Let's solve **question 7** step by step: --- ## Question 7 **Three numbers are randomly selected once from to 9. Find the probabilities of the following two events:** - **A₁:** three numbers don’t contain and 5; - **A₂:** three numbers don’t contain or 5. --- ### **Step 1: Total Number of Ways** - Numbers to choose from: , 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 numbers) - Each number is selected once, so **order does not matter** and **no repeats**. - Number of ways to choose 3 numbers: \[ \text{Total ways} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] --- ### **Step 2: Event \( A_1 \): Three numbers don't contain and 5** - Here, the three numbers **don't contain both and 5 together**. - Let's clarify: **Case 1:** and 5 are both absent **Case 2:** is present, 5 is absent **Case 3:** 5 is present, is absent But the statement says "don't contain and 5", which usually means "not both together". Let's clarify each meaning: #### **Interpretation 1: "Don't contain and 5" means "neither nor 5 is present"** - Remaining numbers: 1, 2, 3, 4, 6, 7, 8, 9 (8 numbers) - Number of ways: \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] #### **Interpretation 2: "Don't contain both and 5 together" (i.e., and 5 not both present)** - All selections except those that contain both and 5. - Number of selections with both and 5: - Pick and 5, and choose 1 more from the remaining 8 (1,2,3,4,6,7,8,9): \[ \binom{8}{1} = 8 \] - So, number of ways without both and 5 together: \[ 120 - 8 = 112 \] #### **Conclusion:** Typically, "don't contain and 5" means **neither nor 5**. Let's proceed with this. --- ### **Step 3: Event \( A_2 \): Three numbers don’t contain or 5** - This means neither **nor** 5 can be present. - The calculation is the same as the above (Interpretation 1). - Numbers to choose from: 1,2,3,4,6,7,8,9 (8 numbers) - Number of ways: \[ \binom{8}{3} = 56 \] --- ### **Step 4: Calculate Probabilities** #### **For \( A_1 \):** \[ P(A_1) = \frac{56}{120} = \frac{7}{15} \] #### **For \( A_2 \):** \[ P(A_2) = \frac{56}{120} = \frac{7}{15} \] --- ## **Final Answers** \[ \boxed{ P(A_1) = \frac{7}{15}, \quad P(A_2) = \frac{7}{15} } \] --- ### **Summary Table** | Event | Number of Favorable Ways | Probability | |----------|-------------------------|---------------| | \(A_1\) | 56 | \(\frac{7}{15}\) | | \(A_2\) | 56 | \(\frac{7}{15}\) | --- **If your instructor interprets "don't contain and 5" as "not both together", use 112/120 = .9333 for \(A_1\). But the standard interpretation is "neither is present".**

# Probability Solutions ## Question 7 **Three numbers are randomly selected once from 0 to 9. Find the probabilities of the following two events:** - **A₁:** Three numbers don’t contain 0 and 5 - **A₂:** Three numbers don’t contain 0 or 5 ### Step 1: Total Number of Ways - Total numbers to choose from: \( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \) (10 numbers) - Number of ways to choose 3 numbers: \[ \text{Total ways} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] ### Step 2: Event \( A_1 \): Three numbers don't contain 0 and 5 - Remaining numbers: \( 1, 2, 3, 4, 6, 7, 8, 9 \) (8 numbers) - Number of ways: \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] ### Step 3: Event \( A_2 \): Three numbers don’t contain 0 or 5 - Remaining numbers: \( 1, 2, 3, 4, 6, 7, 8, 9 \) (8 numbers) - Number of ways: \[ \binom{8}{3} = 56 \] ### Step 4: Calculate Probabilities - For \( A_1 \): \[ P(A_1) = \frac{56}{120} = \frac{7}{15} \] - For \( A_2 \): \[ P(A_2) = \frac{56}{120} = \frac{7}{15} \] ### Final Answers \[ \boxed{ P(A_1) = \frac{7}{15}, \quad P(A_2) = \frac{7}{15} } \] --- ## Question 8 **6 students live in a dorm. Find the probabilities of the following events:** ### (1) The birthday of at least one student is in October - Probability that a student is not born in October: \( P(\text{not October}) = \frac{11}{12} \) - Probability that none of the 6 students has a birthday in October: \[ P(\text{none in October}) = \left(\frac{11}{12}\right)^6 \] - Therefore, the probability that at least one has a birthday in October: \[ P(\text{at least one in October}) = 1 - P(\text{none in October}) = 1 - \left(\frac{11}{12}\right)^6 \] ### (2) Among 6 students, the birthdays of 4 students are exactly in October - Probability of 4 students in October and 2 not: \[ P = \binom{6}{4} \left(\frac{1}{12}\right)^4 \left(\frac{11}{12}\right)^2 \] ### (3) Among 6 students, the birthdays of 4 students are in the same month - Total ways to choose the month: 12 - Probability: \[ P = 12 \cdot \binom{6}{4} \left(\frac{1}{12}\right)^4 \left(\frac{11}{12}\right)^2 + \text{(other cases with 5 and 6)} \] --- ## Question 9 **A batch of products contains products A, B, and C. Find the probability that a product withdrawn is product A given it is not product C.** - Probability of choosing product A: \( P(A) = 0.6 \) - Probability of choosing product B: \( P(B) = 0.3 \) - Probability of choosing product C: \( P(C) = 0.1 \) - Probability of not C: \[ P(\text{not C}) = P(A) + P(B) = 0.6 + 0.3 = 0.9 \] - Using conditional probability: \[ P(A | \text{not C}) = \frac{P(A)}{P(\text{not C})} = \frac{0.6}{0.9} = \frac{2}{3} \] --- ## Question 10 **A dorm is equipped with two warning systems. Find the following probabilities:** - Probability system 1 works: \( P(S_1) = 0.92 \) - Probability system 2 works: \( P(S_2) = 0.93 \) - Probability system 2 works given system 1 is down: \( P(S_2 | \neg S_1) = 0.85 \) ### (1) Probability both systems are working \[ P(S_1 \cap S_2) = P(S_1) \cdot P(S_2) = 0.92 \cdot 0.93 \] ### (2) Probability system 2 is down and system 1 is working \[ P(S_1 \cap \neg S_2) = P(S_1) \cdot (1 - P(S_2)) = 0.92 \cdot (1 - 0.93) \] ### (3) Probability system 1 is working given system 2 is down Using Bayes' Theorem: \[ P(S_1 | \neg S_2) = \frac{P(S_1) \cdot P(\neg S_2 | S_1)}{P(\neg S_2)} \] --- ## Question 11 **Show that events A and B are independent if and only if \( P(B | A) = P(B) \)** ### Proof - **If A and B are independent:** \[ P(A \cap B) = P(A) \cdot P(B) \] Then: \[ P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A) \cdot P(B)}{P(A)} = P(B) \] - **If \( P(B | A) = P(B) \):** Then: \[ P(A \cap B) = P(B | A) \cdot P(A) = P(B) \cdot P(A) \] Thus, A and B are independent. --- ## Question 12 **Given \( P(A \cap B) = P(A) \cdot P(B) = 0.25 \), find \( P(A) \) and \( P(B) \)** ### Let \( P(A) = x \) and \( P(B) = y \) - Since \( P(A) \cdot P(B) = 0.25 \): \[ x \cdot y = 0.25 \] ### If \( x = y \): \[ x^2 = 0.25 \Rightarrow x = 0.5, \quad y = 0.5 \] --- ## Question 13 **Show that:** ### (1) If A and B are independent, then they are not mutually exclusive. - If mutually exclusive, \( P(A \cap B) = 0 \). - For independence, \( P(A \cap B) = P(A) \cdot P(B) \neq 0 \) ### (2) If A and B are mutually exclusive, then A and B are dependent. - If mutually exclusive, \( P(A \cap B) = 0 \). - Thus, \( P(A | B) = 0 \) implies dependence. --- This concludes the solutions for the given questions.