Difference between revisions of "Talk:2010 AIME I Problems/Problem 4"

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This can also be solved with generating functions.
 
  
Let <math>x^0 = 1</math> represent heads and <math>x</math> represent tails.
 
 
The generating functions for these coins are <math>(7+7x)</math>,<math>(7+7x)</math>,and <math>(8+6x)</math> in order. (weighted)
 
 
The product is <math>98(3+10x+11x^2+4x^3)</math>.
 
 
The sum of the coefficients squared is 784 and the sum of the squares of each coefficient is 246.
 
The probability is then <math> \frac{246}{784} = \frac{123}{392}</math>.
 
 
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<math>123 + 392 = 515</math>
 

Latest revision as of 16:47, 18 November 2011