Difference between revisions of "2012 AMC 12A Problems/Problem 11"
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<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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+ | ==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 | ||
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+ | ~IceMatrix | ||
== See Also == | == See Also == |
Revision as of 20:40, 20 October 2021
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
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
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.