(i) Let the probability of landing on a 3 be \(p = \frac{1}{3}\). We need to find the probability of getting at least two 3s in four spins. This is given by:

\(P(\geq 2) = 1 - P(0, 1) = 1 - \left(\frac{2}{3}\right)^4 - 4 \cdot \binom{4}{1} \left(\frac{1}{3}\right) \left(\frac{2}{3}\right)^3\)

Alternatively, calculate directly:

\(P(2, 3, 4) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 + \binom{4}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right) + \left(\frac{1}{3}\right)^4\)

Thus, \(P(\geq 2) = \frac{11}{27}\) or 0.407.

(ii) To find the probability that the sum of the four scores is 5, consider the combination (1, 1, 1, 2). There are 4 permutations of this combination:

\(P(\text{sum is 5}) = P(1, 1, 1, 2) \times 4 = \left(\frac{1}{3}\right)^4 \times 4\)

Thus, \(P(\text{sum is 5}) = \frac{4}{81}\) or 0.0494.

The probability that Kersley is either asleep or studying at noon on any given day is the sum of the probabilities of these two independent events:

\(P(A) = 0.2 + 0.6 = 0.8\).

We need to find the probability that Kersley is either asleep or studying on at least 6 days out of 7. This is a binomial probability problem with parameters \(n = 7\) and \(p = 0.8\).

The probability of Kersley being asleep or studying on exactly 6 days is given by:

\(P(6) = \binom{7}{6} (0.8)^6 (0.2)^1\).

The probability of Kersley being asleep or studying on all 7 days is:

\(P(7) = (0.8)^7\).

Thus, the probability that Kersley is either asleep or studying on at least 6 days is:

\(P(6 \text{ or } 7) = P(6) + P(7)\).

Calculating these:

\(P(6) = \binom{7}{6} (0.8)^6 (0.2)^1 = 7 \times 0.262144 \times 0.2 = 0.3670016\).

\(P(7) = (0.8)^7 = 0.2097152\).

Therefore,

\(P(6 \text{ or } 7) = 0.3670016 + 0.2097152 = 0.5767168\).

Rounding to three decimal places, the probability is \(0.577\).

(i) We use the binomial probability formula:

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

where \(n = 12\), \(p = 0.65\), and \(1-p = 0.35\).

Calculate for \(r = 8, 9, 10\):

\(\binom{12}{8} (0.65)^8 (0.35)^4 + \binom{12}{9} (0.65)^9 (0.35)^3 + \binom{12}{10} (0.65)^{10} (0.35)^2\)

\(= 0.541\)

(ii) The probability that the fourth passenger is the first to carry a backpack is calculated by considering the first three passengers do not carry a backpack and the fourth does:

\(P(\overline{R}\overline{R}\overline{R}R) = (0.35)^3 \times 0.65\)

\(= 0.0279\)

First, find the probability of throwing a 4. Since the probabilities for scores 5 and 6 are 0.2 each, the total probability for scores 1 to 4 is:

\(1 - 0.4 = 0.6\)

Since scores 1 to 4 have equal probabilities, the probability of throwing a 4 is:

\(\frac{0.6}{4} = 0.15\)

Let \(p = 0.15\) be the probability of throwing a 4, and \(q = 1 - p = 0.85\) be the probability of not throwing a 4.

We need to find the probability of getting a score of 4 on not more than 1 of the 3 throws. This is the sum of the probabilities of getting a score of 4 on exactly 0 or 1 throw:

\(P(0) = q^3 = (0.85)^3\)

\(P(1) = \binom{3}{1} p q^2 = 3 \times 0.15 \times (0.85)^2\)

Thus, the total probability is:

\(P(0) + P(1) = (0.85)^3 + 3 \times 0.15 \times (0.85)^2\)

Calculating these values:

\((0.85)^3 = 0.614125\)

\(3 \times 0.15 \times (0.85)^2 = 0.325875\)

Adding these gives:

\(0.614125 + 0.325875 = 0.939\)

Let the probability that the score is greater than 2 be denoted by \(p\). The score is greater than 2 if the difference between the numbers on the two spinners is 3 or 4. The possible pairs are (1,4), (1,5), (2,5), (4,1), (5,1), (5,2). There are 6 such pairs out of a total of 25 possible outcomes (since each spinner has 5 sides), so \(p = \frac{6}{25}\).

We need to find the probability that the score is greater than 2 on at least 3 occasions out of 9 spins. This is a binomial probability problem where \(X \sim \text{Binomial}(9, \frac{6}{25})\).

The probability that the score is greater than 2 on at least 3 occasions is:

\(P(X \geq 3) = 1 - P(X = 0, 1, 2)\)

Calculate \(P(X = 0, 1, 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{9}{0} \left(\frac{6}{25}\right)^0 \left(\frac{19}{25}\right)^9\)

\(P(X = 1) = \binom{9}{1} \left(\frac{6}{25}\right)^1 \left(\frac{19}{25}\right)^8\)

\(P(X = 2) = \binom{9}{2} \left(\frac{6}{25}\right)^2 \left(\frac{19}{25}\right)^7\)

Substitute and calculate:

\(P(X \geq 3) = 1 - (0.08459 + 0.2404 + 0.3037)\)

\(P(X \geq 3) = 1 - 0.62869 = 0.37131\)

Thus, the probability that the score is greater than 2 on at least 3 occasions is approximately 0.371.