1. **Problem (e):** Find the probability that a person rides both Daifong and Torbellino. Since the probability of riding Torbellino is $0.61$ and the probability of riding Daifong is independent, we need the probability $P(D \cap T) = P(D) \times P(T)$. However, the value of $P(D)$ is not given, so we cannot find $P(D \cap T)$ without it. --- 2. **Problem (f):** Find the number $n$ of people such that the probability at most 500 ride Torbellino is about $0.693$. Assuming the number of people who ride Torbellino follows a Binomial distribution $X \sim \text{Binomial}(n, 0.61)$. Using the normal approximation to the binomial: $$X \approx N(\mu = np, \sigma = \sqrt{np(1-p)})$$ The probability that at most 500 ride Torbellino is: $$P(X \leq 500) \approx P\left(Z \leq \frac{500 + 0.5 - np}{\sqrt{np(1-p)}}\right) = 0.693$$ From standard normal tables, $\Phi^{-1}(0.693) \approx 0.5$. Hence, $$0.5 = \frac{500.5 - 0.61 n}{\sqrt{0.61 n (1-0.61)}} = \frac{500.5 - 0.61 n}{\sqrt{0.2379 n}}$$ Multiply both sides: $$0.5 \sqrt{0.2379 n} = 500.5 - 0.61 n$$ Square both sides: $$0.25 \times 0.2379 n = (500.5 - 0.61 n)^2$$ $$0.059475 n = (500.5 - 0.61 n)^2$$ Expand: $$(500.5 - 0.61 n)^2 = 500.5^2 - 2 \times 500.5 \times 0.61 n + (0.61)^2 n^2 = 250500.25 - 610.61 n + 0.3721 n^2$$ Rearranged: $$0.3721 n^2 - 610.61 n + 250500.25 - 0.059475 n = 0$$ $$0.3721 n^2 - 610.669475 n + 250500.25 = 0$$ Use quadratic formula: $$n = \frac{610.669475 \pm \sqrt{610.669475^2 - 4 \times 0.3721 \times 250500.25}}{2 \times 0.3721}$$ Calculate discriminant: $$610.669475^2 - 4 \times 0.3721 \times 250500.25 \approx 373,019.9$$ Square root: $$\sqrt{373,019.9} \approx 610.79$$ Calculate roots: $$n = \frac{610.669475 \pm 610.79}{0.7442}$$ Possible values: - $$n_1 = \frac{610.669475 + 610.79}{0.7442} \approx 1639.7$$ - $$n_2 = \frac{610.669475 - 610.79}{0.7442} \approx -0.16$$ (discard negative) Therefore, the value of $n \approx 1640$. --- **Final answers:** - (e) Cannot determine $P(D \cap T)$ without $P(D)$. - (f) $n \approx 1640$ people.