(i) The probability that a visitor sees all the animals is the product of the individual probabilities:

\(0.9 \times 0.95 \times 0.85 \times 0.1 = 0.0727\).

(ii) Let the random variable \(X\) be the number of visitors who see lions. \(X\) follows a binomial distribution \(X \sim B(12, 0.1)\). We need to find \(P(X < 3)\), which is \(P(0) + P(1) + P(2)\).

\(P(0) = (0.9)^{12}\)

\(P(1) = \binom{12}{1} (0.1)(0.9)^{11}\)

\(P(2) = \binom{12}{2} (0.1)^2(0.9)^{10}\)

Thus, \(P(X < 3) = (0.9)^{12} + \binom{12}{1} (0.1)(0.9)^{11} + \binom{12}{2} (0.1)^2(0.9)^{10} = 0.889\).

(iii) Let \(Y\) be the number of people who see zebras. \(Y\) follows a binomial distribution \(Y \sim B(50, 0.85)\). The mean \(E(Y)\) is given by \(np\) and the variance \(Var(Y)\) is given by \(np(1-p)\).

Mean: \(50 \times 0.85 = 42.5\)

Variance: \(50 \times 0.85 \times 0.15 = 6.375\).