\int_0^{0.2} w^5 dw = 6 \cdot \frac{w^6}{6} \Big|_0^{0.2} = (0.2)^6 = \frac{1}{15625}
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- Second integral:
- \int_{0.8}^1 30 \cdot \frac{(1 - u)^5}{5} du = 6 \int_{0.8}^1 (1 - u)^5 du
- Let $ w = 1 - u $, $ dw = -du $, limits $ w = 0.2 $ to $ 0 $:
- Total: $ \frac{1}{15625} + \frac{1}{15625} = \frac{2}{15625} $
- Thus, the probability that all 6 vehicles arrive within some 2-minute interval is:
- \boxed{\frac{2}{15625}}
