Explanation:
This is a Poisson distribution problem where:
We need P(K > 1) = 1 - P(K ≤ 1) = 1 - [P(K=0) + P(K=1)]
Poisson probability formula: P(K=k) = (λ^k * e^(-λ)) / k!
Calculations:
Final probability: P(K > 1) = 1 - (0.2019 + 0.3230) = 1 - 0.5249 = 0.4751 = 47.51%
Therefore, the nearest probability is 47.51%, which corresponds to option C.
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A certain low-severity administrative (operational) process tends to produce an average of eight errors per week (where each week is five workdays). If this loss frequency process can be characterized by a Poisson distribution, which is NEAREST to the probability that more than one error will be produced tomorrow; i.e., P(K > 1 | λ = 8/5)?
A
20.19%
B
32.30%
C
47.51%
D
66.49%
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