(a) The probability of getting tails on both coins in one throw is \(\frac{1}{4}\). The expected number of throws to get a pair of tails is the reciprocal of this probability, which is \(\frac{1}{\frac{1}{4}} = 4\).

(b) The probability of getting exactly 3 throws to obtain a pair of tails is calculated by getting not tails in the first two throws and tails in the third throw. The probability is \(\left( \frac{3}{4} \right)^2 \times \frac{1}{4} = \frac{9}{64}\).

(c) The probability that fewer than 6 throws are required is the complement of the probability that 6 or more throws are required. The probability of not getting tails in the first 5 throws is \(\left( \frac{3}{4} \right)^5\). Therefore, the probability of fewer than 6 throws is \(1 - \left( \frac{3}{4} \right)^5 = 0.763\).

The probability of landing on a 5 in one spin is \(P(5) = 0.2\) since the spinner is fair and has 5 sides.

We need to find the probability that it takes fewer than 7 spins to get a 5. This is the complement of not getting a 5 in all 6 spins.

The probability of not getting a 5 in one spin is \(1 - 0.2 = 0.8\).

The probability of not getting a 5 in 6 spins is \(0.8^6\).

Therefore, the probability of getting at least one 5 in fewer than 7 spins is:

\(P(X < 7) = 1 - 0.8^6\)

Calculating this gives:

\(1 - 0.8^6 = 1 - 0.262144 = 0.737856\)

Thus, the probability that it takes fewer than 7 spins for George to obtain a 5 is approximately 0.73856.

(a) The possible combinations to get a score of 4 are (1,3), (3,1), and (2,2). There are 3 favorable outcomes. The total number of outcomes when two dice are thrown is 36. Therefore, the probability is \(\frac{3}{36} = \frac{1}{12}\).

(b) The mean of a geometric distribution is given by \(\frac{1}{p}\), where \(p\) is the probability of success. Here, \(p = \frac{1}{12}\), so the mean is \(\frac{1}{\frac{1}{12}} = 12\).

(c) The probability that the first success occurs on the 6th trial is given by \(\left( \frac{11}{12} \right)^5 \times \frac{1}{12}\). Calculating this gives \(0.0539\) or \(\frac{161051}{2985984}\).

(d) The probability that \(X < 8\) is \(1 - \left( \frac{11}{12} \right)^7\). Calculating this gives \(0.456\) or \(\frac{16344637}{35831808}\).

The probability of obtaining a 1 or a 6 on a single throw of a fair die is \(\frac{2}{6} = \frac{1}{3}\).

The probability of not obtaining a 1 or a 6 on a single throw is \(\frac{4}{6} = \frac{2}{3}\).

We need to find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or a 6.

This can be calculated as:

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

Calculating each term:

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^2 = \frac{4}{27}\)

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^3 = \frac{8}{81}\)

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^4 = \frac{16}{243}\)

Adding these probabilities:

\(\frac{4}{27} + \frac{8}{81} + \frac{16}{243} = \frac{76}{243}\)

Thus, the probability is \(\frac{76}{243}\) or approximately 0.313.

(a) The sum of probabilities must equal 1: \(0.28 + p + 2p + 3p = 1\). Simplifying gives \(0.28 + 6p = 1\). Solving for \(p\), we get \(6p = 0.72\), so \(p = 0.12\).

(b) For fair spinners (blue and green), the probability of any score is 0.25. We consider scenarios where the sum of scores is 4 or less:

• R = 1, B = 1, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 1, G = 2: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 2, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 2, B = 1, G = 1: \(0.12 \times (0.25)^2 = 0.0075\)

Adding these probabilities gives \(0.0175 + 0.0175 + 0.0175 + 0.0075 = 0.06\).

(c) First, find \(P(X \text{ is odd}) = 0.28 + 2p = 0.28 + 0.24 = 0.52\).

Scenarios where the product of scores is 4 or less and X is odd:

• R = 1, B = 1, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 1, G = 2: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 2, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 1, G = 3: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 3, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 3, B = 1, G = 1: \(0.24 \times (0.25)^2 = 0.015\)

Total probability for product less than 4 and X odd: \(0.28 \times (0.25)^2 \times 8 + 0.24 \times (0.25)^2 = 0.155\).

Conditional probability: \(\frac{0.155}{0.52} = 0.298\).