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- \boxed{\frac{2}{15625}}
- 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} $.
- This is a classic combinatorics problem: number of ways to place 3 non-adjacent 1s in 8 positions.
- Model: Place 3 ones such that no two are adjacent. To enforce non-adjacency, we can use the "stars and bars with gaps" method.
- Represent placing 3 ones with at least one zero between each. First place 3 ones with a zero between each: 1 0 1 0 1 → uses 5 positions, leaving $ 8 - 5 = 3 $ zeros to distribute freely in the 4 gaps: before first 1, between first and second, between second and third, and after third.
