Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.
Ramesh and Suresh have two fair coins and a third coin that comes up heads with probability \(\frac{4}{7}\),Ramesh flips the three coins, and then Suresh flips the three coins, let \(\frac{m}{n}\) be the probability that Ramesh gets the same number of heads as Suresh, where m and n are relatively prime positive integers. Find m+n.
Series
Probability
Number Theory
Answer: is 107.
AIME, 2010, Question 4
Combinatorics by Brualdi
No heads TTT is \(\frac{1.1.1}{2.2.7}=\frac{3}{28}\)and \((\frac{3}{28})^{2}=\frac{9}{784}\)
One Head HTT THT TTH with \(\frac{3}{28}\) \(\frac {3}{28}\) and \(\frac{4}{28}\) then probability is \(\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}\)=\(\frac{100}{784}\)
Two heads HHT \(\frac{4}{28}\) HTH \(\frac{4}{28}\) THH \(\frac{3}{28}\) then probability is \(\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}\)=\(\frac{121}{784}\).
Three heads HHH is \(\frac{4}{28}\) then probability \(\frac{16}{784}\)
Then sum is \(\frac{9+100+121+16}{784}=\frac{123}{392}\) then 123+392=515.
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