Categories
AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Fair coin Problem | AIME I, 1990 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Fair Coin Problem.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on fair coin.

Fair Coin Problem – AIME I, 1990


A fair coin is to be tossed 10 times. Let i|j, in lowest terms, be the probability that heads never occur on consecutive tosses, find i+j.

  • is 107
  • is 73
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Combinatorics

Algebra

Check the Answer


But try the problem first…

Answer: is 73.

Source
Suggested Reading

AIME I, 1990, Question 9

Elementary Algebra by Hall and Knight

Try with Hints


First hint

5 tails flipped, any less,

by Pigeonhole principle there will be heads that appear on consecutive tosses

Second Hint

(H)T(H)T(H)T(H)T(H)T(H) 5 tails occur there are 6 slots for the heads to be placed but only 5H remaining, \({6 \choose 5}\) possible combination of 6 heads there are

Final Step

\(\sum_{i=6}^{11}{i \choose 11-i}\)=\({6 \choose 5} +{7 \choose 4}+{8 \choose 3}+{9 \choose 2} +{10 \choose 1} +{11 \choose 0}\)=144

there are \(2^{10}\) possible flips of 10 coins

or, probability=\(\frac{144}{1024}=\frac{9}{64}\) or, 9+64=73.

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.