Difference between revisions of "2022 AIME II Problems/Problem 2"
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Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability <math>\frac23</math>. When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability <math>\frac34</math>. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability <math>\frac23</math>. When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability <math>\frac34</math>. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
− | ==Solution | + | ==Solution== |
Let <math>A</math> be Azar, <math>C</math> be Carl, <math>J</math> be Jon, and <math>S</math> be Sergey. The <math>4</math> circles represent the <math>4</math> players, and the arrow is from the winner to the loser with the winning probability as the label. | Let <math>A</math> be Azar, <math>C</math> be Carl, <math>J</math> be Jon, and <math>S</math> be Sergey. The <math>4</math> circles represent the <math>4</math> players, and the arrow is from the winner to the loser with the winning probability as the label. | ||
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For this to happen, <math>A</math> must have won <math>J</math> or <math>S</math> in the semifinal, the probability is <math>\frac34</math>. Therefore, the probability that <math>C</math> won <math>A</math> in the final is <math>\frac34 \cdot \frac13</math>. | For this to happen, <math>A</math> must have won <math>J</math> or <math>S</math> in the semifinal, the probability is <math>\frac34</math>. Therefore, the probability that <math>C</math> won <math>A</math> in the final is <math>\frac34 \cdot \frac13</math>. | ||
− | <math>\textbf{Case 2. | + | <math>\textbf{Case 2.2:}</math> <math>C</math>'s opponent for the final is <math>J</math> or <math>S</math> |
For this to happen, <math>J</math> or <math>S</math> must have won <math>A</math> in the semifinal, the probability is <math>\frac14</math>. Therefore, the probability that <math>C</math> won <math>J</math> or <math>S</math> in the final is <math>\frac14 \cdot \frac34</math>. | For this to happen, <math>J</math> or <math>S</math> must have won <math>A</math> in the semifinal, the probability is <math>\frac14</math>. Therefore, the probability that <math>C</math> won <math>J</math> or <math>S</math> in the final is <math>\frac14 \cdot \frac34</math>. | ||
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In Case 2 the probability of <math>C</math> winning the tournament is <math>\frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34)</math> | In Case 2 the probability of <math>C</math> winning the tournament is <math>\frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34)</math> | ||
− | Adding case 1 and case 2 together we get <math>\frac13 \cdot \frac13 \cdot \frac34 + \frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34) = \frac{29}{96}</math> | + | Adding case 1 and case 2 together we get <math>\frac13 \cdot \frac13 \cdot \frac34 + \frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34) = \frac{29}{96},</math> so the answer is <math>29 + 96 = \boxed{\textbf{125}}</math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== | ||
https://www.youtube.com/watch?v=C14f91P2pYc | https://www.youtube.com/watch?v=C14f91P2pYc | ||
+ | |||
+ | ==Video Solution by Power of Logic== | ||
+ | https://youtu.be/vvCes96cPm4 | ||
==See Also== | ==See Also== |
Latest revision as of 16:15, 24 January 2023
Contents
Problem
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability
. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is
, where
and
are relatively prime positive integers. Find
.
Solution
Let be Azar,
be Carl,
be Jon, and
be Sergey. The
circles represent the
players, and the arrow is from the winner to the loser with the winning probability as the label.
This problem can be solved by using cases.
's opponent for the semifinal is
The probability 's opponent is
is
. Therefore the probability
wins the semifinal in this case is
. The other semifinal game is played between
and
, it doesn't matter who wins because
has the same probability of winning either one. The probability of
winning in the final is
, so the probability of
winning the tournament in case 1 is
's opponent for the semifinal is
or
It doesn't matter if 's opponent is
or
because
has the same probability of winning either one. The probability
's opponent is
or
is
. Therefore the probability
wins the semifinal in this case is
. The other semifinal game is played between
and
or
. In this case it matters who wins in the other semifinal game because the probability of
winning
and
or
is different.
's opponent for the final is
For this to happen, must have won
or
in the semifinal, the probability is
. Therefore, the probability that
won
in the final is
.
's opponent for the final is
or
For this to happen, or
must have won
in the semifinal, the probability is
. Therefore, the probability that
won
or
in the final is
.
In Case 2 the probability of winning the tournament is
Adding case 1 and case 2 together we get so the answer is
.
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=C14f91P2pYc
Video Solution by Power of Logic
See Also
2022 AIME II (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.