Solution
The correct answer is \(\frac{5}{8}\)
Key Points
To determine the probability that a randomly generated bit string of length 4 contains at least two consecutive 0's given that the first bit is 0, we proceed as follows:
1. List all possible bit strings of length 4 that start with 0:
Since the first bit is fixed as 0, there are 2^3 = 8 possible bit strings:
\( \begin{aligned}
&0000, \\
&0001, \\
&0010, \\
&0011, \\
&0100, \\
&0101, \\
&0110, \\
&0111
\end{aligned}\)
2. Identify bit strings with at least two consecutive 0's:
- 0000(contains three consecutive 0's)
- 0001 (contains three consecutive 0's)
- 0010 (contains two consecutive 0's)
- 0011 (contains two consecutive 0's)
- 0100 (contains two consecutive 0's)
3. Count the total number of strings with at least two consecutive 0's:
There are 5 such strings: 0000, 0001, 0010, 0011, 0100.
4. Calculate the probability:
- Total number of possible bit strings (starting with 0) = 8
- Number of favorable outcomes (with at least two consecutive 0's) = 5
- Hence, the probability is: \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{8}\)
Therefore, the probability that a randomly generated bit string of length 4 contains at least two consecutive 0's given that the first bit is 0 is: \(\frac{5}{8}\)
