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