Final answer:
The probability of getting a bit string with at least two consecutive 1s, given that the first bit is 0, is 1/2.
Step-by-step explanation:
The probability that a bit string of length four contains at least two consecutive 1s, given that its first bit is 0, can be found using conditional probability.
Let A be the event that the bit string contains at least two consecutive 1s, and let B be the event that the first bit is 0. We need to find P(A|B).
Given that the first bit is 0, there are 8 bit strings of length four that can be formed. Out of these, 4 bit strings contain at least two consecutive 1s (0011, 0111, 1011, 1111).
The student is asking about the probability of generating a random bit string of length four that begins with a 0 and contains at least two consecutive 1s. Since the first bit is a 0, we have three bits left to consider. The bit strings that satisfy the conditions are 0111, 0011, and 0110. There are 23 = 8 possible bit strings beginning with 0. Therefore, the probability is the number of favorable outcomes divided by the total number of outcomes, which is 3/8.
Therefore, the probability of getting a bit string with at least two consecutive 1s, given that the first bit is 0, is 4/8 = 1/2.