198 lines
8.7 KiB
TeX
198 lines
8.7 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\begin{document}
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\renewcommand{\bar}{\xoverline}
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\chapter{单元作业2}
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\begin{enumerate}
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\item 三人独立地对同一目标进行射击, 各人击中目标的概率分别是0.7, 0.8, 0.6, 求目标被击中的概率.
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&将三人击中目标的事件分别表示为A、B、C， \\
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&则P(A)=0.7, P(B)=0.8, P(C)=0.6, \\
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&P(目标被击中) = P(A\cup B\cup C) \\
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&\because 三人独立射击 \\
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&\therefore A、B、C相互独立 \\
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&\therefore \bar{A}、\bar{B}、\bar{C}相互独立 \\
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\end{aligned}
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$$ $$
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\begin{aligned}
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\therefore P(A\cup B\cup C)&=P(\overline{\bar{A} \bar{B} \bar{C}}) = 1 - P(\bar{A} \bar{B} \bar{C}) \\
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&=1 - P(\bar{A})P(\bar{B})P(\bar{C}) \\
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&=1 - (1-P(A))(1-P(B))(1-P(C)) \\
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&=1-(1-0.7)*(1-0.8)*(1-0.6) \\
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&= 0.976 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 制作某个产品有两个关键工序, 第一道和第二道工序的不合格品的概率分别为3\%和5\%, 假定两道工序互不影响, 试问该产品为不合格品的概率(答案保留至小数点后4位).
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&令A：第一道工序不合格，B：第二道工序不合格 \\
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&则P(A)=0.03，P(B)=0.05 \\
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&则P(该产品为不合格品)=P(A\cup B) \\
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&\because 两道工序互不影响 \\
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&\therefore A与B相互独立 \\
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&\therefore \bar{A}与\bar{B}相互独立 \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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\therefore P(A\cup B)&=P(\overline{\bar{A}\bar{B}})=1-P(\bar{A}\bar{B}) \\
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&=1-P(\bar{A})P(\bar{B}) \\
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&=1- (1-P(A))(1-P(B)) \\
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&=1-(1-0.03)(1-0.05) \\
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&= 0.0785 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\myitem{甲乙丙三个同学同时独立参加考试, 不及格的概率分别为: 0.2, 0.3, 0.4,}{
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\item 求恰有2位同学不及格的概率；\label{1}
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\item 若已知3位同学中有2位不及格,求其中1位是同学乙的概率. \label{2}
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A、B、C分别表示事件甲不合格、乙不合格、丙不合格 \\
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&则P(A)=0.2, P(B)=0.3, P(C)=0.4 \\
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&P(\bar{A})=0.8, P(\bar{B})=0.7, P(\bar{C})=0.6 \\
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&且A、B、C相互独立 \\
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\end{aligned}
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$$
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\ref{1}
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$$
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\begin{aligned}
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P(恰有2位同学不合格)&=P(AB \bar{C}+A \bar{B} C+\bar{A}BC) \\
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&=P(A)P(B)P(\bar{C})+P(A)P(\bar{B})P(C)+P(\bar{A})P(B)P(C) \\
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&=0.2\times 0.3\times 0.6+0.2\times 0.7\times 0.4+0.8\times 0.3\times 0.4 \\
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&= 0.188 \\
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\end{aligned}
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$$
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\ref{2}
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\setlength{\lineskip}{2.5pt}
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\setlength{\lineskiplimit}{2.5pt}
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$$
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\begin{aligned}
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% \setstretch{1.5}
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P(其中1位是同学乙)&=P(\left. B\ \right|\ 恰有2位同学不合格) \\
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&=\frac{P(B\cap 恰有2位同学不合格)}{P(恰有2位同学不合格)} \\
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&=\frac{P(AB \bar{C} + \bar{A} BC)}{P(AB \bar{C}+A \bar{B}C+\bar{A}BC)} \\
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&=\frac{P(A)P(B)P(\bar{C})+P(\bar{A})P(B)P(C)}{P(A)P(B)P(\bar{C})+P(A)P(\bar{B})P(C)+P(\bar{A})P(B)P(C)} \\
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&=\frac{0.2\times 0.3\times 0.6+0.8\times 0.3\times 0.4}{0.2\times 0.3\times 0.6+0.2\times 0.7\times 0.4+0.8\times 0.3\times 0.4} \\
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&= \frac{33}{47} \\
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&\approx 0.702127659574468 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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}
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\item 甲袋中装有2个白球和4个黑球, 乙袋中装有3个白球和2个黑球, 现随机地从乙袋中取出一球放入甲袋, 然后从甲袋中随机取出一球, 试求从甲袋中取得的球是白球的概率.
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示从乙袋中取出的是白球，B表示从甲袋中取出的是白球 \\
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则P(B) &= P(A)P(B|A)+P(\bar{A})P(B|\bar{A}) \\
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&=\frac{3}{3+2}\times \frac{2+1}{2+4+1}+\frac{2}{3+2}\times \frac{2}{2+4+1} \\
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&= \frac{13}{35} \\
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&\approx 0.371428571428571 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设n只罐子的每一只中装有4个白球和6个黑球, 另有一只罐子中装有5个白球和5个黑球.从这n+1个罐子中随机地选择一只罐子, 从中任取两个球, 结果发现两个都是黑球. 已知在此条件下, 有5个白球和3个黑球留在选出的罐子中的条件概率是1/7, 求n的值.
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件：选中的是第n+1个罐子 \\
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&设B表示事件：取出的两个都是黑球 \\
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&则事件：有5个白球和3个黑球留在选出的罐子中，\\
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&即选中的是第n+1个罐子，并且取出的两个都是黑球， 可表示为AB\\
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&\therefore P(A|B)=\frac{1}{7}, P(A)=\frac{1}{n+1} \\
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& P(B|A)=\frac{\mathrm{C}_{5}^{2}}{\mathrm{C}_{10}^{2}}=\frac{2}{9} \\
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\therefore P(B)&=P(A)P(B|A)+P(\bar{A})P(B|\bar{A})\\
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&=\frac{1}{n+1}\times \frac{\mathrm{C}_{5}^{2}}{\mathrm{C}_{10}^{2}}+\frac{n}{n+1}\times \frac{\mathrm{C}_{6}^{2}}{\mathrm{C}_{10}^{2}} \\
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&=\frac{3 n + 2}{9 (n + 1)} \\
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% &\because A与B相互独立 \\
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\therefore P(A|B)&=\frac{P(A)P(B|A)}{P(B)} \\
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&=\frac{\frac{1}{n+1}\times \frac{2}{9}}{\frac{3n+2}{9(n+1)}} \\
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&= \frac{2}{3 n + 2} = \frac{1}{7} \\
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&\therefore n=4 \\
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% solve(latex2sympy(r"\frac{2}{3 n + 2} = \frac{1}{7}")) = {n: 4} = []
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% solve(latex2sympy(r"\frac{1}{2} = n"))
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% var["n"]
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设有三张卡片, 第一张两面皆为红色, 第二张两面皆为黄色, 第三张一面是红色一面是黄色. 随机地选择一张卡片并随机地选择其中一面. 如果已知此面是红色, 求另一面也是红色的概率.
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A、B、C分别表示选到了第一、二、三张卡片，设R表示选择的一面是红色。 \\
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&则P(A) = P(B) = P(C) = \frac{1}{3} \\
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&P(R|A)=1, P(R|B)=0, P(R|C)=\frac{1}{2} \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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则P(另一面也是红色) &= P(A|R) \\
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&=\frac{P(A)P(R|A)}{P(A)P(R|A)+P(B)P(R|B)+P(C)P(R|C)} \\
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&=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times 0+\frac{1}{3}\times \frac{1}{2}} \\
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&= \frac{2}{3} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 从装有r个红球和w个白球的盒子中不返回的取出两只, 求事件“第一只为红球, 第二只为白球”的概率.
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示取出的两只球一红一白 \\
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&则P(第一只为红球，第二只为白球) = \frac{P(A)}{\mathrm{P}_{2}^{2}} \\
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&=\frac{\frac{\mathrm{C}_{r}^{1}\mathrm{C}_{w}^{1}}{\mathrm{C}_{r+w}^{2}}}{\mathrm{P}_{2}^{2}} \\
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&=\frac{\frac{\binom{r}{1}\binom{w}{1}}{\binom{r+w}{2}}}{2!} \\
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&= \frac{r w}{(r + w) (r + w - 1)} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 已知$P(A)=\frac{1}{4}, P(B|A)=\frac{1}{3},P(A|B)=\frac{1}{2}$,求概率$P(A\cup B)$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&P(AB) = P(B|A)P(A) = \frac{1}{3}\times \frac{1}{4} = \frac{1}{12} \\
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&P(B)=\frac{P(AB)}{P(A|B)}=\frac{\frac{1}{12}}{\frac{1}{2}} = \frac{1}{6} \\
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&\therefore P(A\cup B)=P(A)+P(B)-P(AB) \\
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&=\frac{1}{4}+\frac{1}{6} - \frac{1}{12} \\
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&= \frac{1}{3} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$P(A)=P(B)=\frac{1}{3},P(A|B)=\frac{1}{6}$，求概率$P(\bar{A}|\bar{B})$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&P(AB)=P(B)P(A|B)=\frac{1}{3}\times \frac{1}{6} = \frac{1}{18} \\
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&P(\bar{A} \bar{B})=P(\overline{A\cup B}) = 1 - P(A\cup B)=1 - (P(A)+P(B)-P(AB))\\
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&=1 - (\frac{1}{3}+\frac{1}{3}-\frac{1}{18}) = \frac{7}{18}\\
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&P(\bar{B}) = 1-P(B) = 1 - \frac{1}{3} = \frac{2}{3} \\
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&\therefore P(\bar{A}|\bar{B})=\frac{P(\bar{A} \bar{B})}{P(\bar{B})}=\frac{\frac{7}{18}}{\frac{2}{3}} = \frac{7}{12} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\end{document}
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