Difference between revisions of "2009 AIME I Problems/Problem 3"
Aryabhata000 (talk | contribs) (→Solution) |
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25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ | 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ | ||
25(1-p)^2&=p^2 \\ | 25(1-p)^2&=p^2 \\ | ||
| − | + | 25p^2-50p+25=p^2 \\ | |
| − | + | 24p^2-50p+25=0 \\ | |
| − | |||
p&=\frac {5}{6}\end{align*}</cmath> | p&=\frac {5}{6}\end{align*}</cmath> | ||
Revision as of 11:00, 3 August 2019
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
The probability of three heads and five tails is 
and the probability of five heads and three tails is 
.
Therefore, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
