Explanation

This is a Poisson distribution problem where:

• Average errors per week = 8
• Workdays per week = 5
• Average errors per day (λ) = 8/5 = 1.6
• We want P(K > 1) = 1 - P(K ≤ 1)

Using the Poisson probability formula: $P(K = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

First calculate P(K = 0): $P(K = 0) = \frac{e^{-1.6} \cdot 1.6^0}{0!} = e^{-1.6} ≈ 0.2019$

Then calculate P(K = 1): $P(K = 1) = \frac{e^{-1.6} \cdot 1.6^1}{1!} = 1.6 \cdot e^{-1.6} ≈ 1.6 × 0.2019 ≈ 0.3230$

Now P(K ≤ 1) = P(K = 0) + P(K = 1) ≈ 0.2019 + 0.3230 = 0.5249

Therefore, P(K > 1) = 1 - P(K ≤ 1) ≈ 1 - 0.5249 = 0.4751 = 47.51%

However, looking at the options, 47.51% is option C, but the question asks for the probability that more than one error will be produced, which is P(K > 1) = 47.51%. But the answer choices show 32.30% as option B. Let me recalculate:

Actually, P(K > 1) = 1 - [P(K = 0) + P(K = 1)] = 1 - (0.2019 + 0.3230) = 1 - 0.5249 = 0.4751 = 47.51%

But the question says "which is NEAREST to the probability that more than one error will be produced" and the answer is given as 32.30%. Let me check if there's a misinterpretation.

Wait, I think the confusion is that P(K = 1) = 32.30%, but we want P(K > 1). Let me verify the calculation:

λ = 1.6 P(K = 0) = e^{-1.6} = 0.2019 P(K = 1) = 1.6 × e^{-1.6} = 0.3230 P(K > 1) = 1 - (0.2019 + 0.3230) = 0.4751 = 47.51%

Given the options, 47.51% is option C, but the correct answer should be B (32.30%). This suggests there might be an error in the question or my interpretation. Let me recalculate more precisely:

P(K = 0) = e^{-1.6} = 0.2018965 P(K = 1) = 1.6 × e^{-1.6} = 0.3230344 P(K > 1) = 1 - (0.2018965 + 0.3230344) = 0.4750691 = 47.51%

Since the question asks for P(K > 1) and the correct calculation gives 47.51%, but the answer key shows 32.30% as the correct answer, there might be a discrepancy in the source material. Based on standard Poisson calculations, the correct answer should be 47.51% (option C).