Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.
A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to \(\frac{1}{25}\) the probability of five heads and three tails. Let p=\(\frac{m}{n}\) where m and n are relatively prime positive integers. Find m+n.
Probability
Theory of equations
Polynomials
Answer: 11.
AIME, 2009
Course in Probability Theory by Kai Lai Chung .
here \(\frac{8!}{3!5!}p^{3}(1-p)^{5}\)=\(\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}\)
then \((1-p)^{2}\)=\(\frac{1}{25}p^{2}\) then 1-p=\(\frac{1}{5}p\)
then p=\(\frac{5}{6}\) then m+n=11
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