\boxed{\frac{2}{15625}}
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- \int_0^{0.2} w^5 dw = 6 \cdot \frac{w^6}{6} \Big|_0^{0.2} = (0.2)^6 = \frac{1}{15625}
- Total: $ \frac{1}{15625} + \frac{1}{15625} = \frac{2}{15625} $
- Thus, the probability that all 6 vehicles arrive within some 2-minute interval is:
- Question: A programmer is testing a neural network that generates binary strings of length 8, where each bit is independently 1 with probability $ \frac{1}{3} $ and 0 with probability $ \frac{2}{3} $. What is the probability that the string contains exactly three 1s, with no two 1s adjacent?
- Solution: We want the probability that a binary string of length 8 has exactly three 1s, no two of which are adjacent, with each 1 occurring independently with probability $ \frac{1}{3} $, and 0s with $ \frac{2}{3} $.
- First, compute the number of binary strings of length 8 with exactly 3 non-adjacent 1s.
