Let's solve **Question 7** step by step: --- ## **Given Data** Three numbers are randomly selected from to 9 (total 10 digits, selection is **with replacement** unless otherwise stated). Let: - \( A_1 \): the event that the three numbers do **not** contain or 5 - \( A_2 \): the event that the three numbers do **not** contain **or** 5 We are to find \( P(A_1) \) and \( P(A_2) \). --- ## **Definitions Used** - **Probability** of an event \( A \): \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \) - If numbers are chosen **with replacement**, each selection is independent. --- ## **Step-by-Step Solution** ### **Total Number of Possible Outcomes** Each number is chosen independently from to 9, so there are: \[ 10 \times 10 \times 10 = 100 \] total possible outcomes. --- ### **Event \( A_1 \): Three numbers don’t contain **and** 5** This means: none of the three numbers is or 5. - The available digits are: 1, 2, 3, 4, 6, 7, 8, 9 (**exclude and 5**) - Total available digits = 8 - Each of the three numbers must be selected from these 8 digits Number of favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] Probability: \[ P(A_1) = \frac{512}{100} = .512 \] --- ### **Event \( A_2 \): Three numbers don’t contain **or** 5** This means: none of the three numbers is , and none is 5 (i.e., numbers don't contain or don't contain 5). But from the wording, it is likely asking: **the three numbers do not contain either or 5** (i.e., all three numbers are not or all three are not 5). But for clarity, let's solve **both interpretations**: #### **Case 1: "Don't contain or 5" = Don't contain , and don't contain 5 (same as \(A_1\))** In this case, \[ P(A_2) = P(A_1) = .512 \] #### **Case 2: Don't contain or don't contain 5 (at least one of or 5 is missing in the selected numbers)** Let’s use inclusion-exclusion principle: Let: - \( B \): three numbers do not contain - \( C \): three numbers do not contain 5 We need \( P(B \cup C) \). - \( P(B) \): none of the three numbers is → each chosen from 1-9 (9 choices), so \( 9 \times 9 \times 9 = 729 \) - \( P(C) \): none of the three numbers is 5 → each chosen from -4, 6-9 (9 choices), so \( 9 \times 9 \times 9 = 729 \) - \( P(B \cap C) \): none of the three numbers is or 5 → each chosen from 1-4, 6-9 (8 choices), so \( 8 \times 8 \times 8 = 512 \) By inclusion-exclusion: \[ P(B \cup C) = \frac{729 + 729 - 512}{100} = \frac{946}{100} = .946 \] --- ## **Final Answers** - \( \boxed{P(A_1) = .512} \) - \( \boxed{P(A_2) = .946} \) (if interpreted as "do not contain or do not contain 5") If \(A_2\) is interpreted as "do not contain and do not contain 5", then \(P(A_2) = P(A_1) = .512\). --- ### **Summary Table** | Event | Probability | |--------------|-------------| | \(A_1\) | .512 | | \(A_2\) | .946 | --- **If you need a solution for another question, let me know!**

# Probability Questions Solution Let's solve the following questions step by step. --- ## **Question 7: Three Numbers from 0 to 9** ### Given Data - Three numbers are randomly selected from 0 to 9. - Let: - \( A_1 \): Event that the three numbers don’t contain 0 and 5. - \( A_2 \): Event that the three numbers don’t contain 0 or 5. ### Definitions Used - Probability of an event \( A \): \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] ### Total Number of Possible Outcomes Each number is chosen independently from 0 to 9 (10 total digits): \[ 10 \times 10 \times 10 = 1000 \] ### Event \( A_1 \): Three numbers don’t contain 0 and 5 - Available digits: 1, 2, 3, 4, 6, 7, 8, 9 (8 digits). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_1) = \frac{512}{1000} = 0.512 \] ### Event \( A_2 \): Three numbers don’t contain 0 or 5 - Available digits: 1, 2, 3, 4, 6, 7, 8, 9 (8 digits). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_2) = \frac{512}{1000} = 0.512 \] ### Final Answers for Question 7 - \( \boxed{P(A_1) = 0.512} \) - \( \boxed{P(A_2) = 0.512} \) --- ## **Question 8: Birthday Probabilities of 6 Students** ### Given Data - 6 students live in a dorm. ### (1) Probability that at least one birthday is in October - Complement probability: No birthdays in October. - Probability of a birthday not in October = \( \frac{11}{12} \). - Probability of none in October: \[ \left( \frac{11}{12} \right)^6 \] - Probability: \[ P(\text{at least one in October}) = 1 - \left( \frac{11}{12} \right)^6 \approx 0.4323 \] ### (2) Probability that 4 birthdays are exactly in October - Number of ways to choose 4 students: \[ \binom{6}{4} = 15 \] - Probability for 4 in October and 2 not: \[ \left( \frac{1}{12} \right)^4 \left( \frac{11}{12} \right)^2 \] - Total probability: \[ P = 15 \times \left( \frac{1}{12} \right)^4 \times \left( \frac{11}{12} \right)^2 \approx 0.0027 \] ### (3) Probability that 4 birthdays are in the same month - Choose a month (12 choices). - Probability: \[ 12 \times \left( \frac{1}{12} \right)^4 \left( \frac{11}{12} \right)^2 \times \binom{6}{4} = 12 \times \frac{1}{20736} \times 15 \approx 0.0086 \] ### Final Answers for Question 8 - (1) \( \boxed{0.4323} \) - (2) \( \boxed{0.0027} \) - (3) \( \boxed{0.0086} \) --- ## **Question 9: Product Probabilities** ### Given Data - Products A, B, C account for 60%, 30%, and 10% of total products respectively. ### Probability that a withdrawn product is A given it is not C - Let \( P(A) = 0.6 \), \( P(B) = 0.3 \), \( P(C) = 0.1 \). - Products not C: A and B. - Probability of A given not C: \[ P(A| \text{not } C) = \frac{P(A)}{P(A) + P(B)} = \frac{0.6}{0.6 + 0.3} = \frac{0.6}{0.9} = \frac{2}{3} \approx 0.6667 \] ### Final Answer for Question 9 - \( \boxed{0.6667} \) --- ## **Question 10: Warning Systems** ### Given Data - Probability of System 1 working = 0.92 - Probability of System 2 working = 0.93 - Probability that System 2 works if System 1 is down = 0.85. ### (1) Probability that both systems are working \[ P(\text{1 and 2 working}) = P(1) \times P(2) = 0.92 \times 0.93 = 0.8576 \] ### (2) Probability that System 2 is down and System 1 is working \[ P(1 \text{ working}) \times P(2 \text{ down}) = 0.92 \times (1 - 0.93) = 0.92 \times 0.07 = 0.0644 \] ### (3) Probability that System 1 is working given System 2 is down Using Bayes' theorem: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{P(1 \text{ working}) \times P(2 \text{ down} | 1 \text{ working})}{P(2 \text{ down})} \] - \( P(2 \text{ down}) = P(2 \text{ down} | 1 \text{ working}) \cdot P(1) + P(2 \text{ down} | 1 \text{ down}) \cdot P(1 \text{ down}) \) - \( P(2 \text{ down} | 1 \text{ working}) = 1 - 0.93 = 0.07 \) - \( P(2 \text{ down} | 1 \text{ down}) = 1 - 0.85 = 0.15 \) - \( P(2 \text{ down}) = 0.07 \cdot 0.92 + 0.15 \cdot 0.08 = 0.0644 + 0.012 = 0.0764 \) Then: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{0.92 \cdot 0.07}{0.0764} \approx 0.8425 \] ### Final Answers for Question 10 - (1) \( \boxed{0.8576} \) - (2) \( \boxed{0.0644} \) - (3) \( \boxed{0.8425} \) --- ## **Question 11: Independent Events** ### Given Data Let \( 0 < P(A) < 1 \). ### Proof - Events A and B are independent if: \[ P(B | A) = P(B) \] - By definition of conditional probability: \[ P(B | A) = \frac{P(A \cap B)}{P(A)} \] - Thus, if \( P(B | A) = P(B) \), it implies \( P(A \cap B) = P(A) \cdot P(B) \). ### Final Answer for Question 11 The independence condition holds: - \( \boxed{\text{A and B are independent if and only if } P(B | A) = P(B)} \) --- ## **Question 12: Independent Events Probabilities** ### Given Data - \( P(A) = P(B) = 0.25 \) ### Solution Since \( A \) and \( B \) are independent: - \( P(A \cap B) = P(A) \cdot P(B) = 0.25 \cdot 0.25 = 0.0625 \) ### Final Answers for Question 12 - \( \boxed{P(A) = 0.25} \) - \( \boxed{P(B) = 0.25} \) --- ## **Question 13: Independence and Mutual Exclusivity** ### Given Data Let \( P(A) > 0 \) and \( P(B) > 0 \). ### (1) Independence implies not mutually exclusive - If A and B are independent: \[ P(A \cap B) = P(A) \cdot P(B) > 0 \] - Thus, they cannot be mutually exclusive. ### (2) Mutual exclusivity implies dependence - If A and B are mutually exclusive, then: \[ P(A \cap B) = 0 \] - Hence, they are dependent events. ### Final Answers for Question 13 - (1) \( \boxed{\text{If A and B are independent, they are not mutually exclusive.}} \) - (2) \( \boxed{\text{If A and B are mutually exclusive, they are dependent.}} \) --- If you have any further questions or need additional clarification, feel free to ask!

# Probability Questions Solution Let's provide a comprehensive solution for the given questions along with the final answers for each. --- ## **Question 7: Three Numbers from 0 to 9** ### Given Data - Three numbers are randomly selected from 0 to 9. - Let: - \( A_1 \): Event that the three numbers don’t contain 0 and 5. - \( A_2 \): Event that the three numbers don’t contain 0 or 5. ### Definitions Used - Probability of an event \( A \): \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] ### Total Number of Possible Outcomes Each number is chosen independently from 0 to 9 (10 total digits): \[ 10 \times 10 \times 10 = 1000 \] ### Event \( A_1 \): Three numbers don’t contain 0 and 5 - Available digits: 1, 2, 3, 4, 6, 7, 8, 9 (8 digits). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_1) = \frac{512}{1000} = 0.512 \] ### Event \( A_2 \): Three numbers don’t contain 0 or 5 - Available digits: 1, 2, 3, 4, 6, 7, 8, 9 (8 digits). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_2) = \frac{512}{1000} = 0.512 \] ### Final Answers for Question 7 - \( \boxed{P(A_1) = 0.512} \) - \( \boxed{P(A_2) = 0.512} \) --- ## **Question 8: Birthday Probabilities of 6 Students** ### Given Data - 6 students live in a dorm. ### (1) Probability that at least one birthday is in October - Complement probability: No birthdays in October. - Probability of a birthday not in October = \( \frac{11}{12} \). - Probability of none in October: \[ \left( \frac{11}{12} \right)^6 \approx 0.4943 \] - Probability: \[ P(\text{at least one in October}) = 1 - \left( \frac{11}{12} \right)^6 \approx 1 - 0.4943 \approx 0.5057 \] ### (2) Probability that 4 birthdays are exactly in October - Number of ways to choose 4 students: \[ \binom{6}{4} = 15 \] - Probability for 4 in October and 2 not: \[ \left( \frac{1}{12} \right)^4 \left( \frac{11}{12} \right)^2 \] - Total probability: \[ P = 15 \times \left( \frac{1}{12} \right)^4 \times \left( \frac{11}{12} \right)^2 \approx 15 \times \frac{1}{20736} \times \frac{121}{144} \approx 0.0008 \] ### (3) Probability that 4 birthdays are in the same month - Choose a month (12 choices). - Probability: \[ 12 \times \left( \frac{1}{12} \right)^4 \left( \frac{11}{12} \right)^2 \times \binom{6}{4} = 12 \times \frac{1}{20736} \times 15 \approx 0.0086 \] ### Final Answers for Question 8 - (1) \( \boxed{0.5057} \) - (2) \( \boxed{0.0008} \) - (3) \( \boxed{0.0086} \) --- ## **Question 9: Product Probabilities** ### Given Data - Products A, B, C account for 60%, 30%, and 10% of total products respectively. ### Probability that a withdrawn product is A given it is not C - Let \( P(A) = 0.6 \), \( P(B) = 0.3 \), \( P(C) = 0.1 \). - Products not C: A and B. - Probability of A given not C: \[ P(A| \text{not } C) = \frac{P(A)}{P(A) + P(B)} = \frac{0.6}{0.6 + 0.3} = \frac{0.6}{0.9} = \frac{2}{3} \approx 0.6667 \] ### Final Answer for Question 9 - \( \boxed{0.6667} \) --- ## **Question 10: Warning Systems** ### Given Data - Probability of System 1 working = 0.92 - Probability of System 2 working = 0.93 - Probability that System 2 works if System 1 is down = 0.85. ### (1) Probability that both systems are working \[ P(\text{1 and 2 working}) = P(1) \times P(2) = 0.92 \times 0.93 = 0.8576 \] ### (2) Probability that System 2 is down and System 1 is working \[ P(1 \text{ working}) \times P(2 \text{ down}) = 0.92 \times (1 - 0.93) = 0.92 \times 0.07 = 0.0644 \] ### (3) Probability that System 1 is working given System 2 is down Using Bayes' theorem: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{P(1 \text{ working}) \times P(2 \text{ down} | 1 \text{ working})}{P(2 \text{ down})} \] - \( P(2 \text{ down}) = P(2 \text{ down} | 1 \text{ working}) \cdot P(1) + P(2 \text{ down} | 1 \text{ down}) \cdot P(1 \text{ down}) \) - \( P(2 \text{ down} | 1 \text{ working}) = 1 - 0.93 = 0.07 \) - \( P(2 \text{ down} | 1 \text{ down}) = 1 - 0.85 = 0.15 \) - \( P(2 \text{ down}) = 0.07 \cdot 0.92 + 0.15 \cdot 0.08 = 0.0644 + 0.012 = 0.0764 \) Then: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{0.92 \cdot 0.07}{0.0764} \approx 0.8425 \] ### Final Answers for Question 10 - (1) \( \boxed{0.8576} \) - (2) \( \boxed{0.0644} \) - (3) \( \boxed{0.8425} \) --- ## **Question 11: Independent Events** ### Given Data Let \( 0 < P(A) < 1 \). ### Proof - Events A and B are independent if: \[ P(B | A) = P(B) \] - By definition of conditional probability: \[ P(B | A) = \frac{P(A \cap B)}{P(A)} \] - Thus, if \( P(B | A) = P(B) \), it implies \( P(A \cap B) = P(A) \cdot P(B) \). ### Final Answer for Question 11 The independence condition holds: - \( \boxed{\text{A and B are independent if and only if } P(B | A) = P(B)} \) --- ## **Question 12: Independent Events Probabilities** ### Given Data - \( P(A) = P(B) = 0.25 \) ### Solution Since \( A \) and \( B \) are independent: - \( P(A \cap B) = P(A) \cdot P(B) = 0.25 \cdot 0.25 = 0.0625 \) ### Final Answers for Question 12 - \( \boxed{P(A) = 0.25} \) - \( \boxed{P(B) = 0.25} \) --- ## **Question 13: Independence and Mutual Exclusivity** ### Given Data Let \( P(A) > 0 \) and \( P(B) > 0 \). ### (1) Independence implies not mutually exclusive - If A and B are independent: \[ P(A \cap B) = P(A) \cdot P(B) > 0 \] - Thus, they cannot be mutually exclusive. ### (2) Mutual exclusivity implies dependence - If A and B are mutually exclusive, then: \[ P(A \cap B) = 0 \] - Hence, they are dependent events. ### Final Answers for Question 13 - (1) \( \boxed{\text{If A and B are independent, they are not mutually exclusive.}} \) - (2) \( \boxed{\text{If A and B are mutually exclusive, they are dependent.}} \) --- If you have further questions or need additional assistance, feel free to ask!

# Final Answers Summary for All Questions Here are the final answers to all the questions presented: ## **Question 7: Three Numbers from 0 to 9** - \( P(A_1) = 0.512 \) - \( P(A_2) = 0.512 \) ## **Question 8: Birthday Probabilities of 6 Students** - (1) Probability that at least one birthday is in October: \( \boxed{0.5057} \) - (2) Probability that 4 birthdays are exactly in October: \( \boxed{0.0008} \) - (3) Probability that 4 birthdays are in the same month: \( \boxed{0.0086} \) ## **Question 9: Product Probabilities** - Probability that a withdrawn product is A given it is not C: \( \boxed{0.6667} \) ## **Question 10: Warning Systems** - (1) Probability that both systems are working: \( \boxed{0.8576} \) - (2) Probability that System 2 is down and System 1 is working: \( \boxed{0.0644} \) - (3) Probability that System 1 is working given System 2 is down: \( \boxed{0.8425} \) ## **Question 11: Independent Events** - A and B are independent if and only if \( P(B | A) = P(B) \) ## **Question 12: Independent Events Probabilities** - \( P(A) = 0.25 \) - \( P(B) = 0.25 \) ## **Question 13: Independence and Mutual Exclusivity** - (1) If A and B are independent, they are not mutually exclusive. - (2) If A and B are mutually exclusive, they are dependent. --- If there are any further questions or clarifications needed, please let me know!

# Comprehensive Probability Solutions This document presents unique and plagiarism-free solutions for various probability questions, along with final answers compiled at the end. --- ## **Question 7: Three Numbers from 0 to 9** ### Given Data - Selection of three numbers from the digits 0 to 9. - Define: - \( A_1 \): Three numbers do not contain 0 and 5. - \( A_2 \): Three numbers do not contain 0 or 5. ### Definitions - Probability of an event \( A \): \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] ### Total Outcomes All combinations of three numbers from 0 to 9: \[ 10 \times 10 \times 10 = 1000 \] ### Event \( A_1 \): No 0 and 5 - Valid digits: 1, 2, 3, 4, 6, 7, 8, 9 (8 options). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_1) = \frac{512}{1000} = 0.512 \] ### Event \( A_2 \): No 0 or 5 - Valid digits: 1, 2, 3, 4, 6, 7, 8, 9 (same as above). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_2) = \frac{512}{1000} = 0.512 \] ### Final Answers for Question 7 - \( \boxed{P(A_1) = 0.512} \) - \( \boxed{P(A_2) = 0.512} \) --- ## **Question 8: Birthday Probabilities of 6 Students** ### Given Data - 6 students living in a dorm. ### (1) Probability of at least one birthday in October - Probability of a birthday not in October: \( \frac{11}{12} \). - Probability of none in October: \[ \left( \frac{11}{12} \right)^6 \approx 0.4943 \] - Thus, probability of at least one in October: \[ P(\text{at least one in October}) = 1 - 0.4943 \approx 0.5057 \] ### (2) Probability that exactly 4 birthdays are in October - Ways to choose 4 students: \[ \binom{6}{4} = 15 \] - Probability for 4 in October and 2 not: \[ = 15 \times \left( \frac{1}{12} \right)^4 \times \left( \frac{11}{12} \right)^2 \approx 0.0008 \] ### (3) Probability that 4 birthdays are in the same month - Choose a month (12 options): \[ 12 \times \left( \frac{1}{12} \right)^4 \times \left( \frac{11}{12} \right)^{2} \times \binom{6}{4} \approx 0.0086 \] ### Final Answers for Question 8 - (1) \( \boxed{0.5057} \) - (2) \( \boxed{0.0008} \) - (3) \( \boxed{0.0086} \) --- ## **Question 9: Product Probabilities** ### Given Data - Products A, B, C account for 60%, 30%, and 10% of total products respectively. ### Probability that a withdrawn product is A given it is not C - Let \( P(A) = 0.6 \), \( P(B) = 0.3 \), \( P(C) = 0.1 \). - Probability of A given not C: \[ P(A | \text{not } C) = \frac{P(A)}{P(A) + P(B)} = \frac{0.6}{0.6 + 0.3} = \frac{0.6}{0.9} = \frac{2}{3} \approx 0.6667 \] ### Final Answer for Question 9 - \( \boxed{0.6667} \) --- ## **Question 10: Warning Systems** ### Given Data - System 1 working probability = 0.92 - System 2 working probability = 0.93 - Probability that System 2 works if System 1 is down = 0.85. ### (1) Probability that both systems are working \[ P(\text{1 and 2 working}) = P(1) \times P(2) = 0.92 \times 0.93 = 0.8576 \] ### (2) Probability that System 2 is down and System 1 is working \[ P(1 \text{ working}) \times P(2 \text{ down}) = 0.92 \times (1 - 0.93) = 0.92 \times 0.07 = 0.0644 \] ### (3) Probability that System 1 is working given System 2 is down Using Bayes' theorem: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{P(1 \text{ working}) \times P(2 \text{ down} | 1 \text{ working})}{P(2 \text{ down})} \] - \( P(2 \text{ down}) = 0.07 \cdot 0.92 + 0.15 \cdot 0.08 = 0.0764 \) Then: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{0.92 \cdot 0.07}{0.0764} \approx 0.8425 \] ### Final Answers for Question 10 - (1) \( \boxed{0.8576} \) - (2) \( \boxed{0.0644} \) - (3) \( \boxed{0.8425} \) --- ## **Question 11: Independent Events** ### Given Data Let \( 0 < P(A) < 1 \). ### Proof - Events A and B are independent if: \[ P(B | A) = P(B) \] - Using the definition of conditional probability: \[ P(B | A) = \frac{P(A \cap B)}{P(A)} \] - Thus, if \( P(B | A) = P(B) \), then \( P(A \cap B) = P(A) \cdot P(B) \). ### Final Answer for Question 11 - \( \boxed{\text{A and B are independent if and only if } P(B | A) = P(B)} \) --- ## **Question 12: Independent Events Probabilities** ### Given Data - \( P(A) = P(B) = 0.25 \) ### Solution Since \( A \) and \( B \) are independent: \[ P(A \cap B) = P(A) \cdot P(B) = 0.25 \times 0.25 = 0.0625 \] ### Final Answers for Question 12 - \( \boxed{P(A) = 0.25} \) - \( \boxed{P(B) = 0.25} \) --- ## **Question 13: Independence and Mutual Exclusivity** ### Given Data Let \( P(A) > 0 \) and \( P(B) > 0 \). ### (1) Independence implies not mutually exclusive - If A and B are independent: \[ P(A \cap B) = P(A) \cdot P(B) > 0 \] - Therefore, they cannot be mutually exclusive. ### (2) Mutual exclusivity implies dependence - If A and B are mutually exclusive, then: \[ P(A \cap B) = 0 \] - Hence, they are dependent events. ### Final Answers for Question 13 - (1) \( \boxed{\text{If A and B are independent, they are not mutually exclusive.}} \) - (2) \( \boxed{\text{If A and B are mutually exclusive, they are dependent.}} \) --- # **Final Answers Summary** - **Question 7:** - \( P(A_1) = 0.512 \) - \( P(A_2) = 0.512 \) - **Question 8:** - (1) \( \boxed{0.5057} \) - (2) \( \boxed{0.0008} \) - (3) \( \boxed{0.0086} \) - **Question 9:** - \( \boxed{0.6667} \) - **Question 10:** - (1) \( \boxed{0.8576} \) - (2) \( \boxed{0.0644} \) - (3) \( \boxed{0.8425} \) - **Question 11:** - \( \boxed{P(B | A) = P(B)} \) - **Question 12:** - \( P(A) = 0.25 \) - \( P(B) = 0.25 \) - **Question 13:** - (1) \( \boxed{\text{If A and B are independent, they are not mutually exclusive.}} \) - (2) \( \boxed{\text{If A and B are mutually exclusive, they are dependent.}} \) If you have any additional questions or need further assistance, please let me know!

# Unique Solutions for Probability Questions This document presents a unique approach to solving the given probability questions, along with a summary of final answers. --- ## **Question 7: Selecting Three Numbers from 0 to 9** ### Given Data - Three numbers are selected from the digits 0 to 9. - Define: - \( A_1 \): The event that the three numbers do not include 0 and 5. - \( A_2 \): The event that the three numbers do not include either 0 or 5. ### Definitions - The probability of an event \( A \) is defined as: \[ P(A) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} \] ### Total Outcomes Each number can be chosen independently, leading to: \[ 10 \times 10 \times 10 = 1000 \] ### Event \( A_1 \): Excluding 0 and 5 - Valid choices are: 1, 2, 3, 4, 6, 7, 8, 9 (total of 8 options). - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_1) = \frac{512}{1000} = 0.512 \] ### Event \( A_2 \): Excluding 0 or 5 - Valid choices remain the same: 1, 2, 3, 4, 6, 7, 8, 9. - Favorable outcomes: \[ 8 \times 8 \times 8 = 512 \] - Probability: \[ P(A_2) = \frac{512}{1000} = 0.512 \] ### Final Answers for Question 7 - \( \boxed{P(A_1) = 0.512} \) - \( \boxed{P(A_2) = 0.512} \) --- ## **Question 8: Birthday Probabilities for 6 Students** ### Given Data - 6 students reside in a dormitory. ### (1) Probability that at least one birthday is in October - The probability that a birthday is not in October is \( \frac{11}{12} \). - To find the probability of none having a birthday in October: \[ \left( \frac{11}{12} \right)^6 \approx 0.4943 \] - Therefore, the probability of at least one birthday in October: \[ P(\text{at least one in October}) = 1 - 0.4943 \approx 0.5057 \] ### (2) Probability that exactly 4 birthdays are in October - Number of ways to choose 4 students: \[ \binom{6}{4} = 15 \] - Probability of 4 having birthdays in October and 2 not: \[ = 15 \times \left( \frac{1}{12} \right)^4 \times \left( \frac{11}{12} \right)^2 \approx 0.0008 \] ### (3) Probability that 4 birthdays are in the same month - Choose a month (12 choices): \[ 12 \times \left( \frac{1}{12} \right)^4 \times \left( \frac{11}{12} \right)^2 \times \binom{6}{4} \approx 0.0086 \] ### Final Answers for Question 8 - (1) \( \boxed{0.5057} \) - (2) \( \boxed{0.0008} \) - (3) \( \boxed{0.0086} \) --- ## **Question 9: Probabilities of Products A, B, and C** ### Given Data - Products A, B, and C account for 60%, 30%, and 10% of total products, respectively. ### Probability of withdrawing product A given it is not product C - Let \( P(A) = 0.6 \), \( P(B) = 0.3 \), \( P(C) = 0.1 \). - The probability of selecting A given that it is not C: \[ P(A | \text{not } C) = \frac{P(A)}{P(A) + P(B)} = \frac{0.6}{0.6 + 0.3} = \frac{0.6}{0.9} = \frac{2}{3} \approx 0.6667 \] ### Final Answer for Question 9 - \( \boxed{0.6667} \) --- ## **Question 10: Warning Systems Functionality** ### Given Data - Probability of System 1 functioning = 0.92 - Probability of System 2 functioning = 0.93 - Probability that System 2 works if System 1 is down = 0.85. ### (1) Probability that both systems are operational \[ P(\text{Both systems working}) = P(1) \times P(2) = 0.92 \times 0.93 = 0.8576 \] ### (2) Probability that System 2 is down while System 1 is operational \[ P(1 \text{ working}) \times P(2 \text{ down}) = 0.92 \times (1 - 0.93) = 0.92 \times 0.07 = 0.0644 \] ### (3) Probability that System 1 is operational given System 2 is down Using Bayes' theorem: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{P(1 \text{ working}) \times P(2 \text{ down} | 1 \text{ working})}{P(2 \text{ down})} \] - Calculate \( P(2 \text{ down}) \): \[ P(2 \text{ down}) = P(2 \text{ down} | 1 \text{ working}) \cdot P(1) + P(2 \text{ down} | 1 \text{ down}) \cdot P(1 \text{ down}) \] \[ = 0.07 \cdot 0.92 + 0.15 \cdot 0.08 = 0.0644 + 0.012 = 0.0764 \] - Therefore: \[ P(1 \text{ working} | 2 \text{ down}) = \frac{0.92 \cdot 0.07}{0.0764} \approx 0.8425 \] ### Final Answers for Question 10 - (1) \( \boxed{0.8576} \) - (2) \( \boxed{0.0644} \) - (3) \( \boxed{0.8425} \) --- ## **Question 11: Independence of Events** ### Given Data Assuming \( 0 < P(A) < 1 \). ### Proof of Independence - Events A and B are defined to be independent if: \[ P(B | A) = P(B) \] - By the definition of conditional probability: \[ P(B | A) = \frac{P(A \cap B)}{P(A)} \] - Thus, if \( P(B | A) = P(B) \), it follows that \( P(A \cap B) = P(A) \cdot P(B) \). ### Final Answer for Question 11 - \( \boxed{\text{A and B are independent if and only if } P(B | A) = P(B)} \) --- ## **Question 12: Values of Probabilities for Independent Events** ### Given Data - Both \( P(A) \) and \( P(B) \) are equal to 0.25. ### Calculation Since \( A \) and \( B \) are independent: \[ P(A \cap B) = P(A) \cdot P(B) = 0.25 \times 0.25 = 0.0625 \] ### Final Answers for Question 12 - \( \boxed{P(A) = 0.25} \) - \( \boxed{P(B) = 0.25} \) --- ## **Question 13: Relationship Between Independence and Mutual Exclusivity** ### Given Data Assuming \( P(A) > 0 \) and \( P(B) > 0 \). ### (1) Independence Implies Non-Mutual Exclusivity - If A and B are independent: \[ P(A \cap B) = P(A) \cdot P(B) > 0 \] - Therefore, they cannot be mutually exclusive. ### (2) Mutual Exclusivity Implies Dependence - If A and B are mutually exclusive, then: \[ P(A \cap B) = 0 \] - Hence, they are dependent events. ### Final Answers for Question 13 - (1) \( \boxed{\text{If A and B are independent, they are not mutually exclusive.}} \) - (2) \( \boxed{\text{If A and B are mutually exclusive, they are dependent.}} \) --- # **Final Answers Summary** - **Question 7:** - \( P(A_1) = 0.512 \) - \( P(A_2) = 0.512 \) - **Question 8:** - (1) \( \boxed{0.5057} \) - (2) \( \boxed{0.0008} \) - (3) \( \boxed{0.0086} \) - **Question 9:** - \( \boxed{0.6667} \) - **Question 10:** - (1) \( \boxed{0.8576} \) - (2) \( \boxed{0.0644} \) - (3) \( \boxed{0.8425} \) - **Question 11:** - \( \boxed{P(B | A) = P(B)} \) - **Question 12:** - \( P(A) = 0.25 \) - \( P(B) = 0.25 \) - **Question 13:** - (1) \( \boxed{\text{If A and B are independent, they are not mutually exclusive.}} \) - (2) \( \boxed{\text{If A and B are mutually exclusive, they are dependent.}} \) If you have further inquiries or require clarification on any of the solutions, feel free to ask!