{"id":830,"date":"2026-09-28T20:36:12","date_gmt":"2026-09-28T11:36:12","guid":{"rendered":"https:\/\/wuhanqing.cn\/wordpress\/?p=830"},"modified":"2026-09-28T20:36:12","modified_gmt":"2026-09-28T11:36:12","slug":"quiz-1-probability-and-random-variables-2026-09-28-2","status":"publish","type":"post","link":"https:\/\/wuhanqing.cn\/wordpress\/2026\/09\/28\/quiz-1-probability-and-random-variables-2026-09-28-2\/","title":{"rendered":"QUIZ 1 | Probability and Random Variables | 2026.09.28"},"content":{"rendered":"
\n
\n QUIZ 1<\/span>
\n Probability and Random Variables<\/span>
\n 2026.09.28<\/span>\n <\/div>\n<\/div>\n

**Problem 1. [20 Points]** A fair six-sided die has its faces labeled $\\{1, 2, 2, 3, 3, 3\\}$. The die is rolled eight times independently.<\/p>\n

i. (10 pts) Find the probability that the number 1 appears exactly twice and the number 2 appears exactly three times.<\/p>\n

ii. (10 pts) Given that the number 3 appears exactly four times, find the conditional probability that the number 1 appears exactly twice.<\/p>\n

**\u3010\u4e2d\u6587\u7ffb\u8bd1\u3011**
\n\u4e00\u679a\u5747\u5300\u7684\u516d\u9762\u9ab0\u5b50\uff0c\u5176\u516d\u4e2a\u9762\u4e0a\u7684\u6570\u5b57\u5206\u522b\u4e3a $\\{1, 2, 2, 3, 3, 3\\}$\u3002\u72ec\u7acb\u6295\u63b7\u8be5\u9ab0\u5b50\u516b\u6b21\u3002
\ni. (10\u5206) \u6c42\u6570\u5b57 1 \u6070\u597d\u51fa\u73b0\u4e24\u6b21\uff0c\u4e14\u6570\u5b57 2 \u6070\u597d\u51fa\u73b0\u4e09\u6b21\u7684\u6982\u7387\u3002
\nii. (10\u5206) \u5df2\u77e5\u6570\u5b57 3 \u6070\u597d\u51fa\u73b0\u4e86\u56db\u6b21\uff0c\u6c42\u6570\u5b57 1 \u6070\u597d\u51fa\u73b0\u4e24\u6b21\u7684\u6761\u4ef6\u6982\u7387\u3002<\/p>\n

---<\/p>\n

### **\u3010\u7b2c\u4e00\u9898\u89e3\u6790\u3011**<\/p>\n

**\u7406\u8bba\u80cc\u666f**\uff1a\u672c\u9898\u8003\u67e5\u591a\u9879\u8bd5\u9a8c\uff08Multinomial Trials\uff09\u3001\u591a\u9879\u5206\u5e03\uff08Multinomial Distribution\uff09\u516c\u5f0f\u4ee5\u53ca\u5df2\u77e5\u90e8\u5206\u7ed3\u679c\u6761\u4ef6\u4e0b\u7684\u6761\u4ef6\u6982\u7387\u66f4\u65b0\u3002<\/p>\n

\u9996\u5148\uff0c\u6839\u636e\u6837\u672c\u7a7a\u95f4 $S = \\{1, 2, 2, 3, 3, 3\\}$ \u4e2d\u5404\u9762\u51fa\u73b0\u7684\u5360\u6bd4\uff0c\u8ba1\u7b97\u5355\u6b21\u6295\u63b7\u5f97\u5230\u5404\u4e2a\u6570\u5b57\u7684\u5148\u9a8c\u6982\u7387\uff1a
\n- $P(1) = \\frac{1}{6}$
\n- $P(2) = \\frac{2}{6} = \\frac{1}{3}$
\n- $P(3) = \\frac{3}{6} = \\frac{1}{2}$<\/p>\n

---<\/p>\n

#### **\u7b2c (i) \u95ee**
\n\u5728 $n = 8$ \u6b21\u72ec\u7acb\u6295\u63b7\u4e2d\uff0c\u8981\u6c42\u6570\u5b57 1 \u51fa\u73b0 $n_1 = 2$ \u6b21\uff0c\u6570\u5b57 2 \u51fa\u73b0 $n_2 = 3$ \u6b21\u3002\u56e0\u4e3a\u9ab0\u5b50\u9762\u53ea\u6709 1\u30012\u30013 \u4e09\u79cd\u6570\u5b57\uff0c\u56e0\u6b64\u5269\u4f59\u7684 $n_3 = 8 - 2 - 3 = 3$ \u6b21\u5fc5\u7136\u5168\u90e8\u662f\u6570\u5b57 3\u3002<\/p>\n

\u5e94\u7528\u591a\u9879\u5206\u5e03\u6982\u7387\u516c\u5f0f\uff1a
\n$$P(N_1=2, N_2=3, N_3=3) = \\frac{n!}{n_1! n_2! n_3!} p_1^{n_1} p_2^{n_2} p_3^{n_3}$$<\/p>\n

\u5177\u4f53\u4ee3\u5165\u8ba1\u7b97\uff1a
\n1. **\u591a\u9879\u5f0f\u7ec4\u5408\u7cfb\u6570**\uff1a
\n $$\\frac{8!}{2! \\cdot 3! \\cdot 3!} = \\frac{40320}{2 \\times 6 \\times 6} = 560$$
\n2. **\u5404\u5355\u6b21\u6982\u7387\u8fde\u4e58**\uff1a
\n $$\\left(\\frac{1}{6}\\right)^2 \\times \\left(\\frac{1}{3}\\right)^3 \\times \\left(\\frac{1}{2}\\right)^3 = \\frac{1}{36} \\times \\frac{1}{27} \\times \\frac{1}{8} = \\frac{1}{7776}$$
\n3. **\u6700\u7ec8\u4e58\u79ef\u4e0e\u7ea6\u5206**\uff1a
\n $$P = 560 \\times \\frac{1}{7776} = \\frac{560}{7776} = \\frac{140}{1944} = \\frac{35}{486} \\approx 7.20\\%$$<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$\\frac{35}{486}$ \uff08\u6216\u7ea6 $7.20\\%$\uff09\u3002<\/p>\n

---<\/p>\n

#### **\u7b2c (ii) \u95ee**
\n\u5df2\u77e5\u6570\u5b57 3 \u6070\u597d\u51fa\u73b0\u4e86 4 \u6b21\uff0c\u8fd9\u610f\u5473\u7740\u5728\u6392\u9664\u8fd9 4 \u6b21\u540e\uff0c\u5269\u4e0b\u7684 $8 - 4 = 4$ \u6b21\u6295\u63b7\u7ed3\u679c\u53ea\u80fd\u662f\u6570\u5b57 1 \u6216\u6570\u5b57 2\u3002<\/p>\n

\u6211\u4eec\u53ef\u4ee5\u5728\u6392\u9664\u6570\u5b57 3 \u7684\u6761\u4ef6\u4e0b\u91cd\u65b0\u5f52\u4e00\u5316\u76f8\u5bf9\u6982\u7387\uff1a
\n- $P(1 \\mid 3^c) = \\frac{1\/6}{1\/2} = \\frac{1}{3}$
\n- $P(2 \\mid 3^c) = \\frac{1\/3}{1\/2} = \\frac{2}{3}$<\/p>\n

\u6b64\u65f6\u95ee\u9898\u7b80\u5316\u4e3a\uff1a\u5728\u8fd9\u5269\u4f59\u7684 4 \u6b21\u72ec\u7acb\u6295\u63b7\u4e2d\uff0c\u6570\u5b57 1 \u6070\u597d\u51fa\u73b0 2 \u6b21\u7684\u4e8c\u9879\u5206\u5e03\u6982\u7387 $\\text{Bin}(4, 1\/3)$\uff1a
\n$$P = \\binom{4}{2} \\left(\\frac{1}{3}\\right)^2 \\left(\\frac{2}{3}\\right)^{4-2} = 6 \\times \\frac{1}{9} \\times \\frac{4}{9} = \\frac{24}{81} = \\frac{8}{27} \\approx 29.63\\%$$<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$\\frac{8}{27}$ \uff08\u6216\u7ea6 $29.63\\%$\uff09\u3002<\/p>\n

---<\/p>\n

**Problem 2. [20 Points]** Two events $A$ and $B$ satisfy<\/p>\n

$$P[A] = 0.5, \\quad P[B] = 0.4, \\quad P[A \\cup B] = 0.7$$<\/p>\n

i. (10 pts) Determine whether $A$ and $B$ are independent.<\/p>\n

ii. (10 pts) Find $P[B \\mid A^c]$.<\/p>\n

**\u3010\u4e2d\u6587\u7ffb\u8bd1\u3011**
\n\u4e24\u4e2a\u4e8b\u4ef6 $A$ \u548c $B$ \u6ee1\u8db3 $P[A] = 0.5, P[B] = 0.4, P[A \\cup B] = 0.7$\u3002
\ni. (10\u5206) \u5224\u65ad\u4e8b\u4ef6 $A$ \u548c $B$ \u662f\u5426\u76f8\u4e92\u72ec\u7acb\u3002
\nii. (10\u5206) \u6c42 $P[B \\mid A^c]$\u3002<\/p>\n

---<\/p>\n

### **\u3010\u7b2c\u4e8c\u9898\u89e3\u6790\u3011**<\/p>\n

**\u7406\u8bba\u80cc\u666f**\uff1a\u672c\u9898\u8003\u67e5\u5bb9\u65a5\u539f\u7406\uff08Inclusion-Exclusion Principle\uff09\u3001\u4e8b\u4ef6\u72ec\u7acb\u6027\uff08Event Independence\uff09\u7684\u5145\u8981\u5b9a\u4e49\u4ee5\u53ca\u6761\u4ef6\u6982\u7387\u4e0e\u8865\u96c6\u516c\u5f0f\u3002<\/p>\n

---<\/p>\n

#### **\u7b2c (i) \u95ee**
\n\u6839\u636e\u6982\u7387\u7684\u5bb9\u65a5\u539f\u7406\u516c\u5f0f\uff1a
\n$$P[A \\cup B] = P[A] + P[B] - P[A \\cap B]$$<\/p>\n

\u4ee3\u5165\u5df2\u77e5\u6761\u4ef6\u6c42\u89e3\u4ea4\u96c6\u6982\u7387 $P[A \\cap B]$\uff1a
\n$$0.7 = 0.5 + 0.4 - P[A \\cap B] \\implies P[A \\cap B] = 0.2$$<\/p>\n

\u6982\u7387\u8bba\u4e2d\uff0c\u4e24\u4e2a\u4e8b\u4ef6 $A$ \u548c $B$ \u76f8\u4e92\u72ec\u7acb\u5f53\u4e14\u4ec5\u5f53 $P[A \\cap B] = P[A]P[B]$\u3002\u8ba1\u7b97\u8fb9\u7f18\u6982\u7387\u7684\u4e58\u79ef\uff1a
\n$$P[A]P[B] = 0.5 \\times 0.4 = 0.2$$<\/p>\n

\u7531\u4e8e\u8ba1\u7b97\u5f97\u51fa $P[A \\cap B] = P[A]P[B] = 0.2$\uff0c\u6545\u4e8b\u4ef6 $A$ \u4e0e\u4e8b\u4ef6 $B$ **\u76f8\u4e92\u72ec\u7acb**\u3002<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a\u4e8b\u4ef6 $A$ \u548c $B$ \u76f8\u4e92\u72ec\u7acb\u3002<\/p>\n

---<\/p>\n

#### **\u7b2c (ii) \u95ee**
\n\u6839\u636e\u6761\u4ef6\u6982\u7387\u7684\u57fa\u672c\u5b9a\u4e49\uff1a
\n$$P[B \\mid A^c] = \\frac{P[B \\cap A^c]}{P[A^c]}$$<\/p>\n

\u5229\u7528\u96c6\u5408\u5206\u89e3\uff0c\u4e8b\u4ef6 $B$ \u53ef\u4ee5\u5212\u5206\u4e3a\u4e24\u4e2a\u4e0d\u76f8\u4ea4\u5b50\u96c6\u7684\u5e76\u96c6 $B = (A \\cap B) \\cup (A^c \\cap B)$\uff1a
\n$$P[B \\cap A^c] = P[B] - P[A \\cap B] = 0.4 - 0.2 = 0.2$$<\/p>\n

\u5206\u6bcd\u90e8\u5206\u8865\u96c6\u6982\u7387\u4e3a\uff1a
\n$$P[A^c] = 1 - P[A] = 1 - 0.5 = 0.5$$<\/p>\n

\u4ee3\u5165\u8ba1\u7b97\u5f97\uff1a
\n$$P[B \\mid A^c] = \\frac{0.2}{0.5} = 0.4$$<\/p>\n

*\uff08\u6ce8\uff1a\u7531\u4e8e\u7b2c (i) \u95ee\u5df2\u8bc1\u660e $A$ \u548c $B$ \u76f8\u4e92\u72ec\u7acb\uff0c\u56e0\u6b64 $A^c$ \u4e0e $B$ \u4e5f\u76f8\u4e92\u72ec\u7acb\uff0c\u5f97\u77e5 $A$ \u672a\u53d1\u751f\u5e76\u4e0d\u6539\u53d8\u5bf9 $B$ \u53d1\u751f\u6982\u7387\u7684\u8ba4\u77e5\uff0c\u56e0\u6b64\u53ef\u4ee5\u76f4\u63a5\u5f97\u51fa $P[B \\mid A^c] = P[B] = 0.4$\u3002\uff09*<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$0.4$\u3002<\/p>\n

---<\/p>\n

**Problem 3. [20 Points]** A communication system attempts to transmit a data packet until the transmission succeeds, with a maximum of four attempts. The probability of success on each attempt is $0.8$, and the outcomes of different attempts are independent. No further attempts are made once a transmission succeeds.<\/p>\n

i. (10 pts) Find the probability that the packet is successfully transmitted within four attempts.<\/p>\n

ii. (10 pts) Given that the packet is successfully transmitted, find the probability that the first successful transmission occurs on the third attempt.<\/p>\n

**\u3010\u4e2d\u6587\u7ffb\u8bd1\u3011**
\n\u4e00\u4e2a\u901a\u4fe1\u7cfb\u7edf\u5c1d\u8bd5\u4f20\u8f93\u4e00\u4e2a\u6570\u636e\u5305\u76f4\u5230\u4f20\u8f93\u6210\u529f\uff0c\u6700\u591a\u5c1d\u8bd5\u56db\u6b21\u3002\u6bcf\u6b21\u5c1d\u8bd5\u6210\u529f\u7684\u6982\u7387\u662f 0.8\uff0c\u5e76\u4e14\u4e0d\u540c\u5c1d\u8bd5\u7684\u7ed3\u679c\u662f\u76f8\u4e92\u72ec\u7acb\u7684\u3002\u4e00\u65e6\u4f20\u8f93\u6210\u529f\uff0c\u5c31\u4e0d\u4f1a\u8fdb\u884c\u8fdb\u4e00\u6b65\u7684\u5c1d\u8bd5\u3002
\ni. (10\u5206) \u6c42\u6570\u636e\u5305\u5728\u56db\u6b21\u5c1d\u8bd5\u5185\u6210\u529f\u4f20\u8f93\u7684\u6982\u7387\u3002
\nii. (10\u5206) \u5df2\u77e5\u6570\u636e\u5305\u88ab\u6210\u529f\u4f20\u8f93\uff0c\u6c42\u7b2c\u4e00\u6b21\u6210\u529f\u4f20\u8f93\u53d1\u751f\u5728\u7b2c\u4e09\u6b21\u5c1d\u8bd5\u7684\u6982\u7387\u3002<\/p>\n

---<\/p>\n

### **\u3010\u7b2c\u4e09\u9898\u89e3\u6790\u3011**<\/p>\n

**\u7406\u8bba\u80cc\u666f**\uff1a\u672c\u9898\u8003\u67e5\u72ec\u7acb\u4f2f\u52aa\u5229\u8bd5\u9a8c\u5e8f\u5217\uff08Bernoulli Trials\uff09\u3001\u622a\u65ad\u51e0\u4f55\u5206\u5e03\u4ee5\u53ca\u8d1d\u53f6\u65af\u5b9a\u7406\uff08\u6761\u4ef6\u6982\u7387\uff09\u3002<\/p>\n

---<\/p>\n

#### **\u7b2c (i) \u95ee**
\n\u8bbe\u5355\u6b21\u4f20\u8f93\u6210\u529f\u6982\u7387 $p = 0.8$\uff0c\u5931\u8d25\u6982\u7387 $q = 1 - p = 0.2$\u3002<\/p>\n

\u201c\u56db\u6b21\u5c1d\u8bd5\u5185\u6210\u529f\u4f20\u8f93\u201d\u7684\u5bf9\u7acb\u4e8b\u4ef6\u662f\u201c\u56db\u6b21\u5c1d\u8bd5\u5168\u90e8\u5931\u8d25\u201d\u3002\u56e0\u4e3a\u5404\u6b21\u5c1d\u8bd5\u76f8\u4e92\u72ec\u7acb\uff0c\u56db\u6b21\u7686\u5931\u8d25\u7684\u8054\u5408\u6982\u7387\u4e3a\uff1a
\n$$P(\\text{\u5168\u5931\u8d25}) = q^4 = (0.2)^4 = 0.0016$$<\/p>\n

\u5229\u7528\u5bf9\u7acb\u4e8b\u4ef6\u6982\u7387\u516c\u5f0f\u8ba1\u7b97\u56db\u6b21\u5185\u81f3\u5c11\u6210\u529f\u4e00\u6b21\u7684\u6982\u7387\uff1a
\n$$P(\\text{\u56db\u6b21\u5185\u6210\u529f}) = 1 - 0.0016 = 0.9984 \\quad (99.84\\%)$$<\/p>\n

*\uff08\u53e6\u89e3\uff1a\u76f4\u63a5\u6309\u4e92\u65a5\u8def\u5f84\u76f8\u52a0\uff1a$0.8 + 0.2(0.8) + 0.2^2(0.8) + 0.2^3(0.8) = 0.8(1 + 0.2 + 0.04 + 0.008) = 0.9984$\u3002\uff09*<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$0.9984$ \uff08\u6216 $99.84\\%$\uff09\u3002<\/p>\n

---<\/p>\n

#### **\u7b2c (ii) \u95ee**
\n\u8bbe\u4e8b\u4ef6 $S$ \u4e3a\u6570\u636e\u5305\u5728\u56db\u6b21\u5185\u6210\u529f\u4f20\u8f93\uff08\u7531\u7b2c (i) \u95ee\u77e5 $P[S] = 0.9984$\uff09\u3002
\n\u8bbe\u4e8b\u4ef6 $S_3$ \u4e3a\u7b2c\u4e00\u6b21\u6210\u529f\u53d1\u751f\u5728\u7b2c\u4e09\u6b21\u5c1d\u8bd5\u3002<\/p>\n

\u4e8b\u4ef6 $S_3$ \u8981\u6c42\u524d\u4e24\u6b21\u5c1d\u8bd5\u5931\u8d25\u4e14\u7b2c\u4e09\u6b21\u5c1d\u8bd5\u6210\u529f\uff1a
\n$$P[S_3] = q \\times q \\times p = (0.2) \\times (0.2) \\times (0.8) = 0.032$$<\/p>\n

\u7531\u4e8e $S_3$ \u662f\u4e8b\u4ef6 $S$ \u7684\u4e00\u4e2a\u5b50\u96c6\uff08\u5373 $S_3 \\subseteq S$\uff09\uff0c\u5b83\u4eec\u7684\u4ea4\u96c6\u4e3a $S_3 \\cap S = S_3$\u3002\u4ee3\u5165\u8d1d\u53f6\u65af\/\u6761\u4ef6\u6982\u7387\u516c\u5f0f\uff1a
\n$$P[S_3 \\mid S] = \\frac{P[S_3 \\cap S]}{P[S]} = \\frac{P[S_3]}{P[S]} = \\frac{0.032}{0.9984} = \\frac{320}{9984} = \\frac{5}{156} \\approx 3.21\\%$$<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$\\frac{5}{156}$ \uff08\u6216\u7ea6 $3.21\\%$\uff09\u3002<\/p>\n

---<\/p>\n

**Problem 4. [20 Points]** A factory has three production lines, $L_1, L_2$, and $L_3$, which produce $20\\%$, $30\\%$, and $50\\%$ of the total products, respectively. The defective rates of the three production lines are $1\\%$, $2\\%$, and $4\\%$, respectively. One product is randomly selected from the factory's total production.<\/p>\n

i. (10 pts) Find the probability that the selected product is defective.<\/p>\n

ii. (10 pts) Given that the selected product is not defective, find the probability that it was produced by either $L_1$ or $L_2$.<\/p>\n

**\u3010\u4e2d\u6587\u7ffb\u8bd1\u3011**
\n\u4e00\u5bb6\u5de5\u5382\u6709\u4e09\u6761\u751f\u4ea7\u7ebf $L_1, L_2$ \u548c $L_3$\uff0c\u5206\u522b\u751f\u4ea7\u4e86\u603b\u4ea7\u54c1\u7684 20%, 30% \u548c 50%\u3002\u8fd9\u4e09\u6761\u751f\u4ea7\u7ebf\u7684\u6b21\u54c1\u7387\u5206\u522b\u4e3a 1%, 2% \u548c 4%\u3002\u4ece\u5de5\u5382\u7684\u603b\u4ea7\u54c1\u4e2d\u968f\u673a\u9009\u53d6\u4e00\u4ef6\u4ea7\u54c1\u3002
\ni. (10\u5206) \u6c42\u9009\u51fa\u7684\u4ea7\u54c1\u662f\u6b21\u54c1\u7684\u6982\u7387\u3002
\nii. (10\u5206) \u5df2\u77e5\u9009\u51fa\u7684\u4ea7\u54c1\u4e0d\u662f\u6b21\u54c1\uff0c\u6c42\u5b83\u662f $L_1$ \u6216 $L_2$ \u751f\u4ea7\u7684\u6982\u7387\u3002<\/p>\n

---<\/p>\n

### **\u3010\u7b2c\u56db\u9898\u89e3\u6790\u3011**<\/p>\n

**\u7406\u8bba\u80cc\u666f**\uff1a\u672c\u9898\u8003\u67e5**\u5168\u6982\u7387\u516c\u5f0f (Law of Total Probability, LTP)** \u4e0e **\u8d1d\u53f6\u65af\u5b9a\u7406 (Bayes' Theorem)** \u5728\u5212\u5206\uff08Partition\uff09\u4e0a\u7684\u5e94\u7528\u3002<\/p>\n

\u8bbe $L_1, L_2, L_3$ \u6784\u6210\u672c\u5de5\u5382\u751f\u4ea7\u7ebf\u7684\u5212\u5206\uff0c\u5148\u9a8c\u6982\u7387\u4e3a\uff1a
\n$$P[L_1] = 0.20, \\quad P[L_2] = 0.30, \\quad P[L_3] = 0.50$$<\/p>\n

\u8bbe $D$ \u4e3a\u9009\u51fa\u7684\u4ea7\u54c1\u4e3a\u6b21\u54c1\uff0c\u6761\u4ef6\u6b21\u54c1\u7387\u5206\u522b\u4e3a\uff1a
\n$$P[D \\mid L_1] = 0.01, \\quad P[D \\mid L_2] = 0.02, \\quad P[D \\mid L_3] = 0.04$$<\/p>\n

---<\/p>\n

#### **\u7b2c (i) \u95ee**
\n\u6839\u636e\u5168\u6982\u7387\u516c\u5f0f (LTP)\uff1a
\n$$P[D] = \\sum_{i=1}^{3} P[L_i] P[D \\mid L_i]$$<\/p>\n

\u4ee3\u5165\u6570\u503c\u8ba1\u7b97\u5168\u5c40\u6b21\u54c1\u7387\uff1a
\n$$P[D] = (0.20 \\times 0.01) + (0.30 \\times 0.02) + (0.50 \\times 0.04) = 0.002 + 0.006 + 0.020 = 0.028 \\quad (2.8\\%)$$<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$0.028$ \uff08\u6216 $2.8\\%$\uff09\u3002<\/p>\n

---<\/p>\n

#### **\u7b2c (ii) \u95ee**
\n\u8bbe $D^c$ \u4e3a\u9009\u51fa\u7684\u4ea7\u54c1\u4e0d\u662f\u6b21\u54c1\u3002\u9009\u4e2d\u975e\u6b21\u54c1\u7684\u5168\u5c40\u6982\u7387\u4e3a\uff1a
\n$$P[D^c] = 1 - P[D] = 1 - 0.028 = 0.972$$<\/p>\n

\u6211\u4eec\u8981\u6c42\u7684\u662f\u6761\u4ef6\u6982\u7387 $P[L_1 \\cup L_2 \\mid D^c]$\u3002\u7531\u4e8e $L_1$ \u548c $L_2$ \u662f\u4e92\u65a5\u7684\u751f\u4ea7\u7ebf\uff1a
\n$$P[L_1 \\cup L_2 \\mid D^c] = \\frac{P[(L_1 \\cup L_2) \\cap D^c]}{P[D^c]} = \\frac{P[L_1 \\cap D^c] + P[L_2 \\cap D^c]}{P[D^c]}$$<\/p>\n

\u5206\u522b\u8ba1\u7b97 $L_1$ \u548c $L_2$ \u751f\u4ea7\u51fa\u5408\u683c\u54c1\u7684\u8054\u5408\u6982\u7387\uff1a
\n- $P[L_1 \\cap D^c] = P[L_1] P[D^c \\mid L_1] = 0.20 \\times (1 - 0.01) = 0.20 \\times 0.99 = 0.198$
\n- $P[L_2 \\cap D^c] = P[L_2] P[D^c \\mid L_2] = 0.30 \\times (1 - 0.02) = 0.30 \\times 0.98 = 0.294$<\/p>\n

\u5c06\u5206\u5b50\u76f8\u52a0\u5e76\u9664\u4ee5\u975e\u6b21\u54c1\u603b\u6982\u7387 $P[D^c]$\uff1a
\n$$P[L_1 \\cup L_2 \\mid D^c] = \\frac{0.198 + 0.294}{0.972} = \\frac{0.492}{0.972} = \\frac{492}{972} = \\frac{41}{81} \\approx 50.62\\%$$<\/p>\n

**\u6b63\u786e\u7b54\u6848**\uff1a$\\frac{41}{81}$ \uff08\u6216\u7ea6 $50.62\\%$\uff09\u3002<\/p>\n","protected":false},"excerpt":{"rendered":"\r\n \r\n QUIZ 1\r\n Probability and Random Variables\r\n 2026.09.28\r\n \r\n\r\n\r\n**Problem 1. [20 Points]** A fair six-sided die has its faces labeled $\\{1, 2, 2, 3, 3, 3\\}$. The die is rolled eight times independently.\r\n\r\ni. (10 pts) Find the probability that the number 1 appears exactly twice and the number 2 appears exactly three times.\r\n\r\nii. (10 pts) Given that the number 3 appears exactly four times, find the conditional probability that the number 1 appears exactly twice.\r\n\r\n**\u3010\u4e2d\u6587\u7ffb\u8bd1\u3011**\r\n\u4e00\u679a\u5747\u5300\u7684\u516d\u9762\u9ab0\u5b50\uff0c\u5176\u516d\u4e2a\u9762\u4e0a\u7684\u6570\u5b57\u5206\u522b\u4e3a $\\{1, 2, 2, 3, 3, 3\\}$\u3002\u72ec\u7acb\u6295\u63b7\u8be5\u9ab0\u5b50\u516b\u6b21\u3002\r\ni. (10\u5206) \u6c42\u6570\u5b57 1 \u6070\u597d\u51fa\u73b0\u4e24\u6b21\uff0c\u4e14\u6570\u5b57 2 \u6070\u597d\u51fa\u73b0\u4e09\u6b21\u7684\u6982\u7387\u3002\r\nii. (10\u5206) \u5df2\u77e5\u6570\u5b57 3 \u6070\u597d\u51fa\u73b0\u4e86\u56db\u6b21\uff0c\u6c42\u6570\u5b57 1 \u6070\u597d\u51fa\u73b0\u4e24\u6b21\u7684\u6761\u4ef6\u6982\u7387\u3002\r\n\r\n---\r\n\r\n**\u3010\u7b2c\u4e00\u9898\u89e3\u6790\u3011**\r\n\r\n**\u7406\u8bba\u80cc\u666f**\uff1a\u672c\u9898\u8003...","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"emotion":"","emotion_color":"","title_style":"","license":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-830","post","type-post","status-publish","format-standard","hentry","category-1"],"_links":{"self":[{"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts\/830","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/comments?post=830"}],"version-history":[{"count":2,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts\/830\/revisions"}],"predecessor-version":[{"id":832,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts\/830\/revisions\/832"}],"wp:attachment":[{"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/media?parent=830"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/categories?post=830"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/tags?post=830"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}