Difference between revisions of "2009 AIME I Problems/Problem 3"
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<cmath>p = \frac {5} {6}</cmath> | <cmath>p = \frac {5} {6}</cmath> | ||
<cmath>5 + 6 = \boxed{11}</cmath> | <cmath>5 + 6 = \boxed{11}</cmath> | ||
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| + | == See also == | ||
| + | {{AIME box|year=2009|n=I|num-b=2|num-a=4}} | ||
Revision as of 19:47, 19 March 2009
Problem
A coin that comes up heads with probability 
and tails with probability 
independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to 
of the probability of five heads and three tails. Let 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
If we let the odds of a tails 
equal 
, then the probability of three heads and five tails is 
The probability of five heads and three tails is 
![\[25{p^3}{t^5} = {p^5}{t^3}\]](http://latex.artofproblemsolving.com/4/b/a/4ba1985da4959a7ecdda2d1e60841f091c4ca89f.png)
![\[25{t^2} = {p^2}\]](http://latex.artofproblemsolving.com/8/1/b/81bc64a9ee6fbf2e68ea27e126d13f71e09b6487.png)
![\[5t = p\]](http://latex.artofproblemsolving.com/3/7/f/37f7b058d54fdc37061dd2272ad7a1151fd13193.png)
![\[5(1 - p) = p\]](http://latex.artofproblemsolving.com/7/a/e/7aea3da60219be77ea261a138400e633f49ce6b4.png)
![\[5 - 5p = p\]](http://latex.artofproblemsolving.com/a/0/5/a05c5089bfa12660b991fb6862b169eea49ae832.png)
![\[5 = 6p\]](http://latex.artofproblemsolving.com/0/a/c/0ac38ba1f0c4e29040330202264e751314957453.png)
![\[p = \frac {5} {6}\]](http://latex.artofproblemsolving.com/0/e/c/0ecbd5351614a18d24439a4ad42c0655053a6bb8.png)
![\[5 + 6 = \boxed{11}\]](http://latex.artofproblemsolving.com/b/f/e/bfed68cbfc0e1324f5fac69d3f24bd8f5687cc6c.png)
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
