Difference between revisions of "2019 AIME I Problems/Problem 2"
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-eric2020 | -eric2020 | ||
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| + | ==Solution 5== | ||
| + | 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*19/2}=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> | ||
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| + | -josephwidjaja | ||
==Video Solution == | ==Video Solution == | ||
Revision as of 10:49, 7 June 2020
Contents
Problem 2
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
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 2
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 3
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 4
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 5
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
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
Video Solution 2
https://youtu.be/TSKcjht8Rfk?t=488
~IceMatrix
Video Solution 3
https://www.youtube.com/watch?v=nbtIBP6Auig&t=460s
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. 
