Difference between revisions of "2019 AIME I Problems/Problem 2"
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− | We can see that if B chooses 20, J | + | 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*19/2}{20*19}=\frac{9}{20}</math>. Therefore, the answer is <math>9+20=\boxed{029}</math> |
-eric2020 | -eric2020 |
Revision as of 11:51, 26 December 2019
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 asks 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
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
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.