Difference between revisions of "2009 AIME I Problems/Problem 3"
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==Video Solution 2== | ==Video Solution 2== |
Revision as of 14:26, 29 December 2020
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
Video Solution
~IceMatrixx
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.