159 lines
6.7 KiB
TeX
159 lines
6.7 KiB
TeX
\documentclass[全部作业]{subfiles}
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\begin{document}
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\chapter{单元作业1}
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\begin{enumerate}
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\item 证明$P(AB)\ge P(A)+P(B)-1$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设\Omega 表示全集，则A \in \Omega, B \in \Omega \\
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&\therefore A\cup B\in \Omega \\
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&\therefore P(A\cup B)\le P(\Omega) \\
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&\therefore P(A)+P(B)-P(AB)\le 1 \\
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&\therefore P(AB)\ge P(A)+P(B)-1 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 一副标准扑克牌52张一张一张轮流分发给给4名游戏者，求每人恰好得到1张A的概率。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件“每人恰好得到一张A”，由盒子模型可知： \\
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&P(A) = \frac{\mathrm{P}_{4}^{4}}{4^{4}} = \frac{4!}{256} = \frac{3}{32} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$P(A)=0.7,P(A-B)=0.3$，求概率$P(\overline{AB})$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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P(\overline{AB})&=1-P(AB)=1-P(A-(A-B)) \\
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&=1-(P(A)-P(A-B))=1-(0.7-0.3)=0.6 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$P(A)=a,P(B)=b,P(A\cup B)=c$，求概率$P(\overline{\bar{A}\cup \bar{B}})$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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P(\overline{\bar{A}\cup \bar{B}})=P(A\cap B)=P(A)+P(B)-P(A\cup B)=a+b-c
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$A,B\in \mathcal{F}$，证明$P(A)=P(AB)+P(A \bar{B}), \quad P(A\triangle B)=P(A)+P(B)-2P(AB)$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设\Omega 表示全集 \\
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P(A)&=P(A\cap \Omega)=P(A\cap (B\cup \bar{B})) \\
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&=P(A(B+\bar{B}))=P(AB+A \bar{B})=P(AB)+P(A \bar{B}) \\
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P(A \triangle B)&=P((A\backslash B)\cup (B\backslash A))=P((A-(AB))+(B-(AB))) \\
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&=P(A)-P(AB)+P(B)-P(AB)=P(A)+P(B)-2P(AB) \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$\Omega =(-\infty, \infty), A=\{ x\in \Omega: 1\le x\le 5 \}, B=\{ x\in \Omega:3<x<7 \},C=\{ x\in \Omega:x<0 \}$，求下列事件$\bar{A}, A\cup B,B \bar{C}, \bar{A}\cap \bar{B}\cap \bar{C}, (A\cup B)C$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&\bar{A}=\{ x\in \Omega: x < 1 或x>5 \} \\
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&A\cup B=\{ x\in \Omega: 1\le x<7 \} \\
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&B \bar{C}=\{ x\in \Omega:3<x<7 \} \\
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&\bar{A}\cap \bar{B}\cap \bar{C}=\{ x\in \Omega:0\le x<1或x\ge 7 \} \\
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&(A\cup B)C=\varnothing \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设\textit{I}是任意指标集，$\{ A_i, i\in I \}$是一事件类，证明$\overline{\displaystyle \bigcup_{i \in I} A_i}=\displaystyle \bigcap_{i \in I} \overline{A_i}$, \quad $\overline{\displaystyle \bigcap_{i \in I} A_i}=\displaystyle \bigcup_{i \in I} \overline{A_i}$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&\forall x\in \overline{\bigcup_{i \in I} A_i},\ 即x \notin \bigcup_{i \in I} A_i, 即 \\
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&\forall i \in I,x \notin A_i, \\
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&\therefore \forall i \in I, x\in \overline{A_i} \\
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&\therefore x\in \bigcap_{i \in I} \overline{A_i} \\
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\end{aligned}
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$$
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\begin{equation}
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\therefore \overline{\bigcup_{i \in I} A_i} \subseteq \bigcap_{i \in I} \overline{A_i} \tag{1}\label{eq:7.1}
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\end{equation}
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$$
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\begin{aligned}
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&\forall x\in \bigcap_{i \in I} \overline{A_i}, 即 \\
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&\forall i \in I,x\in \overline{A_i} \\
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&\therefore \forall i\in I,x \notin A_i \\
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&\therefore x\notin \bigcup_{i \in I} A_i \\
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&\therefore x\in \overline{\bigcup_{i \in I} A_i} \\
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\end{aligned}
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$$
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\begin{equation}
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\therefore \bigcap_{i \in I} \overline{A_i} \subseteq \overline{\bigcup_{i \in I} A_i} \tag{2}\label{eq:7.2}
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\end{equation}
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$$
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\begin{aligned}
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&由\eqref{eq:7.1}\eqref{eq:7.2}可知，\overline{\bigcup_{i \in I} A_i}=\bigcap_{i \in I} \overline{A_i} \\
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&由对偶公式可知, \overline{\overline{\bigcup_{i \in I} A_i}}=\overline{\bigcap_{i \in I} \overline{A_i}} \\
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&\therefore \overline{\bigcap_{i \in I} \overline{A_i}}=\bigcup_{i \in I} A_i \\
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&\because \{ \overline{A_i}|i\in I \}也是一事件类 \\
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&\therefore 将\overline{A_i}替换为A_i后可得 \\
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&\overline{\bigcap_{i \in I} A_i}=\bigcup_{i \in I} \overline{A_i} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 从装有10双不同尺码或不同样式的皮鞋的箱子中, 任取4只, 求恰能成1双的概率。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件“恰能成1双”，B表示事件“恰能成2双”，C表示事件“不能成双”， \\
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&则由不返回抽样模型可知： \\
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P(A)&=1-P(B)-P(C)=1-\frac{\mathrm{C}_{10}^{2}}{\mathrm{C}_{20}^{4}}-\frac{\mathrm{C}_{20}^{1}\mathrm{C}_{18}^{1}\mathrm{C}_{16}^{1}\mathrm{C}_{14}^{1}}{\mathrm{P}_{20}^{4}} \\
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&=1-\frac{\binom{10}{2}}{\binom{20}{4}}-\frac{\binom{20}{1}\binom{18}{1}\binom{16}{1}\binom{14}{1}}{\binom{20}{4}\times 4!} = \frac{96}{323} \approx 0.297213622291022 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 现从有15名男生和30名女生的班级中随机挑选10名同学参加某项课外活动, 求在被挑选的同学中恰好有3名男生的概率。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件“在被挑选的同学中恰好有3名男生”，则 \\
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&P(A)=\frac{\mathrm{C}_{15}^{3}\mathrm{C}_{30}^{7}}{\mathrm{C}_{45}^{10}}=\frac{\binom{15}{3}\binom{30}{7}}{\binom{45}{10}} = \frac{3958500}{13633279} \approx 0.290355680390609 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$\mathcal{F}$是$\Omega$上的事件域，$A,B\in \mathcal{F}$，证明$A\cup B,AB\in \mathcal{F}$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A_1=A,A_2=B,\forall n>2,A_n=\varnothing \\
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&\because \mathcal{F}是\Omega上的事件域,\forall n\geqslant 1,A_n\in \mathcal{F} \\
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&\therefore \bigcup_{n=1}^{\infty}A_n=A\cup B\in \mathcal{F} \\
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&又\because A,B\in \mathcal{F} \\
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&\therefore \bar{A},\bar{B}\in \mathcal{F} \\
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&同理，\bar{A}\cup \bar{B}\in \mathcal{F} \\
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&\therefore \overline{\bar{A}\cup \bar{B}}=AB\in \mathcal{F} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\end{document} |