1. **Problem statement:** In a university, 55% of students live in dormitories. A sample of 5 students is selected randomly. We use the binomial distribution to find probabilities related to students living in dormitories. 2. **Define variables:** Let $X$ be the number of students in the sample who live in dormitories. Then $X \sim \text{Binomial}(n=5, p=0.55)$. 3. **(a) Probability exactly 3 live in dormitories:** $$P(X=3) = \binom{5}{3} (0.55)^3 (0.45)^2$$ Calculate: $$\binom{5}{3} = 10$$ $$P(X=3) = 10 \times (0.55)^3 \times (0.45)^2 = 10 \times 0.166375 \times 0.2025 = 0.3369$$ 4. **(b) Probability no students live in dormitories:** $$P(X=0) = \binom{5}{0} (0.55)^0 (0.45)^5 = 1 \times 1 \times (0.45)^5 = 0.0185$$ 5. **(c) Probability more than 3 students do not live in dormitories:** Number of students not living in dormitories is $Y = 5 - X$. We want $P(Y > 3) = P(Y=4) + P(Y=5)$. Since $Y = 5 - X$, this is equivalent to $P(X < 1) = P(X=0)$. From (b), $$P(Y > 3) = P(X=0) = 0.0185$$ 6. **(d) Expected number of students not living in dormitories:** Expected value of $X$ is $E(X) = np = 5 \times 0.55 = 2.75$. Expected number not living in dormitories is $$E(Y) = 5 - E(X) = 5 - 2.75 = 2.25$$ **Final answers:** (a) $0.3369$ (b) $0.0185$ (c) $0.0185$ (d) $2.25$