Difference between revisions of "2012 AMC 12A Problems/Problem 11"
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<math> \textbf{(A)}\ \frac{5}{72}\qquad\textbf{(B)}\ \frac{5}{36}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ 1 </math> | <math> \textbf{(A)}\ \frac{5}{72}\qquad\textbf{(B)}\ \frac{5}{36}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ 1 </math> | ||
− | == Solution == | + | == Solution 1 == |
If <math>m</math> is the probability Mel wins and <math>c</math> is the probability Chelsea wins, <math>m=2c</math> and <math>m+c=\frac12</math>. From this we get <math>m=\frac13</math> and <math>c=\frac16</math>. For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is <math>\frac{1}{2^3\cdot3^2\cdot6}=\frac{1}{432}</math>. Multiply this by the number of permutations (orders they can win) which is <math>\frac{6!}{3!2!1!}=60.</math> | If <math>m</math> is the probability Mel wins and <math>c</math> is the probability Chelsea wins, <math>m=2c</math> and <math>m+c=\frac12</math>. From this we get <math>m=\frac13</math> and <math>c=\frac16</math>. For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is <math>\frac{1}{2^3\cdot3^2\cdot6}=\frac{1}{432}</math>. Multiply this by the number of permutations (orders they can win) which is <math>\frac{6!}{3!2!1!}=60.</math> | ||
<cmath>\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}</cmath> | <cmath>\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}</cmath> | ||
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+ | == Solution 2 == | ||
+ | |||
+ | The probability that Alex wins is <math>\frac{1}{2}</math> , the probability that Mel wins is <math>\frac{1}{3}</math> , and the probability that Chelsea wins is <math>\frac{1}{6}</math> . There are <math>\binom{6}{3}</math> ways to pick the three games that Alex wins, and <math>\binom{3}{2}</math> ways to pick 2 out of the 3 remaining games that Mel wins - theres <math>\binom{1}{1}</math> ways to see that Chelsea wins. Therefore, the number of ways is | ||
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+ | <cmath>\frac{1}{2^3}\times\frac{1}{3^2}\times\frac{1}{6}\times\binom{6}{3}\times\binom{3}{2} = \boxed{\textbf{(B)}\ \frac{5}{36}}</cmath> | ||
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+ | ~ youtube.com/@indianmathguy | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | I preview the concept that will be used in the first part of this video, and show why it's important "Don't Memorize, Understand". I adapt the solution method of an older problem to inform the solution of this one: https://youtu.be/PO3XZaSchJc | ||
+ | |||
+ | ~IceMatrix | ||
== See Also == | == See Also == |
Latest revision as of 10:16, 8 June 2024
Problem
Alex, Mel, and Chelsea play a game that has rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is
, and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?
Solution 1
If is the probability Mel wins and
is the probability Chelsea wins,
and
. From this we get
and
. For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is
. Multiply this by the number of permutations (orders they can win) which is
Solution 2
The probability that Alex wins is , the probability that Mel wins is
, and the probability that Chelsea wins is
. There are
ways to pick the three games that Alex wins, and
ways to pick 2 out of the 3 remaining games that Mel wins - theres
ways to see that Chelsea wins. Therefore, the number of ways is
~ youtube.com/@indianmathguy
Video Solution by TheBeautyofMath
I preview the concept that will be used in the first part of this video, and show why it's important "Don't Memorize, Understand". I adapt the solution method of an older problem to inform the solution of this one: https://youtu.be/PO3XZaSchJc
~IceMatrix
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.