To solve these problems, we first need to determine the probability of obtaining a score of 8 or more with two dice. The possible scores are 8, 9, 10, 11, and 12. The combinations for these scores are:

• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 9: (3,6), (4,5), (5,4), (6,3)
• 10: (4,6), (5,5), (6,4)
• 11: (5,6), (6,5)
• 12: (6,6)

There are 36 possible outcomes when two dice are thrown. The number of favorable outcomes for a score of 8 or more is 15. Thus, the probability of obtaining a score of 8 or more in one throw is:

\(p = \frac{15}{36} = \frac{5}{12} \approx 0.4167\)

The probability of not obtaining a score of 8 or more is:

\(1 - p = \frac{21}{36} = \frac{7}{12} \approx 0.5833\)

(a) The probability that it takes exactly 5 throws to obtain a score of 8 or more is given by the geometric distribution:

\(P(X = 5) = (1-p)^4 \times p = \left(\frac{21}{36}\right)^4 \times \frac{15}{36}\)

Calculating this gives:

\(P(X = 5) = \frac{12005}{248832} \approx 0.0482\)

(b) The probability that it takes no more than 4 throws to obtain a score of 8 or more is:

Method 1:

\(P(X \leq 4) = 1 - \left(\frac{21}{36}\right)^4\)

Calculating this gives:

\(P(X \leq 4) = \frac{18335}{20736} \approx 0.884\)

Method 2:

Using the cumulative probability:

\(P(X \leq 4) = \frac{15}{36} + \frac{15}{36} \times \frac{21}{36} + \frac{15}{36} \times \left(\frac{21}{36}\right)^2 + \frac{15}{36} \times \left(\frac{21}{36}\right)^3\)

Calculating this gives the same result:

\(P(X \leq 4) = \frac{18335}{20736} \approx 0.884\)

(a) The probability of obtaining a pair of heads (HH) in one throw is \(\frac{1}{4}\). The expected number of throws \(E(X)\) is the reciprocal of this probability, so \(E(X) = \frac{1}{\frac{1}{4}} = 4\).

(b) The probability that exactly 5 throws are required is given by the geometric distribution formula: \(P(X = 5) = \left( \frac{3}{4} \right)^4 \left( \frac{1}{4} \right) = 0.0791\).

(c) The probability that fewer than 7 throws are required is \(P(X < 7) = 1 - \left( \frac{3}{4} \right)^6\). Calculating this gives \(1 - \left( \frac{3}{4} \right)^6 = 0.822\).

(a) To find \(P(X = 4)\), we need the probability that the first 3 spins do not result in a 2, and the 4th spin does. The probability of not getting a 2 on a single spin is \(\frac{4}{5}\), and the probability of getting a 2 is \(\frac{1}{5}\). Therefore,

\(P(X = 4) = \left( \frac{4}{5} \right)^3 \left( \frac{1}{5} \right) = (0.8)^3 (0.2) = 0.1024 = \frac{64}{625}.\)

(b) To find \(P(X < 6)\), we calculate the probability of getting a 2 within the first 5 spins. This is the complement of not getting a 2 in the first 5 spins:

\(P(X < 6) = 1 - (0.8)^5.\)

Calculating this gives:

\(P(X < 6) = 1 - 0.8^5 = 1 - 0.32768 = 0.67232.\)

Let the probability of receiving an elephant be \(\frac{1}{5}\) and the probability of not receiving an elephant be \(\frac{4}{5}\).

We need to find the probability that the first elephant is received before the 6th Monday, i.e., \(P(X < 6)\).

Method 1:

Calculate \(P(X \leq 5) = 1 - (0.8)^5\).

\(P(X \leq 5) = 1 - 0.8^5 = 0.672\).

Method 2:

Calculate \(P(X < 6) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)\).

\(P(X = 1) = \frac{1}{5}\).

\(P(X = 2) = \left( \frac{4}{5} \right) \times \frac{1}{5}\).

\(P(X = 3) = \left( \frac{4}{5} \right)^2 \times \frac{1}{5}\).

\(P(X = 4) = \left( \frac{4}{5} \right)^3 \times \frac{1}{5}\).

\(P(X = 5) = \left( \frac{4}{5} \right)^4 \times \frac{1}{5}\).

Summing these probabilities gives \(0.672\).

(b) The probability that the 5th person is the first person not in favour is given by the geometric distribution formula:

\((0.8)^4 \times (0.2) = 0.08192\)

This is because the first four people are in favour (probability 0.8 each), and the 5th person is not in favour (probability 0.2).

(c) The probability that the 7th person is the second person not in favour is calculated using the negative binomial distribution:

\((0.8)^5 \times (0.2)^2 \times 6 = 0.0786\)

This accounts for 5 people in favour, 2 not in favour, and the combination factor of choosing which two out of seven are not in favour.