2019 AIME I Problems/Problem 2
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
The probability that is by symmetry. because . 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 | |
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All AIME Problems and Solutions |
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