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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



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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AMC 8 Combinatorics Math Olympiad Probability

Probability | AMC 8, 2010 | Problem no. 24

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

Probability | AMC-8, 2007 |Problem 24

A bag contains four pieces of paper, each labeled with one of the digits 1,2,3 or 4.  with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

  • \(\frac{3}{4}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)

Key Concepts



Number counting

Check the Answer

Answer: \(\frac{1}{2}\)

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

Try with Hints

there are two ways that The combination of digits that give multiples of 3

Can you now finish the problem ……….

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4)

can you finish the problem……..

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4) . The number of ways to choose three digits out of four is 4. Therefore, the probability is \(\frac{1}{2}\)

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Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

Combinations- AIME, 2009

A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

  • 110
  • 420
  • 430
  • 111

Key Concepts


Theory of equations


Check the Answer

Answer: 420.

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

Try with Hints

Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420

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