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Probability in Games | AIME I, 1999 | Question 13

Try this beautiful problem from American Invitational Mathematics Examination, AIME, 1999 based on Probability in Games. You may use sequential hints.

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.

Probability in Games – AIME I, 1999 Question 13


Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find \(log_{2}n\)

  • 10
  • 742
  • 30
  • 11

Key Concepts


Probability

Theory of equations

Combinations

Check the Answer


But try the problem first…

Answer: 742.

Source
Suggested Reading

AIME, 1999 Q13

Course in Probability Theory by Kai Lai Chung .

Try with Hints


First hint

\({40 \choose 2}\)=780 pairings with \(2^{780}\) outcomes

Second hint

no two teams win the same number of games=40! required probability =\(\frac{40!}{2^{780}}\)

Final Step

the number of powers of 2 in 40!=[\(\frac{40}{2}\)]+[\(\frac{40}{4}\)]+[\(\frac{40}{8}\)]+[\(\frac{40}{16}\)]+[\(\frac{40}{32}\)]=20+10+5+2+1=38 then 780-38=742.

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