expressed in the form
where
and
are relatively prime positive integers. Find
.
Solution 1
because
, so the probability that
is
by symmetry.
The probability that is
because there are 19 pairs:
.
The probability that is
Solution 2
By symmetry, the desired probability is equal to the probability that is at most
, which is
where
is the probability that
and
differ by
(no zero, because the two numbers are distinct). There are
total possible combinations of
and
, and
ones that form
, so
. Therefore the answer is
.
Solution 3
This problem is essentially asking how many ways there are to choose distinct elements from a
element set such that no
elements are adjacent. Using the well-known formula
, there are
ways. Dividing
by
, our desired probability is
. Thus, our answer is
.
-Fidgetboss_4000
Solution 4
Create a grid using graph paper, with columns for the values of
from
to
and
rows for the values of
from
to
. Since
cannot equal
, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since
must be at least
, we can mark the line where
. Now we sum the number of squares that are on this line and below it. We get
. Then we find the number of total squares, which is
. Finally, we take the ratio
, which simplifies to
. Our answer is
.
Solution 5
We can see that if chooses
,
can choose from
through
such that
. If
chooses
,
has choices
~
. By continuing this pattern,
will choose
and
will have
option. Summing up the total, we get
as the total number of solutions. The total amount of choices is
(B and J must choose different numbers), so the probability is
. Therefore, the answer is
-eric2020
Solution 6
Similar to solution 4, we can go through the possible values of to find all the values of
that makes
. If
chooses
, then
can choose anything from
to
. If
chooses
, then
can choose anything from
to
. By continuing this pattern, we can see that there is
possible solutions. The amount of solutions is, therefore,
. Now, because
and
must be different, we have
possible choices, so the probability is
. Therefore, the final answer is
-josephwidjaja
Solution 7 (Official MAA)
There are equally likely pairs
. In
of these pairs
, the numbers differ by less than 2, so the probability that the numbers differ by at least 2 is
. Then
holds in exactly half of these cases, so it has probability
. The requested sum is
.
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
~Shreyas S
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.