(a) The probability of obtaining two tails in one throw is \(\frac{1}{4}\). The probability of not obtaining two tails is \(\frac{3}{4}\). For the first time on the 7th throw, we need 6 failures followed by 1 success:

\(\left( \frac{3}{4} \right)^6 \times \frac{1}{4} = \frac{729}{16384} \approx 0.0445\)

(b) The probability that it takes more than 9 throws is the probability of failing 9 times:

\(\left( \frac{3}{4} \right)^9 = \frac{19683}{262144} \approx 0.0751\)

(a) To find the probability that the score is 4, we need to consider the possible combinations of dice rolls that sum to 4. The possible combinations are (1, 1, 2), (1, 2, 1), and (2, 1, 1). Each die has a probability of \(\frac{1}{6}\) for each face. Therefore, the probability for each combination is \(\left( \frac{1}{6} \right)^3\). Since there are 3 such combinations, the total probability is:

\(3 \times \left( \frac{1}{6} \right)^3 = \frac{3}{216} = \frac{1}{72}\)

(b) The probability of obtaining a score of 18 in a single throw is \(\left( \frac{1}{6} \right)^3 = \frac{1}{216}\) because all dice must show a 6. To find the probability that this occurs for the first time on the 5th throw, we use the geometric distribution formula:

\(P(18 \text{ on 5th throw}) = \left( \frac{215}{216} \right)^4 \times \frac{1}{216}\)

Calculating this gives:

\(\left( \frac{215}{216} \right)^4 \approx 0.981\)

\(0.981 \times \frac{1}{216} \approx 0.00454\)

(a) The mean of a geometric distribution with success probability \(p\) is given by \(\frac{1}{p}\). For a fair die, the probability of rolling a 5 is \(\frac{1}{6}\). Thus, the mean is \(\frac{1}{\frac{1}{6}} = 6\).

(b) The probability that a 5 is first obtained after the 3rd throw but before the 8th throw is calculated as:

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

Alternatively, it can be calculated as:

\(\left( \frac{5}{6} \right)^3 - \left( \frac{5}{6} \right)^7\)

This results in approximately 0.300.

(c) The probability that a 5 is first obtained in fewer than 10 throws is:

\(1 - \left( \frac{5}{6} \right)^9\)

This results in approximately 0.806.

(a) The probability of not getting a 3 on a single spin is \(\frac{4}{5}\). The probability of getting a 3 on the 8th spin for the first time is \(\left( \frac{4}{5} \right)^7 \times \frac{1}{5}\). This evaluates to \(\frac{16384}{390625}\) or approximately 0.0419.

(b) The probability of getting a 3 for the first time in fewer than 6 spins is the complement of not getting a 3 in the first 5 spins. This is calculated as:

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

Alternatively, it can be calculated using the sum of probabilities for getting a 3 on each of the first 5 spins:

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

This evaluates to \(\frac{2101}{3125}\) or approximately 0.672.

(a) The probability of not obtaining a 6 on a single throw is \(\frac{5}{6}\). The probability of not obtaining a 6 for 8 consecutive throws is \(\left( \frac{5}{6} \right)^8\). Therefore, the probability that obtaining a 6 takes more than 8 throws is \(\left( \frac{5}{6} \right)^8 = 0.233\).

(b) For two dice, the probability of obtaining a pair of 6s on a single throw is \(\frac{1}{36}\). The expected number of throws to get a pair of 6s is the reciprocal of this probability, which is \(\frac{1}{\frac{1}{36}} = 36\).

(c) The probability of obtaining a pair of 6s on the 10th throw is \(\left( \frac{35}{36} \right)^9 \times \frac{1}{36}\). The probability of obtaining a pair of 6s on the 11th throw is \(\left( \frac{35}{36} \right)^{10} \times \frac{1}{36}\). Therefore, the probability that obtaining a pair of 6s takes either 10 or 11 throws is \(\left( \frac{35}{36} \right)^9 \times \frac{1}{36} + \left( \frac{35}{36} \right)^{10} \times \frac{1}{36} = 0.0425\).