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:

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