Three components are connected to form a system as shown in the accompanying diagram. Because the components in the 2-3 subsystem are connected in parallel, that subsystem will function if at least one of the two individual components functions. For the entire system to function, component 1 must function and so must the 2-3 subsystem.

The experiment consists of determining the condition of each component [$S$ (success) for a functioning component and $F$ (failure) for a nonfunctioning component].

A. What outcomes are contained in the event $A$ that exactly two out of the three components function?
B. What outcomes are contained in the event $B$ that at least two of the components function?
C. What outcomes are contained in the event $C$ that the system functions?
D. List outcomes in $C^{c}$, $A\cup C$, $A\cap C$, $B\cup C$, and $B\cap C$.
 

Reference:

Jay L. Devore, Probability and Statistics, For Engineering and The Sciences (Fifth Edtion)

Four universities -- 1,2,3 and 4 -- are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4. Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 (1 beats 2 and 3 beats 4 in first round games, and then 1 beats 3 and 2 beats 4).

A. List all outcomes in $\mathcal{S}$.
B. Let $A$ denote the event that 1 wins the tournament. List outcomes in $A$.
C. Let $B$ denote the event that 2 gets into the championship game. List outcomes in $B$.
D. What are the outcomes in $A\cup B$ and in $A\cap B$? What are the outcomes in $A^{c}$?
 

Reference:

Jay L. Devore, Probability and Statistics, For Engineering and The Sciences (Fifth Edtion)

Definition. Let $\Omega$ be a finite or countable set. Let $p:\Omega\rightarrow[0,1]$ be a function such that \[\sum_{\omega\in\Omega}p_{\omega}=1.\]

Then $(\Omega,p)$ is called a discrete probability space. $\Omega$ is called the sample space and $p_{\omega}$ are called elementary probabilities.

定义: 设 $\Omega$ 是一个有限集或可数集. 设 $p:\Omega\rightarrow[0,1]$ 是 $\Omega$ 上的一个函数, 满足 \[\sum_{\omega\in\Omega}p_{\omega}=1.\]

则 $(\Omega,p)$ 被称为一个离散概率空间. $\Omega$ 被称为样本空间, $p_{\omega}$ 称作为基本概率.

References:

Manjunath Krishnapur, Probability and Statistics

在 $20\times 5$ 的格子上, 有90个白棋, 10个黑棋. 任意放在每个格子中. 求

Q1. 存在一行有3个以上黑棋的概率.

Q2. 存在两行有3个以上黑棋的概率.

Q3. 存在三行有3个以上黑棋的概率.

Remark:

题目来源: David Chen(陈)

证明: 正态分布的峰度是3.

峰度的定义

\[
g_2=\frac{1}{s^4}\sum_{i=1}^{n}(X_i-\bar{X})^4,
\]

其中 $s$ 指标准差, 定义为

\[
s=\biggl[\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2\bigg]^{\frac{1}{2}}.
\]

性质. ($\chi^2$-分布的可加性)

若 $Y_1,Y_2,\ldots,Y_k$ 相互独立且都服从 $\chi^2$-分布, 自由度分别为 $n_1,n_2,\ldots,n_k$. 即

\[
Y_i\sim\chi^2(n_i),\quad i=1,2,\ldots,k,
\]

则

\[
\sum_{i=1}^{k}Y_i\sim\chi^2(n),\quad\text{where}\ n=\sum_{i=1}^{k}n_i.
\]

设 $X$ 和 $Y$ 是分布服从 $N(\mu_1,\sigma_1^2)$ 和 $N(\mu_2,\sigma_2^2)$ 的两个总体. 分别从它们中抽取容量为 $n_1$ 和 $n_2$ 的样本 $\{X_i\}_{i=1}^{n_1}$, $\{Y_j\}_{j=1}^{n_2}$. 假设所有的抽样都是相互独立的, 由此得到的样本 $X_i$ 和 $Y_j$ 都是相互独立的随机变量.

设样本 $\{X_i\}_{i=1}^{n_1}$, $\{Y_j\}_{j=1}^{n_2}$ 的均值分别为 $\bar{X}$ 和 $\bar{Y}$, 样本方差分别为 $s_1^2$ 和 $s_2^2$, 则有

(1)

\[
U:=\frac{(\bar{X}-\bar{Y})-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\ \sim\ N(0,1).
\]

特别地, 若 $\sigma_1=\sigma_2=\sigma$ 时, 有

\[
U:=\frac{(\bar{X}-\bar{Y})-(\mu_1-\mu_2)}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\ \sim\ N(0,1).
\]

(2)

\[
F:=\frac{\frac{s_1^2}{\sigma_1^2}}{\frac{s_2^2}{\sigma_2^2}}\ \sim\ F(n_1-1,n_2-1).
\]

(3)

若 $\sigma_1=\sigma_2$, 则

\[
T:=\frac{(\bar{X}-\bar{Y})-(\mu_1-\mu_2)}{s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\ \sim\ t(n_1+n_2-2),
\]

其中

\[
s=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}.
\]

References:

赵静、但琦 等编 《数学建模与数学实验》（第四版）,  高等教育出版社.  P. 186.

$t$ 分布 $t(n)$

若 $X\sim N(0,1)$, $Y\sim\chi^2(n)$, 且互相独立, 则随机变量 $T:=\frac{X}{\sqrt{\frac{Y}{n}}}$ 服从自由度为 $n$ 的 $t$-分布, 记为 $T\sim t(n)$.

设总体 $X\sim N(\mu,\sigma^2)$, $X_1,X_2,\ldots,X_n$ 是一容量为 $n$ 的样本, 其均值记为 $\bar{X}$, 标准差记为 $s$, 则有

（1）

\[
\bar{X}\sim N(\mu,\frac{\sigma^2}{n}),
\]

或等价的,

\[
\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\sim N(0,1).
\]

（2）

\[
\frac{\sum_{i=1}^{n}(X_i-\mu)^2}{\sigma^2}\sim\chi^2(n-1),
\]

\[
\frac{(n-1)s^2}{\sigma^2}\sim\chi^2(n-1).
\]

（3）

\[
\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}\sim t(n-1).
\]

References:

赵静、但琦 等编 《数学建模与数学实验》 高等教育出版社  P. 186.

Prop 1. 设 $X_1,X_2,\ldots,X_n$ 是 $n$ 个随机变量, $c_i$ 是常数, $i=1,2,\ldots,n$. 则期望满足线性性质:

\[
E(\sum_{i=1}^{n}c_i X_i)=\sum_{i=1}^{n}c_i E(X_i).
\]

若用 $\mu(X)$ 表示随机变量的期望, 则可写为

\[
\mu(\sum_{i=1}^{n}c_i X_i)=\sum_{i=1}^{n}c_i \mu(X_i).
\]

Prop 2. 设 $X$ 和 $Y$ 是两个互相独立的随机变量, 则有

\[
E(XY)=E(X)E(Y)
\]

这个性质可以推广到 $n$ 个互相独立的随机变量的情形:

Cor 3. 若 $X_1,X_2,\ldots,X_n$ 是 $n$ 个相互独立的随机变量, 则

\[
E(X_1 X_2\cdots X_n)=E(X_1)E(X_2)\cdots E(X_n).
\]

或写为

\[
\mu(X_1 X_2\cdots X_n)=\mu(X_1)\mu(X_2)\cdots\mu(X_n).
\]

or

\[
\mu(\prod_{i=1}^{n}X_i)=\prod_{i=1}^{n}\mu(X_i).
\]
 

Note:

Prop 2 的逆命题不成立. 请举出反例.