Let the number of days Khalid rides his bicycle be a binomial random variable with parameters 45 and 0.6. The variance is given by:

\(\text{Var}(B) = 45 \times 0.6 \times 0.4\)

Similarly, let the number of days Khalid rides his skateboard be a binomial random variable with parameters 45 and 0.4. The variance is given by:

\(\text{Var}(S) = 45 \times 0.4 \times 0.6\)

Since \(\text{Var}(B) = \text{Var}(S)\), the variances are the same.

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

(i) Let the probability of a screw being faulty be \(p\). Given the mean number of faulty screws per packet is 1.2, we have:

\(1.2 = 15p\)

Solving for \(p\), we get:

\(p = \frac{1.2}{15} = 0.08\)

The variance \(\text{Var} = npq\), where \(q = 1 - p\).

\(\text{Var} = 15 \times 0.08 \times 0.92 = 1.104\)

(ii) The probability that a packet contains at most 2 faulty screws is:

\(P(0, 1, 2) = (0.92)^{15} + 15C_1(0.08)(0.92)^{14} + 15C_2(0.08)^2(0.92)^{13}\)

Calculating each term:

\((0.92)^{15} = 0.275\)

\(15 \times 0.08 \times (0.92)^{14} = 0.359\)

\(15C_2 \times (0.08)^2 \times (0.92)^{13} = 0.253\)

Adding these probabilities gives:

\(P(0, 1, 2) = 0.275 + 0.359 + 0.253 = 0.887\)

(iii) The probability that there is at least 1 faulty screw in a packet is:

\(P(\text{at least 1 faulty screw}) = 1 - P(0) = 1 - (0.92)^{15}\)

\(= 1 - 0.275 = 0.7137\)

The probability that there are exactly 7 packets with at least 1 faulty screw is:

\(P(\text{exactly 7 packets}) = 8C_7(0.7137)^7(0.2863)\)

\(= 0.216\)

(i) Let the probability that a random integer is less than or equal to 4 be p. Since there are 4 integers (1, 2, 3, 4) less than or equal to 4 out of 9, p = 4/9. The probability that at least 2 of the 5 integers are less than or equal to 4 is given by:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - \left(\frac{5}{9}\right)^5 - \binom{5}{1} \left(\frac{4}{9}\right) \left(\frac{5}{9}\right)^4\)

\(= 0.735\)

(ii) We know that the mean \(np = 96\) and the variance \(npq = 32\), where \(q = 1 - p\).

From \(np = 96\), we have \(p = \frac{96}{n}\).

From \(npq = 32\), we have \(np(1-p) = 32\).

Substituting \(p = \frac{96}{n}\) into \(npq = 32\), we get:

\(n \left(\frac{96}{n}\right) \left(1 - \frac{96}{n}\right) = 32\)

\(96 \left(1 - \frac{96}{n}\right) = 32\)

\(96 - \frac{9216}{n} = 32\)

\(\frac{9216}{n} = 64\)

\(n = 144\)

Since \(p = \frac{96}{144} = \frac{2}{3}\), \(k\) is such that \(P(\leq k) = \frac{2}{3}\). Therefore, \(k = 6\).

The average number of 5s in 20 throws is given as 4.8. This implies that the expected value for the number of 5s is 4.8, which can be expressed as:

\(20p = 4.8\)

Solving for \(p\), we get:

\(p = \frac{4.8}{20} = 0.24\)

We need to find the probability that the number of 5s is less than 3 in 20 throws. This is a binomial distribution problem where \(n = 20\) and \(p = 0.24\).

The probability of getting exactly \(k\) successes in a binomial distribution is given by:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

We need to calculate \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Calculating each term:

\(P(X = 0) = \binom{20}{0} (0.24)^0 (0.76)^{20} = (0.76)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.24)^1 (0.76)^{19} = 20 \times 0.24 \times (0.76)^{19}\)

\(P(X = 2) = \binom{20}{2} (0.24)^2 (0.76)^{18} = \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)

Summing these probabilities:

\(P(X < 3) = (0.76)^{20} + 20 \times 0.24 \times (0.76)^{19} + \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)

Calculating the above gives:

\(P(X < 3) = 0.109\)