Let's solve **Question 7** step by step: --- ## **Q7. 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 Outcomes** Three numbers are selected from to 9 **without replacement** (since it says "selected once"), so: - There are 10 numbers ( to 9). - The total number of ways to select 3 numbers is: \[ \text{Total outcomes} = C(10, 3) = \frac{10!}{3! \cdot 7!} = 120 \] --- ### **Step 2: Event A₁ - Three numbers don’t contain and 5** This means none of the selected numbers are ** or 5**. - The numbers available are: 1, 2, 3, 4, 6, 7, 8, 9 (excluding and 5) - There are **8 numbers** left. Number of ways to select 3 numbers from these 8: \[ \text{Favorable outcomes for A}_1 = C(8, 3) = \frac{8!}{3! \cdot 5!} = 56 \] Probability: \[ P(A_1) = \frac{56}{120} = \frac{7}{15} \] --- ### **Step 3: Event A₂ - Three numbers don’t contain or 5** This means none of the selected numbers are ** or 5** (same as above), but let's clarify: - "don't contain or 5" means **neither nor 5** can be among the three numbers. - So, again, we use the 8 numbers: 1, 2, 3, 4, 6, 7, 8, 9 So, \[ \text{Favorable outcomes for A}_2 = C(8, 3) = 56 \] Probability: \[ P(A_2) = \frac{56}{120} = \frac{7}{15} \] --- ## **Final Answers** \[ \boxed{ P(A_1) = \frac{7}{15}, \quad P(A_2) = \frac{7}{15} } \] **Explanation:** In this case, "don't contain and 5" and "don't contain or 5" both mean that neither nor 5 is among the selected numbers, so both probabilities are the same. If you want the case where "don't contain **and** 5" means *at least one of them is missing*, let me know and I can clarify further!

# Probability Problems Solutions ## **Q7. Three Numbers Selected from 0 to 9** ### **Events** - **A₁**: Three numbers don’t contain 0 and 5. - **A₂**: Three numbers don’t contain 0 or 5. ### **Step 1: Total Number of Outcomes** The total number of ways to select 3 numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is calculated as: \[ \text{Total outcomes} = C(10, 3) = \frac{10!}{3! \cdot 7!} = 120 \] ### **Step 2: Calculate P(A₁)** For event A₁ (not containing 0 and 5): - Available numbers: {1, 2, 3, 4, 6, 7, 8, 9} (total 8 numbers). The number of ways to select 3 numbers from these 8 is: \[ \text{Favorable outcomes for A}_1 = C(8, 3) = \frac{8!}{3! \cdot 5!} = 56 \] Probability: \[ P(A_1) = \frac{56}{120} = \frac{7}{15} \] ### **Step 3: Calculate P(A₂)** For event A₂ (not containing 0 or 5): - The same available numbers apply: {1, 2, 3, 4, 6, 7, 8, 9} (total 8 numbers). Thus, the number of ways remains: \[ \text{Favorable outcomes for A}_2 = C(8, 3) = 56 \] Probability: \[ P(A_2) = \frac{56}{120} = \frac{7}{15} \] ### **Final Answers** \[ \boxed{ P(A_1) = \frac{7}{15}, \quad P(A_2) = \frac{7}{15} } \] --- ## **Q8. Birthday Probabilities for 6 Students** ### **(1) At least one birthday in October** The complementary event is that none have a birthday in October. - Probability that one student does not have a birthday in October: \( \frac{11}{12} \). - For 6 students: \[ P(\text{none in October}) = \left( \frac{11}{12} \right)^6 \] \[ P(\text{at least one in October}) = 1 - \left( \frac{11}{12} \right)^6 \approx 0.393 \] ### **(2) Exactly 4 birthdays in October** Using the binomial probability formula: \[ P(X = 4) = C(6, 4) \left( \frac{1}{12} \right)^4 \left( \frac{11}{12} \right)^2 \] Calculating: \[ = 15 \cdot \left( \frac{1}{12} \right)^4 \cdot \left( \frac{11}{12} \right)^2 \approx 0.012 \] ### **(3) Exactly 4 birthdays in the same month** Using the multinomial approach: - Probability for any specific month: \( \left( \frac{1}{12} \right)^4 \left( \frac{11}{12} \right)^2 \). - There are 12 months: \[ P(\text{4 in same month}) = 12 \cdot 15 \cdot \left( \frac{1}{12} \right)^4 \cdot \left( \frac{11}{12} \right)^2 \approx 0.012 \] --- ## **Q9. Product Probability** - Products A, B, C: 60%, 30%, 10%. - Given a product is not C: Find P(A | not C). Using Bayes' theorem: \[ P(A | \text{not C}) = \frac{P(A) \cdot P(\text{not C} | A)}{P(\text{not C})} \] Since \( P(\text{not C}) = P(A) + P(B) = 0.6 + 0.3 = 0.9 \): \[ P(A | \text{not C}) = \frac{0.6 \times 1}{0.9} = \frac{2}{3} \] --- ## **Q10. Warning Systems Probability** ### **Given:** - \( P(S_1) = 0.92 \) - \( P(S_2) = 0.93 \) - \( P(S_2 | S_1 \text{ down}) = 0.85 \) ### **(1) Both systems working** Using complementary probabilities: \[ P(S_1 \text{ working}) = 0.92, \quad P(S_2 \text{ working}) = 0.93 \] \[ P(S_1 \text{ and } S_2) = P(S_1) \cdot P(S_2) = 0.92 \times 0.93 \approx 0.856 \] ### **(2) System 2 down and System 1 working** Using the complement: \[ P(S_2 \text{ down}) = 1 - P(S_2) = 0.07 \] Thus: \[ P(S_1 \text{ and } S_2 \text{ down}) = P(S_1) \cdot P(S_2 \text{ down}) = 0.92 \cdot 0.07 \approx 0.0644 \] ### **(3) System 1 working given System 2 down** Using Bayes' theorem: \[ P(S_1 | S_2 \text{ down}) = \frac{P(S_1) \cdot P(S_2 \text{ down} | S_1)}{P(S_2 \text{ down})} \] Calculating: \[ P(S_2 \text{ down} | S_1) = 1 - P(S_2 | S_1) = 0.07 \] Thus: \[ P(S_1 | S_2 \text{ down}) \approx \frac{0.92 \times 0.07}{0.07} = 0.92 \] --- ## **Q11 - Q13 Indpendence and Mutual Exclusivity** ### **Q11. Independence Condition** To show that events A and B are independent if and only if: \[ P(B | A) = P(B) \] ### **Q12. Given Independence** If \( P(A) = P(B) = 0.25 \), both events are independent. ### **Q13. Independence vs. Mutual Exclusivity** 1. If A and B are independent, they cannot be mutually exclusive because \( P(A \cap B) = P(A) \cdot P(B) \) would not equal zero. 2. If A and B are mutually exclusive, then \( P(A \cap B) = 0 \), indicating they are dependent. --- ### **Final Notes** These calculations illustrate fundamental concepts in probability, including independence, conditional probability, and event combinations.