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

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Repeatedly Flipping a Fair Coin.

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.

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