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Probability Biased and Unbiased | AIME I, 2010 Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability Biased and Unbiased.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.

Probability Biased and Unbiased – AIME 2010


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.

  • is 107
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Probability

Number Theory

Check the Answer


Answer: is 107.

AIME, 2010, Question 4

Combinatorics by Brualdi

Try with Hints


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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