How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

Learn MoreContent

[hide]

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

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

Integers

Probability

Algebra

But try the problem first...

Answer: is 37.

Source

Suggested Reading

AIME I, 1995, Question 15

Elementary Number Theory by David Burton

First hint

Let A be head flipped

B be tail flipped

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

Second Hint

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

Final Step

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL