INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

May 10, 2020

Repeatedly Flipping a Fair Coin | AIME I, 1995| Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 repeatedly flipping a fair coin.

Flipping a Fair Coin - AIME I, 1995


Let p be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before on encounters a run of 2 tails. Given that p can be written in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 37
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Probability

Algebra

Check the Answer


Answer: is 37.

AIME I, 1995, Question 15

Elementary Number Theory by David Burton

Try with Hints


Let A be head flipped

B be tail flipped

outcomes are AAAAA, BAAAAA, BB. ABB, AABB, AAABB, AAAABB

with probabilities \(\frac{1}{32}\), \(\frac{1}{64}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\), \(\frac{1}{64}\)

with five heads AAAAA, BAAAAA sum =\(\frac{3}{64}\) and sum of outcomes=\(\frac{34}{64}\)

or, m=3, n=34

or, m+n=37.

Subscribe to Cheenta at Youtube


Leave a Reply

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

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com