(i) To find \(P(2 \leq X \leq 6)\), we use the geometric distribution formula. The probability of passing on the \(n\)-th attempt is \((1-p)^{n-1}p\), where \(p = 0.3\).

\(P(2 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) = 1 - (0.7)^6 - (0.7)\)

Calculating each term:

\((0.7)^6 = 0.117649\)

\((0.7) = 0.7\)

\(P(2 \leq X \leq 6) = 1 - 0.117649 - 0.7 = 0.582351\)

Thus, \(P(2 \leq X \leq 6) \approx 0.582\).

(ii) The expected value \(E(X)\) for a geometric distribution is given by \(\frac{1}{p}\).

\(E(X) = \frac{1}{0.3} = 3.3333\)

Thus, \(E(X) = 3 \frac{1}{3}\).

(a) The probability of obtaining a score of 17 or more in one throw is given by:

\(P(17 \text{ or } 18) = \frac{4}{216} = \frac{1}{54}\)

To find the probability that this score is first obtained on the 6th throw, we use the geometric distribution:

\(P(X = 6) = \left(\frac{53}{54}\right)^5 \times \frac{1}{54}\)

Calculating this gives:

\(P(X = 6) = 0.0169\)

(b) To find the probability that a score of 17 or more is obtained in fewer than 8 throws, we use the complement rule:

\(P(X < 8) = 1 - \left(\frac{53}{54}\right)^7\)

Calculating this gives:

\(P(X < 8) = 0.123\)

An alternative method involves summing probabilities for each throw from 1 to 7:

\(P(X < 8) = \left(\frac{1}{54}\right) + \left(\frac{53}{54}\right)\left(\frac{1}{54}\right) + \left(\frac{53}{54}\right)^2\left(\frac{1}{54}\right) + \ldots + \left(\frac{53}{54}\right)^6\left(\frac{1}{54}\right)\)

This also results in:

\(P(X < 8) = 0.123\)

(a) The probability of obtaining a 4 on a single throw is \(\frac{1}{6}\). The probability of not obtaining a 4 is \(\frac{5}{6}\). For the first 7 throws, Ramesh does not get a 4, and on the 8th throw, he gets a 4. The probability is:

\(\left( \frac{5}{6} \right)^7 \times \frac{1}{6} = \frac{78125}{1679616} \approx 0.0465\)

(b) The probability that it takes no more than 5 throws to obtain a 4 is the complement of not obtaining a 4 in all 5 throws:

\(P(X < 6) = 1 - \left( \frac{5}{6} \right)^5\)

Alternatively, it can be calculated as the sum of probabilities of obtaining a 4 on the 1st, 2nd, 3rd, 4th, or 5th throw:

\(\frac{1}{6} + \left( \frac{5}{6} \right) \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^3 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^4 \left( \frac{1}{6} \right)\)

Calculating gives:

\(\frac{4651}{7776} \approx 0.598\)

(a) The probability that the first lemon chocolate is the 7th is given by the geometric distribution:

\(\left( \frac{4}{5} \right)^6 \times \frac{1}{5} = \frac{4096}{78125} \approx 0.0524\).

(b) The probability that the first lemon chocolate is after at least 6 checks is:

\(1 - \left( \frac{4}{5} \right)^5 = \frac{4096}{15625} \approx 0.262\).

(c) The probability that none of Petra's chocolates is orange is:

\(\frac{10}{15} \times \frac{9}{14} \times \frac{8}{13} = \frac{24}{91} \approx 0.2637\).

(d) The probability that each chocolate has a different flavour is:

\(\frac{7}{15} \times \frac{5}{14} \times \frac{3}{13} = \frac{3}{13} \approx 0.231\).

(e) Given none are orange, the probability that at least 2 are strawberry is:

\(\frac{7}{15} \times \frac{6}{14} \times \frac{5}{13} + \frac{7}{15} \times \frac{6}{14} \times \frac{3}{13} + \frac{7}{15} \times \frac{6}{14} \times \frac{3}{13} = \frac{49}{60} \approx 0.817\).

(b) The probability that the first wet day is 8 October means that the first 7 days are dry and the 8th day is wet. The probability of a dry day is 1 - 0.16 = 0.84. Therefore, the probability is given by:

\((0.84)^7 \times 0.16 = 0.0472\)

(c) We need to find the probability that in exactly 1 out of 4 years, the first wet day is 8 October. Using the result from part (b), the probability for one year is 0.0472. The probability that this happens in exactly 1 out of 4 years is given by the binomial probability formula:

\(4 \times 0.0472 \times (1 - 0.0472)^3 = 0.163\)