2009 AIME I Problems/Problem 3
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