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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.

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


First hint

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 \(\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}\)=\(\frac{100}{784}\)

Second Hint

Two heads HHT \(\frac{4}{28}\) HTH \(\frac{4}{28}\) THH \(\frac{3}{28}\) then probability is \(\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}\)=\(\frac{121}{784}\).

Final Step

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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