1. **State the problem:** We have three circuits installed in a computer. Each circuit can be defective (D) or good (G). The probability a circuit is defective is 0.12, so the probability it is good is $1 - 0.12 = 0.88$. 2. **List the sample space:** The sample space consists of all possible combinations of three circuits, each either D or G: $$\{DDD, DDG, DGD, DGG, GDD, GDG, GGD, GGG\}$$ 3. **Find the probability all three circuits are good:** Since each circuit is independent, $$P(\text{all good}) = P(G) \times P(G) \times P(G) = 0.88^3 = 0.681472$$ Rounded to four decimal places, this is 0.6815. 4. **Find the probability the computer functions:** The computer functions if at least two circuits are good. This means either exactly two or all three are good. - Probability exactly two are good: Number of ways to have exactly two good circuits is 3 (choose which one is defective). Probability for one such arrangement (e.g., GGD) is: $$0.88^2 \times 0.12 = 0.092928$$ Multiply by 3: $$3 \times 0.092928 = 0.278784$$ - Probability all three are good is 0.681472 (from step 3). - Total probability computer functions: $$0.681472 + 0.278784 = 0.960256$$ Rounded to four decimal places, 0.9603. 5. **Determine if all three defective is unusual at 0.05 cutoff:** Probability all three defective: $$P(DDD) = 0.12^3 = 0.001728$$ Since 0.001728 < 0.05, it is unusual for all three circuits to be defective. **Final answers:** - Probability all three good: 0.6815 - Probability computer functions: 0.9603 - All three defective is unusual (probability 0.0017 < 0.05)