1. **Problem Statement:** We have a binary code word of length 6 bits, each bit is either 0 or 1. We want to find the probabilities for three cases: (a) The code word has exactly three 0s and three 1s. (b) The code word begins with two 0s. (c) The code word ends with three 1s. 2. **Total number of possible code words:** Each bit can be 0 or 1 independently, so total code words = $2^6 = 64$. --- ### (a) Probability of exactly three 0s and three 1s 3. The number of ways to choose which 3 positions out of 6 are zeros is given by the binomial coefficient: $$\binom{6}{3} = \frac{6!}{3!3!} = 20$$ 4. Each such choice corresponds to a unique code word with 3 zeros and 3 ones. 5. Probability = $\frac{\text{number of favorable outcomes}}{\text{total outcomes}} = \frac{20}{64} = \frac{5}{16}$. --- ### (b) Probability the code word begins with two 0s 6. Fix the first two bits as 0s. The remaining 4 bits can be anything (0 or 1), so number of such code words = $2^4 = 16$. 7. Probability = $\frac{16}{64} = \frac{1}{4}$. --- ### (c) Probability the code word ends with three 1s 8. Fix the last three bits as 1s. The first 3 bits can be anything, so number of such code words = $2^3 = 8$. 9. Probability = $\frac{8}{64} = \frac{1}{8}$. --- **Final answers:** - (a) $\frac{5}{16}$ - (b) $\frac{1}{4}$ - (c) $\frac{1}{8}$