Difference between revisions of "2012 AMC 12A Problems/Problem 11"
(Created page with "== Problem == Alex, Mel, and Chelsea play a game that has <math>6</math> rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For ea...") |
(→See Also) |
||
| Line 14: | Line 14: | ||
{{AMC12 box|year=2012|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2012|ab=A|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Revision as of 08:56, 4 July 2013
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 
![\[\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}\]](http://latex.artofproblemsolving.com/3/3/b/33b582e1d346e7d58fb28683bacf4579ee7b8dbc.png)
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. 
