Recall the tree diagram example from the basic counting lecture. Suppose we perform the following procedure to get a bit string of length 4 with no consecutive 1s.

Generate one bit at a time.

If there is a previous bit and it is 1, then the next bit is 0 with probability 1.

If there is no previous bit or it is 0, then the next bit is equally likely to be 0 or 1.

(a) What is the probability that the generated bit string has more 0s than 1s?

(b) What is the expected number of zeros in the generated bit string?

(c) Suppose we change the last instruction from above so that the probability that the next bit is 0 is a constant p, where 0 p 1. What value of p would result in the probability that the generated bit string has 3 leading 0s to be exactly 8 /27 ?