Probability and Random Variables
2026.09.28
**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.
i. (10 pts) Find the probability that the number 1 appears exactly twice and the number 2 appears exactly three times.
ii. (10 pts) Given that the number 3 appears exactly four times, find the conditional probability that the number 1 appears exactly twice.
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### **Solution for Problem 1**
**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**.
First, determine the single-trial probabilities for each face label based on the equiprobable sample space of a 6-sided die:
- $P(1) = \frac{1}{6}$
- $P(2) = \frac{2}{6} = \frac{1}{3}$
- $P(3) = \frac{3}{6} = \frac{1}{2}$
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#### **Part (i)**
In $n = 8$ 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.
Using the multinomial probability formula:
$$P(N_1=2, N_2=3, N_3=3) = \frac{n!}{n_1! n_2! n_3!} p_1^{n_1} p_2^{n_2} p_3^{n_3}$$
Substitute the values:
1. **Multinomial Coefficient**:
$$\frac{8!}{2! \cdot 3! \cdot 3!} = \frac{40320}{2 \times 6 \times 6} = \frac{40320}{72} = 560$$
2. **Probability Factors**:
$$\left(\frac{1}{6}\right)^2 \times \left(\frac{1}{3}\right)^3 \times \left(\frac{1}{2}\right)^3 = \frac{1}{36} \times \frac{1}{27} \times \frac{1}{8} = \frac{1}{7776}$$
3. **Combined Calculation**:
$$P = 560 \times \frac{1}{7776} = \frac{560}{7776} = \frac{140}{1944} = \frac{35}{486} \approx 0.072016 \quad (7.20\%)$$
**Final Answer**: $\frac{35}{486}$ (or approximately $7.20\%$).
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#### **Part (ii)**
Given 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.
We construct a conditional sample space over these remaining 4 trials with revised conditional probabilities given $3^c$:
- $P(1 \mid 3^c) = \frac{1/6}{1/2} = \frac{1}{3}$
- $P(2 \mid 3^c) = \frac{1/3}{1/2} = \frac{2}{3}$
The problem simplifies to finding the probability that outcome 1 occurs exactly $k = 2$ times in $m = 4$ independent trials following a Binomial distribution $\text{Bin}(4, 1/3)$:
$$P = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{4-2} = 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{24}{81} = \frac{8}{27} \approx 0.2963 \quad (29.63\%)$$
**Final Answer**: $\frac{8}{27}$ (or approximately $29.63\%$).
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**Problem 2. [20 Points]** Two events $A$ and $B$ satisfy
$$P[A] = 0.5, \quad P[B] = 0.4, \quad P[A \cup B] = 0.7$$
i. (10 pts) Determine whether $A$ and $B$ are independent.
ii. (10 pts) Find $P[B \mid A^c]$.
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### **Solution for Problem 2**
**Concept Overview**: This problem tests the **Inclusion-Exclusion Principle**, the definition of **Event Independence**, and **Conditional Probability**.
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#### **Part (i)**
According to the Inclusion-Exclusion Principle:
$$P[A \cup B] = P[A] + P[B] - P[A \cap B]$$
Substitute the given numerical values to find the joint probability $P[A \cap B]$:
$$0.7 = 0.5 + 0.4 - P[A \cap B] \implies P[A \cap B] = 0.2$$
Two events $A$ and $B$ are defined to be independent if and only if $P[A \cap B] = P[A]P[B]$.
Calculating the product of the marginal probabilities:
$$P[A]P[B] = 0.5 \times 0.4 = 0.2$$
Since $P[A \cap B] = P[A]P[B] = 0.2$, events $A$ and $B$ are **statistically independent**.
**Final Answer**: $A$ and $B$ are independent.
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#### **Part (ii)**
By definition of conditional probability:
$$P[B \mid A^c] = \frac{P[B \cap A^c]}{P[A^c]}$$
Using sample space partitioning, event $B$ can be decomposed into two mutually exclusive sets $B = (A \cap B) \cup (A^c \cap B)$:
$$P[B \cap A^c] = P[B] - P[A \cap B] = 0.4 - 0.2 = 0.2$$
The probability of the complement $A^c$ is:
$$P[A^c] = 1 - P[A] = 1 - 0.5 = 0.5$$
Thus:
$$P[B \mid A^c] = \frac{0.2}{0.5} = 0.4$$
*(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 $P[B \mid A^c] = P[B] = 0.4$.)*
**Final Answer**: $0.4$.
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**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.
i. (10 pts) Find the probability that the packet is successfully transmitted within four attempts.
ii. (10 pts) Given that the packet is successfully transmitted, find the probability that the first successful transmission occurs on the third attempt.
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### **Solution for Problem 3**
**Concept Overview**: This problem evaluates sequential independent Bernoulli trials, truncated geometric distributions, and Bayes' Rule.
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#### **Part (i)**
Let $p = 0.8$ be the probability of success per attempt, and $q = 1 - p = 0.2$ be the probability of failure.
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:
$$P(\text{All 4 Fail}) = q^4 = (0.2)^4 = 0.0016$$
Subtracting this from 1 gives the probability of at least one success within 4 attempts:
$$P(\text{Success within 4 attempts}) = 1 - 0.0016 = 0.9984 \quad (99.84\%)$$
*(Alternative method by summation of disjoint success paths: $P = 0.8 + 0.2(0.8) + 0.2^2(0.8) + 0.2^3(0.8) = 0.8(1 + 0.2 + 0.04 + 0.008) = 0.9984$.)*
**Final Answer**: $0.9984$ (or $99.84\%$).
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#### **Part (ii)**
Let $S$ denote the event that transmission succeeds within 4 attempts ($P[S] = 0.9984$).
Let $S_3$ denote the event that the first successful transmission occurs on the 3rd attempt.
The event $S_3$ requires failure on the 1st attempt, failure on the 2nd attempt, and success on the 3rd attempt:
$$P[S_3] = q \times q \times p = (0.2) \times (0.2) \times (0.8) = 0.032$$
Since $S_3 \subseteq S$, their intersection is $S_3 \cap S = S_3$. Applying Bayes' formula:
$$P[S_3 \mid S] = \frac{P[S_3 \cap S]}{P[S]} = \frac{P[S_3]}{P[S]} = \frac{0.032}{0.9984} = \frac{320}{9984} = \frac{5}{156} \approx 0.032051 \quad (3.21\%)$$
**Final Answer**: $\frac{5}{156}$ (or approximately $3.21\%$).
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**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.
i. (10 pts) Find the probability that the selected product is defective.
ii. (10 pts) Given that the selected product is not defective, find the probability that it was produced by either $L_1$ or $L_2$.
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### **Solution for Problem 4**
**Concept Overview**: This problem tests the **Law of Total Probability (LTP)** and **Bayes' Theorem** applied over a partition of the sample space.
Let $L_1, L_2, L_3$ represent the partition of production lines with prior probabilities:
$$P[L_1] = 0.20, \quad P[L_2] = 0.30, \quad P[L_3] = 0.50$$
Let $D$ denote the event that a product is defective. The conditional defect rates are:
$$P[D \mid L_1] = 0.01, \quad P[D \mid L_2] = 0.02, \quad P[D \mid L_3] = 0.04$$
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#### **Part (i)**
By the Law of Total Probability:
$$P[D] = \sum_{i=1}^{3} P[L_i] P[D \mid L_i]$$
Substitute the given numbers:
$$P[D] = (0.20 \times 0.01) + (0.30 \times 0.02) + (0.50 \times 0.04) = 0.002 + 0.006 + 0.020 = 0.028 \quad (2.8\%)$$
**Final Answer**: $0.028$ (or $2.8\%$).
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#### **Part (ii)**
Let $D^c$ denote the event that a product is non-defective. The total probability of a non-defective product is:
$$P[D^c] = 1 - P[D] = 1 - 0.028 = 0.972$$
We want to find $P[L_1 \cup L_2 \mid D^c]$. Since $L_1$ and $L_2$ are mutually exclusive production lines:
$$P[L_1 \cup L_2 \mid D^c] = \frac{P[(L_1 \cup L_2) \cap D^c]}{P[D^c]} = \frac{P[L_1 \cap D^c] + P[L_2 \cap D^c]}{P[D^c]}$$
Calculate the joint probabilities for producing non-defective items on $L_1$ and $L_2$:
- $P[L_1 \cap D^c] = P[L_1] P[D^c \mid L_1] = 0.20 \times (1 - 0.01) = 0.20 \times 0.99 = 0.198$
- $P[L_2 \cap D^c] = P[L_2] P[D^c \mid L_2] = 0.30 \times (1 - 0.02) = 0.30 \times 0.98 = 0.294$
Summing the numerators and dividing by $P[D^c]$:
$$P[L_1 \cup L_2 \mid D^c] = \frac{0.198 + 0.294}{0.972} = \frac{0.492}{0.972} = \frac{492}{972} = \frac{41}{81} \approx 0.506173 \quad (50.62\%)$$
**Final Answer**: $\frac{41}{81}$ (or approximately $50.62\%$).
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