Difference between revisions of "2009 AIME I Problems/Problem 3"
(New page: == 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 t...) |
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== Problem == | == Problem == | ||
| − | A coin that comes up heads with probability | + | A coin that comes up heads with probability <math>p > 0</math> and tails with probability <math>1 - p > 0</math> independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to <math>\frac {1}{25}</math> of the probability of five heads and three tails. Let <math>p = \frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. |
== Solution == | == Solution == | ||
Revision as of 20:40, 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 (1-p) equal t, then the probablity of three heads and five tails is: p^3t^5 The probability of five heads and three tails is: p^5t^3
25p^3t^5 = p^5t^3 25t^2 = p^2 5t = p 5(1-p) = p 5 - 5p = p 5 = 6p p = 5/6
