What is the probability that a bit string of length 3, chosen at random, does not have two consecutive 0’s?

There is a total of \(2^3 = 8\) outcomes. The number of outcomes with no two consecutive 0’s may be calculated e.g as the complement of events with at least two consecutive 0’s. Such events may happen in 3 ways (001, 100, 000). Thus the number of events with no two consecutive 0’s is \(8 - 3 = 5\), giving the final probability of \(\frac{5}{8}\)