Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on **Probability of tossing a coin**.

## Probability of tossing a coin – AIME I, 2009 Question 3

A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to \(\frac{1}{25}\) the probability of five heads and three tails. Let p=\(\frac{m}{n}\) where m and n are relatively prime positive integers. Find m+n.

- 10
- 20
- 30
- 11

**Key Concepts**

Probability

Theory of equations

Polynomials

## Check the Answer

Answer: 11.

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

## Try with Hints

here \(\frac{8!}{3!5!}p^{3}(1-p)^{5}\)=\(\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}\)

then \((1-p)^{2}\)=\(\frac{1}{25}p^{2}\) then 1-p=\(\frac{1}{5}p\)

then p=\(\frac{5}{6}\) then m+n=11

## Other useful links

- https://www.cheenta.com/functions-and-equations-pre-rmo-2019/
- https://www.youtube.com/watch?v=65RRPvbATsk