1. **Problem statement:** We have three network packets that can be routed to Link A, Link B, or Link C. Let $X$ be the number of packets routed to Link B. We want to find the probability distribution of $X$, the expected value $E(X)$, and the standard deviation $S.D(X)$. 2. **Understanding the problem:** Each packet can be routed independently to any of the three links. Assuming each link is equally likely, the probability a packet goes to Link B is $p=\frac{1}{3}$. 3. **Distribution of $X$:** Since $X$ counts the number of packets routed to Link B out of 3, $X$ follows a Binomial distribution with parameters $n=3$ and $p=\frac{1}{3}$. 4. **Probability mass function (PMF):** $$P(X=k) = \binom{3}{k} \left(\frac{1}{3}\right)^k \left(1-\frac{1}{3}\right)^{3-k} = \binom{3}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{3-k}$$ for $k=0,1,2,3$. 5. **Calculate probabilities:** - $P(X=0) = \binom{3}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^3 = 1 \times 1 \times \frac{8}{27} = \frac{8}{27}$ - $P(X=1) = \binom{3}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^2 = 3 \times \frac{1}{3} \times \frac{4}{9} = \frac{12}{27} = \frac{4}{9}$ - $P(X=2) = \binom{3}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^1 = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9}$ - $P(X=3) = \binom{3}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^0 = 1 \times \frac{1}{27} \times 1 = \frac{1}{27}$ 6. **Expected value $E(X)$:** For a binomial distribution, $E(X) = np = 3 \times \frac{1}{3} = 1$. 7. **Variance and standard deviation:** Variance $Var(X) = np(1-p) = 3 \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{3}$. Standard deviation $S.D(X) = \sqrt{Var(X)} = \sqrt{\frac{2}{3}}$. **Final answers:** - Probability distribution: $$P(X=0)=\frac{8}{27},\quad P(X=1)=\frac{4}{9},\quad P(X=2)=\frac{2}{9},\quad P(X=3)=\frac{1}{27}$$ - Expected value: $E(X) = 1$ - Standard deviation: $S.D(X) = \sqrt{\frac{2}{3}}$