Question

一枚硬币每次独立抛掷时，正面朝上的概率为 $p \gt 0$ ，反面朝上的概率为 $1 - p \gt 0$ 。将这枚硬币抛掷八次。已知出现三次正面、五次反面的概率，等于出现五次正面、三次反面的概率的 $\frac{1}{25}$ 。设 $p = \frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

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 $m + n.$

答案

答案：11

知识点：二项概率代数变形

难度评级：2150

解答：

题设说明 $\begin{aligned} &\binom{8}{3} p^3 (1-p)^5 \\ &= \frac{1}{25} \binom{8}{5} p^5 (1-p)^3. \end{aligned}$ 因为 $\binom{8}{3} = \binom{8}{5}$ ，且 $p$ 与 $1 - p$ 都为正，除以 $p^3(1-p)^3$ 后得到 $(1-p)^2 = \frac{p^2}{25}$ ，所以 $1 - p = \frac{p}{5}$ 。

因此 $p = \frac{5}{6}$ ，所以 $m + n = 5 + 6 = 11$ 。

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 第 3 题

答案：11

难度评级：2150

解答：

其他年份的第 3 题