Difference between revisions of "2009 AIME I Problems/Problem 3"
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~Jerry_Guo | ~Jerry_Guo | ||
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+ | == Solution 3 == | ||
+ | Rewrite it as : <math>(P)^3</math><math>(1-P)^5</math>=<math>\frac {1}{25}<cmath>(P)^5</cmath>(P)^3</math> | ||
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+ | This can be simplified as <math>24P^2 -50P + 25 = 0</math> | ||
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+ | The can be factored into: <math>(4P-5)(6P-5)</math> | ||
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+ | This yields two solutions: <math>5/4</math>(ignored) and <math> \boxed {5/6}</math> | ||
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==Video Solution== | ==Video Solution== |
Revision as of 13:10, 21 January 2021
Contents
Problem
A coin that comes up heads with probability and tails with probability independently on each flip is flipped times. Suppose that 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 .
Solution 2
We start as shown above. However, as we get to , we square root both sides to get . We can do this because we know that both and are between and , so they are both positive. Now, we have:
Now, we get .
~Jerry_Guo
Solution 3
Rewrite it as : =
This can be simplified as
The can be factored into:
This yields two solutions: (ignored) and
Video Solution
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=P00iOJdQiL4
~Shreyas S
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.