1. **Multiplication Law of Probability:** The multiplication law states that for two events A and B, the probability of both occurring is: $$ P(A \cap B) = P(A) \times P(B|A) $$ If A and B are independent, then: $$ P(A \cap B) = P(A) \times P(B) $$ 2. **Question 4 a):** A quiz has 6 questions with 5 options each. Only 1 is correct, so 4 options are incorrect per question. Number of ways to answer all incorrectly is: $$ 4^{6} = 4096 $$ Correct answer: 4) 4096 3. **Question 4 b):** Number of 4-letter codes from Arabic alphabet (assumed 28 letters): Number of codes: $$ 28^{4} = 614656 $$ Correct answer: 2) 614656 4. **Question 4 c):** Sample space $$ \Omega = \{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} $$ $$ A=\{HHH,HHT,THT\}, B=\{HHH,THH\} $$ Complement of A: $$ \bar{A} = \Omega \setminus A = \{HTH,HTT,THH,TTH,TTT\} $$ Event $$ \bar{A} \cap B = \{THH\} $$ Correct answer: 2) {THH} 5. **Question 4 d):** For mutually exclusive events A and B: $$ A \cap B = \emptyset $$ We want $$ P(\bar{A} \cup B) $$ assuming union (there is ambiguity, assuming union symbol instead of space or '\cap'). Since A and B disjoint, also: $$ P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(\bar{A} \cap B) $$ Since $$ \bar{A} \cap B = B $$ (because B is disjoint from A, so entirely in \bar{A}), $$ P(\bar{A} \cup B) = P(\bar{A}) + P(B) - P(B) = P(\bar{A}) $$ Correct answer: 3) P(\bar{A}) 6. **Question 5:** Given: $$ P(A \cap B) = 0.20, P(A \cup B) = 0.80, P(\bar{B})=0.35 $$ 6a. Find $$ P(A) $$ Recall: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$ Also, $$ P(B) = 1 - P(\bar{B}) = 1 - 0.35 = 0.65 $$ Substitute: $$ 0.80 = P(A) + 0.65 - 0.20 $$ $$ P(A) = 0.80 - 0.65 + 0.20 = 0.35 $$ 6b. Find $$ P(\bar{A} \cap \bar{B}) $$ By De Morgan: $$ \bar{A} \cap \bar{B} = \overline{A \cup B} $$ So: $$ P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) = 1 - 0.80 = 0.20 $$ 6c. Find $$ P(B \setminus A) = P(B) - P(A \cap B) = 0.65 - 0.20 = 0.45 $$ 7. **Question 6 a):** Experiment: Roll a fair die and toss a fair coin. Sample space: $$ \Omega = \{ (d,c) | d = 1..6, c = H,T \} $$ Total outcomes = 6 \times 2 = 12 Probability for each outcome: $$ P(\{(d,c)\}) = \frac{1}{12} $$ 7b. Event A: Even number appears Even numbers: 2,4,6 (3 numbers) Possible outcomes for A: $$ 3 \times 2 = 6 $$ Probability: $$ P(A) = \frac{6}{12} = 0.5 $$ 7c. Event B: Number greater than 4 with head Numbers > 4: 5,6 Heads only Number of outcomes: $$ 2 \times 1 = 2 $$ Probability: $$ P(B) = \frac{2}{12} = \frac{1}{6} \approx 0.1667 $$ 7d. Check independence of A and B: Check if: $$ P(A \cap B) = P(A) \times P(B) $$ $$ A \cap B = \{(6,H),(5,H)\} \cap \text{even numbers} = \{(6,H)\} $$ (only 6 is even and >4 with head) So: $$ P(A \cap B) = \frac{1}{12} \approx 0.0833 $$ $$ P(A) \times P(B) = 0.5 \times \frac{1}{6} = \frac{1}{12} = 0.0833 $$ Since equal, A and B are independent. 8. **Question 7 a):** Four machines with equal production capacity, so probability of selecting product from each machine: $$ P(M_i) = \frac{1}{4} $$ Defective probabilities: $$ P(D|M_1) = 0.07, P(D|M_2)=0.05, P(D|M_3) = 0.03, P(D|M_4) = 0.02 $$ Total defective probability: $$ P(D) = \sum_{i=1}^{4} P(M_i)P(D|M_i) = \frac{1}{4}(0.07 + 0.05 + 0.03 + 0.02) = \frac{1}{4} \times 0.17 = 0.0425 $$ 7b. Find $$ P(M_2|D) $$ using Bayes' theorem: $$ P(M_2|D) = \frac{P(M_2) P(D|M_2)}{P(D)} = \frac{\frac{1}{4} \times 0.05}{0.0425} = \frac{0.0125}{0.0425} \approx 0.2941 $$ **Final answers:** - 4a) 4096 - 4b) 614656 - 4c) {THH} - 4d) P(\bar{A}) - 5a) P(A) = 0.35 - 5b) P(\bar{A} \cap \bar{B}) = 0.20 - 5c) P(B \setminus A) = 0.45 - 6a) \Omega with 12 outcomes and P=1/12 each - 6b) P(A) = 0.5 - 6c) P(B) = 1/6 - 6d) A and B independent - 7a) P(D) = 0.0425 - 7b) P(M_2|D) = 0.2941