Since each configuration has the same probability and they are mutually exclusive:
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- So there are 20 valid configurations (binary strings) with exactly 3 non-adjacent 1s.
- For each such configuration, the probability is:
- \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^5 = \frac{1^3 \cdot 2^5}{3^8} = \frac{32}{6561}
- Total probability:
- \cdot \frac{32}{6561} = \frac{640}{6561}
- Thus, the probability is:
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