1. **Problem statement:** We have a web application generating passwords where 15% are rejected for being too weak. We want to find probabilities and statistics related to the number of rejected passwords out of 20 generated. 2. **Model:** This is a binomial distribution problem where $n=20$ (number of trials) and $p=0.15$ (probability of rejection). 3. **Binomial probability formula:** $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $X$ is the number of rejected passwords. 4. **Calculations:** **a.** Given $n=20$, $p=0.15$. **b.** Probability exactly 3 are rejected: $$P(X=3) = \binom{20}{3} (0.15)^3 (0.85)^{17}$$ **c.** Probability at most 4 are rejected: $$P(X \leq 4) = \sum_{k=0}^4 \binom{20}{k} (0.15)^k (0.85)^{20-k}$$ **d.** Probability more than 2 are rejected: $$P(X > 2) = 1 - P(X \leq 2) = 1 - \sum_{k=0}^2 \binom{20}{k} (0.15)^k (0.85)^{20-k}$$ **e.** Expected number of rejected passwords: $$E(X) = np = 20 \times 0.15 = 3$$ **f.** Variance of number of rejections: $$Var(X) = np(1-p) = 20 \times 0.15 \times 0.85 = 2.55$$ **g.** Probability no passwords are rejected: $$P(X=0) = \binom{20}{0} (0.15)^0 (0.85)^{20} = (0.85)^{20}$$ 5. **Summary:** - $P(X=3) = \binom{20}{3} (0.15)^3 (0.85)^{17}$ - $P(X \leq 4) = \sum_{k=0}^4 \binom{20}{k} (0.15)^k (0.85)^{20-k}$ - $P(X > 2) = 1 - \sum_{k=0}^2 \binom{20}{k} (0.15)^k (0.85)^{20-k}$ - $E(X) = 3$ - $Var(X) = 2.55$ - $P(X=0) = (0.85)^{20}$