2019 AIME I Problems/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 .
By symmetry, the desired probability is equal to the probability that is at most , which is where is the probability that and differ by 1 (no zero, because the two numbers are distinct). There are total possible combinations of and , and ones that form , so . Therefore the answer is .
This problem is asks 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 , there are ways. Dividing 171 by 380, our desired probability is . Thus, our answer is . -Fidgetboss_4000
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 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 .
We can see that if B chooses 20, J has options 1-18, such that . 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 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
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