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Talk:2010 AIME I Problems/Problem 4

Revision as of 14:52, 5 March 2011 by Baijiangchen (talk | contribs) (Created page with '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 <ma…')
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This can also be solved with generating functions.

Let $x^0 = 1$ represent heads and $x$ represent tails.

The generating functions for these coins are $(7+7x)$,$(7+7x)$,and $(8+6x)$ in order. (weighted)

The product is $98(3+10x+11x^2+4x^3)$.

The sum of the coefficients squared is 784 and the sum of the squares of each coefficient is 246. The probability is then $\frac{246}{784} = \frac{123}{392}$.

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$123 + 392 = 515$

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