Flaws in a carbon-fiber wing skin occur randomly at an average rate of $\lambda = 0.5$ flaws per square meter according to a Poisson process. The probability that a randomly inspected $4\text{ m}^2$ panel contains exactly 3 flaws is closest to:

(A) 0.180

(B) 0.271

(C) 0.090

(D) 0.323

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles For a spatial Poisson process with parameter rate $\lambda$ per unit area, the number of events $X$ occurring in an inspection region of area $A$ follows a Poisson distribution with parameter $\mu = \lambda A$: $$P(X = k) = \frac{e^{-\mu} \mu^k}{k!}, \quad k = 0, 1, 2, \dots$$ ### 2. Analytical Formulation Given: - Rate $\lambda = 0.5\text{ flaws/m}^2$ - Panel area $A = 4\text{ m}^2$ - Mean number of flaws in panel: $\mu = \lambda A = 0.5 \times 4 = 2.0$ - Number of target flaws $k = 3$ ### 3. Step-by-Step Calculation Substitute into the Poisson probability mass function: $$P(X = 3) = \frac{e^{-2} \cdot 2^3}{3!} = \frac{e^{-2} \cdot 8}{6} = \frac{4}{3} e^{-2}$$ Evaluating with $e^{-2} \approx 0.135335$: $$P(X = 3) = \frac{4}{3} \times 0.135335 = 4 \times 0.045112 \approx 0.18045 \approx 0.180$$ ### 4. Distractor Forensics - **Option A (0.180):** Correct Poisson probability. - **Option B (0.271):** Corresponds to $P(X = 1) = 2 e^{-2} = 0.2707$ or $P(X = 2) = 2 e^{-2} = 0.2707$. - **Option C (0.090):** Omitted factor of 2 or used $k! = 12$. - **Option D (0.323):** Used binomial approximation error. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Remember to scale rate parameter $\lambda$ by the inspection area: $\mu = \lambda A$. - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, 10th Edition, Chapter 24 (Probability and Statistics).

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