{"id":828,"date":"2026-09-28T20:23:58","date_gmt":"2026-09-28T11:23:58","guid":{"rendered":"https:\/\/wuhanqing.cn\/wordpress\/?p=828"},"modified":"2026-09-28T20:23:58","modified_gmt":"2026-09-28T11:23:58","slug":"quiz-1-probability-and-random-variables-2026-09-28","status":"publish","type":"post","link":"https:\/\/wuhanqing.cn\/wordpress\/en\/2026\/09\/28\/quiz-1-probability-and-random-variables-2026-09-28\/","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

---<\/p>\n

### **Solution for Problem 1**<\/p>\n

**Concept Overview**: This problem tests the **Multinomial Distribution** for independent repetitions of a discrete experiment with more than two possible outcomes, as well as **Conditional Probability**.<\/p>\n

First, determine the single-trial probabilities for each face label based on the equiprobable sample space of a 6-sided die:
\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

#### **Part (i)**
\nIn $n = 8$ independent rolls, we require outcome 1 to occur $n_1 = 2$ times, outcome 2 to occur $n_2 = 3$ times, and consequently outcome 3 must occur $n_3 = 8 - 2 - 3 = 3$ times.<\/p>\n

Using the multinomial probability formula:
\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

Substitute the values:
\n1. **Multinomial Coefficient**:
\n $$\\frac{8!}{2! \\cdot 3! \\cdot 3!} = \\frac{40320}{2 \\times 6 \\times 6} = \\frac{40320}{72} = 560$$<\/p>\n

2. **Probability Factors**:
\n $$\\left(\\frac{1}{6}\\right)^2 \\times \\left(\\frac{1}{3}\\right)^3 \\times \\left(\\frac{1}{2}\\right)^3 = \\frac{1}{36} \\times \\frac{1}{27} \\times \\frac{1}{8} = \\frac{1}{7776}$$<\/p>\n

3. **Combined Calculation**:
\n $$P = 560 \\times \\frac{1}{7776} = \\frac{560}{7776} = \\frac{140}{1944} = \\frac{35}{486} \\approx 0.072016 \\quad (7.20\\%)$$<\/p>\n

**Final Answer**: $\\frac{35}{486}$ (or approximately $7.20\\%$).<\/p>\n

---<\/p>\n

#### **Part (ii)**
\nGiven that number 3 appears exactly 4 times in 8 rolls, the remaining $8 - 4 = 4$ rolls can only result in outcome 1 or outcome 2.<\/p>\n

We construct a conditional sample space over these remaining 4 trials with revised conditional probabilities given $3^c$:
\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

The problem simplifies to finding the probability that outcome 1 occurs exactly $k = 2$ times in $m = 4$ independent trials following a Binomial distribution $\\text{Bin}(4, 1\/3)$:
\n$$P = \\binom{4}{2} \\left(\\frac{1}{3}\\right)^2 \\left(\\frac{2}{3}\\right)^{4-2} = 6 \\times \\frac{1}{9} \\times \\frac{4}{9} = \\frac{24}{81} = \\frac{8}{27} \\approx 0.2963 \\quad (29.63\\%)$$<\/p>\n

**Final Answer**: $\\frac{8}{27}$ (or approximately $29.63\\%$).<\/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

---<\/p>\n

### **Solution for Problem 2**<\/p>\n

**Concept Overview**: This problem tests the **Inclusion-Exclusion Principle**, the definition of **Event Independence**, and **Conditional Probability**.<\/p>\n

---<\/p>\n

#### **Part (i)**
\nAccording to the Inclusion-Exclusion Principle:
\n$$P[A \\cup B] = P[A] + P[B] - P[A \\cap B]$$<\/p>\n

Substitute the given numerical values to find the joint probability $P[A \\cap B]$:
\n$$0.7 = 0.5 + 0.4 - P[A \\cap B] \\implies P[A \\cap B] = 0.2$$<\/p>\n

Two events $A$ and $B$ are defined to be independent if and only if $P[A \\cap B] = P[A]P[B]$.
\nCalculating the product of the marginal probabilities:
\n$$P[A]P[B] = 0.5 \\times 0.4 = 0.2$$<\/p>\n

Since $P[A \\cap B] = P[A]P[B] = 0.2$, events $A$ and $B$ are **statistically independent**.<\/p>\n

**Final Answer**: $A$ and $B$ are independent.<\/p>\n

---<\/p>\n

#### **Part (ii)**
\nBy definition of conditional probability:
\n$$P[B \\mid A^c] = \\frac{P[B \\cap A^c]}{P[A^c]}$$<\/p>\n

Using sample space partitioning, event $B$ can be decomposed into two mutually exclusive sets $B = (A \\cap B) \\cup (A^c \\cap B)$:
\n$$P[B \\cap A^c] = P[B] - P[A \\cap B] = 0.4 - 0.2 = 0.2$$<\/p>\n

The probability of the complement $A^c$ is:
\n$$P[A^c] = 1 - P[A] = 1 - 0.5 = 0.5$$<\/p>\n

Thus:
\n$$P[B \\mid A^c] = \\frac{0.2}{0.5} = 0.4$$<\/p>\n

*(Note: Since $A$ and $B$ were proven to be independent in Part (i), the occurrence or non-occurrence of $A$ provides no information about $B$, directly implying $P[B \\mid A^c] = P[B] = 0.4$.)*<\/p>\n

**Final Answer**: $0.4$.<\/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

---<\/p>\n

### **Solution for Problem 3**<\/p>\n

**Concept Overview**: This problem evaluates sequential independent Bernoulli trials, truncated geometric distributions, and Bayes' Rule.<\/p>\n

---<\/p>\n

#### **Part (i)**
\nLet $p = 0.8$ be the probability of success per attempt, and $q = 1 - p = 0.2$ be the probability of failure.<\/p>\n

The complementary event of \"succeeding within 4 attempts\" is \"failing all 4 attempts\". Since individual attempts are independent, the probability of 4 consecutive failures is:
\n$$P(\\text{All 4 Fail}) = q^4 = (0.2)^4 = 0.0016$$<\/p>\n

Subtracting this from 1 gives the probability of at least one success within 4 attempts:
\n$$P(\\text{Success within 4 attempts}) = 1 - 0.0016 = 0.9984 \\quad (99.84\\%)$$<\/p>\n

*(Alternative method by summation of disjoint success paths: $P = 0.8 + 0.2(0.8) + 0.2^2(0.8) + 0.2^3(0.8) = 0.8(1 + 0.2 + 0.04 + 0.008) = 0.9984$.)*<\/p>\n

**Final Answer**: $0.9984$ (or $99.84\\%$).<\/p>\n

---<\/p>\n

#### **Part (ii)**
\nLet $S$ denote the event that transmission succeeds within 4 attempts ($P[S] = 0.9984$).
\nLet $S_3$ denote the event that the first successful transmission occurs on the 3rd attempt.<\/p>\n

The event $S_3$ requires failure on the 1st attempt, failure on the 2nd attempt, and success on the 3rd attempt:
\n$$P[S_3] = q \\times q \\times p = (0.2) \\times (0.2) \\times (0.8) = 0.032$$<\/p>\n

Since $S_3 \\subseteq S$, their intersection is $S_3 \\cap S = S_3$. Applying Bayes' formula:
\n$$P[S_3 \\mid S] = \\frac{P[S_3 \\cap S]}{P[S]} = \\frac{P[S_3]}{P[S]} = \\frac{0.032}{0.9984} = \\frac{320}{9984} = \\frac{5}{156} \\approx 0.032051 \\quad (3.21\\%)$$<\/p>\n

**Final Answer**: $\\frac{5}{156}$ (or approximately $3.21\\%$).<\/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

---<\/p>\n

### **Solution for Problem 4**<\/p>\n

**Concept Overview**: This problem tests the **Law of Total Probability (LTP)** and **Bayes' Theorem** applied over a partition of the sample space.<\/p>\n

Let $L_1, L_2, L_3$ represent the partition of production lines with prior probabilities:
\n$$P[L_1] = 0.20, \\quad P[L_2] = 0.30, \\quad P[L_3] = 0.50$$<\/p>\n

Let $D$ denote the event that a product is defective. The conditional defect rates are:
\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

#### **Part (i)**
\nBy the Law of Total Probability:
\n$$P[D] = \\sum_{i=1}^{3} P[L_i] P[D \\mid L_i]$$<\/p>\n

Substitute the given numbers:
\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

**Final Answer**: $0.028$ (or $2.8\\%$).<\/p>\n

---<\/p>\n

#### **Part (ii)**
\nLet $D^c$ denote the event that a product is non-defective. The total probability of a non-defective product is:
\n$$P[D^c] = 1 - P[D] = 1 - 0.028 = 0.972$$<\/p>\n

We want to find $P[L_1 \\cup L_2 \\mid D^c]$. Since $L_1$ and $L_2$ are mutually exclusive production lines:
\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

Calculate the joint probabilities for producing non-defective items on $L_1$ and $L_2$:
\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

Summing the numerators and dividing by $P[D^c]$:
\n$$P[L_1 \\cup L_2 \\mid D^c] = \\frac{0.198 + 0.294}{0.972} = \\frac{0.492}{0.972} = \\frac{492}{972} = \\frac{41}{81} \\approx 0.506173 \\quad (50.62\\%)$$<\/p>\n

**Final Answer**: $\\frac{41}{81}$ (or approximately $50.62\\%$).<\/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---\r\n\r\n**Solution for Problem 1**\r\n\r\n**Concept Overview**: This problem tests the **Multinomial Distribution** for independent repetitions of a discrete experiment with more than two possible outcomes, as well as **Conditional Probability**.\r\n\r\nFirst, determine the single-trial probabilities fo...","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":[10],"tags":[],"class_list":["post-828","post","type-post","status-publish","format-standard","hentry","category-academics"],"_links":{"self":[{"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts\/828","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=828"}],"version-history":[{"count":1,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts\/828\/revisions"}],"predecessor-version":[{"id":829,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/posts\/828\/revisions\/829"}],"wp:attachment":[{"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/media?parent=828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/categories?post=828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wuhanqing.cn\/wordpress\/wp-json\/wp\/v2\/tags?post=828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}