2015 AIME I Problems/Problem 2
Problem
The nine delegates to the Economic Cooperation Conference include officials from Mexico,
officials from Canada, and
officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is
, where
and
are relatively prime positive integers. Find
.
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=1802
~ pi_is_3.14
Solution
The total number of ways to pick officials from
total is
, which will be our denominator. Now we can consider the number of ways for exactly two sleepers to be from the same country for each country individually and add them to find our numerator:
- There are
different ways to pick
delegates such that
are from Mexico, simply because there are
"extra" delegates to choose to be the third sleeper once both from Mexico are sleeping.
- There are
ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are
"extra" people to select one more sleeper from.
- Lastly, there are
ways to choose for the United States. It is easy to count
different ways to pick
of the
Americans, and each time you do there are
officials left over to choose from.
Thus, the fraction is . Since this does not reduce, the answer is
.
Solution 2
Like in the solution above, there are ways to pick
delegates. We can use casework to find the probability that there aren't exactly
sleepers from a county, then subtract from
.
- If no country has at least
delegates sleeping, then every country must have
delegate sleeping. There are
ways for this to happen.
- If all
sleeping delegates are from Canada, there are
way.
- If all
are from the US, there are
ways.
So, the probability that there are not exactly sleepers from one country is
, and the probability that exactly
are from the same country is
Our answer is
.
See also
2015 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.