--- title: "Independent events; computer errors per week" subject: "Probability and Statistics" chapter: 1 key: "semester_5/probability_and_statistics:independent-events-computer-errors" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/independent-events-computer-errors site: BCT Engineering --- # Independent events; computer errors per week Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability - Asked in: 2071 Chaitra, 6 marks; 2071 Shrawan, 6 marks (asked 2 times) - Video: [Compound probability of independent events | Probability and Statistics | Khan Academy](https://www.youtube.com/watch?v=xSc4oLA9e8o) — Khan Academy - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/independent-events-computer-errors - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1 ## Question Define dependent and independent events with suitable examples (multiplication law of probability for dependent and independent events). The independent probabilities that the three sections of a costing department will encounter a computer error are 0.2, 0.3 and 0.1 per week respectively. What is the probability that there would be: i) At least one computer error per week? ii) One and only one computer error per week? ## Short answer Independent events: occurrence of one does not affect probability of other (e.g., coin toss and die roll). Dependent events: probability changes (e.g., cards without replacement). For computer errors, P(at least one) = 1 − 0.8×0.7×0.9 = 0.496. P(one and only one) = 0.2×0.7×0.9 + 0.8×0.3×0.9 + 0.8×0.7×0.1 = 0.398. ## Answer #### Independent events Two events A and B are **independent** if the occurrence of one does not change the probability of the other: P(B | A) = P(B). Example: tossing a coin and rolling a die. Getting a head does not affect getting a 6. #### Dependent events Two events are **dependent** if the occurrence of one changes the probability of the other. Example: drawing two cards from a pack without replacement. The probability of a second ace depends on whether the first card was an ace. #### Multiplication law - Dependent events: P(A ∩ B) = P(A) · P(B | A) - Independent events: P(A ∩ B) = P(A) · P(B) For three independent events, P(A ∩ B ∩ C) = P(A) · P(B) · P(C). #### Computer errors Let the sections be A, B, C with P(A) = 0.2, P(B) = 0.3, P(C) = 0.1. Probabilities of no error: 0.8, 0.7, 0.9. **i) At least one error** ```text P(at least one) = 1 − P(no error in any section) = 1 − (0.8)(0.7)(0.9) = 1 − 0.504 = 0.496 ``` **ii) One and only one error** ```text Only A: 0.2 × 0.7 × 0.9 = 0.126 Only B: 0.8 × 0.3 × 0.9 = 0.216 Only C: 0.8 × 0.7 × 0.1 = 0.056 Total = 0.126 + 0.216 + 0.056 = 0.398 ``` **Answer:** (i) 0.496, (ii) 0.398. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What defines independent events A and B? A. P(A ∩ B) = 0 B. P(A ∪ B) = 1 C. P(B | A) = 1 D. P(B | A) = P(B) Answer: D. Occurrence of one does not change the probability of the other. (Key point: Independent events) 2. Which is the multiplication law for dependent events? A. P(A ∩ B) = P(A) P(B | A) B. P(A ∩ B) = P(A) + P(B) C. P(A ∩ B) = P(A) P(B) always D. P(A ∩ B) = P(A)/P(B) Answer: A. For dependent events, the second probability is conditional. (Key point: Multiplication law) 3. How is P(at least one error) found, and what is it? A. (0.2)(0.3)(0.1) = 0.006 B. 1 - (0.8)(0.7)(0.9) = 0.496 C. 0.2 + 0.3 + 0.1 = 0.6 D. 1 - 0.504 x 0.5 = 0.748 Answer: B. Use the complement: 1 minus P(no error in any section) = 1 - 0.504. (Key point: At least one error) 4. How is P(exactly one error) found? A. Add 0.2 + 0.3 + 0.1 B. Multiply 0.2 x 0.3 x 0.1 C. Add three cases like 0.2x0.7x0.9, 0.8x0.3x0.9, 0.8x0.7x0.1 = 0.398 D. Use 1 - 0.496 = 0.504 Answer: C. Exactly one section errs while the other two do not; sum of 0.126, 0.216 and 0.056. 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