2019 AIME I Problems/Problem 2
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
Realize that 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
.
Solution 2
This problem is basically asking 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
Solution 3
Simply 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 we know that 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
.
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.