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Difference between revisions of "2015 AIME I Problems/Problem 2"

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==Problem==
 
==Problem==
 
The nine delegates to the Economic Cooperation Conference include <math>2</math> officials from Mexico, <math>3</math> officials from Canada, and <math>4</math> 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
The nine delegates to the Economic Cooperation Conference include <math>2</math> officials from Mexico, <math>3</math> officials from Canada, and <math>4</math> 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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+
==Video Solution For Problems 1-3==
 +
https://www.youtube.com/watch?v=Vhlo_2zx2NY
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==
 
https://youtu.be/IRyWOZQMTV8?t=1802
 
https://youtu.be/IRyWOZQMTV8?t=1802

Revision as of 23:03, 8 January 2023

Problem

The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ 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 $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Video Solution For Problems 1-3

https://www.youtube.com/watch?v=Vhlo_2zx2NY

Video Solution by OmegaLearn

https://youtu.be/IRyWOZQMTV8?t=1802

~ pi_is_3.14

Solution

The total number of ways to pick $3$ officials from $9$ total is $\binom{9}{3} = 84$, 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 $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ "extra" delegates to choose to be the third sleeper once both from Mexico are sleeping.
  • There are $3\times6=18$ ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are $9-3=6$ "extra" people to select one more sleeper from.
  • Lastly, there are $6\times5=30$ ways to choose for the United States. It is easy to count $6$ different ways to pick $2$ of the $4$ Americans, and each time you do there are $9-4=5$ officials left over to choose from.

Thus, the fraction is $\frac{7+18+30}{84} = \frac{55}{84}$. Since this does not reduce, the answer is $55+84=\boxed{139}$.

Solution 2

Like in the solution above, there are $84$ ways to pick $3$ delegates. We can use casework to find the probability that there aren't exactly $2$ sleepers from a county, then subtract from $1$.

  • If no country has at least $2$ delegates sleeping, then every country must have $1$ delegate sleeping. There are $2*3*4=24$ ways for this to happen.
  • If all $3$ sleeping delegates are from Canada, there are $\binom{3}{3} = 1$ way.
  • If all $3$ are from the US, there are $\binom{4}{3} = 4$ ways.

So, the probability that there are not exactly $2$ sleepers from one country is $\frac{24+1+4}{84} = \frac{29}{84}$, and the probability that exactly $2$ are from the same country is $1- \frac{29}{84} = \frac{55}{84}.$ Our answer is $55+84=\boxed{139}$.

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

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