Difference between revisions of "2009 AIME I Problems/Problem 3"
Aryabhata000 (talk | contribs) (→Solution) |
|||
| Line 10: | Line 10: | ||
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
