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Independent events; computer errors 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.

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?

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

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

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.

✓ Answer checked by Claude Sonnet 5.5

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