Question

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

Answer

Answer: 11

Concepts:binomial probability algebraic manipulation

Difficulty rating: 2150

Solution:

The condition says $\begin{aligned} &\binom{8}{3} p^3 (1-p)^5 \\ &= \frac{1}{25} \binom{8}{5} p^5 (1-p)^3. \end{aligned}$ Since $\binom{8}{3} = \binom{8}{5}$ and both $p$ and $1 - p$ are positive, dividing by $p^3(1-p)^3$ leaves $(1-p)^2 = \frac{p^2}{25},$ so $1 - p = \frac{p}{5}.$

Hence $p = \frac{5}{6},$ and $m + n = 5 + 6 = 11.$

2009 AIME I Problem 3

Answer: 11

Difficulty rating: 2150

Solution:

Problem 3 in Other Years