The probability of obtaining a score of 8 or more with two dice is calculated first. The possible outcomes for scores of 8 or more are: (2,6), (3,5), (4,4), (5,3), (6,2), (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). There are 15 favorable outcomes out of 36 total outcomes, so the probability is:

\(P(\text{score} \geq 8) = \frac{15}{36} = \frac{5}{12}\)

We need to find the probability of obtaining a score of 8 or more on fewer than 3 occasions in 8 throws. This is a binomial distribution problem where:

\(n = 8, \quad p = \frac{5}{12}\)

We calculate the probability of obtaining a score of 8 or more on 0, 1, or 2 occasions:

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each term:

\(P(X = 0) = \binom{8}{0} \left(\frac{5}{12}\right)^0 \left(\frac{7}{12}\right)^8 = 0.01341\)

\(P(X = 1) = \binom{8}{1} \left(\frac{5}{12}\right)^1 \left(\frac{7}{12}\right)^7 = 0.07661\)

\(P(X = 2) = \binom{8}{2} \left(\frac{5}{12}\right)^2 \left(\frac{7}{12}\right)^6 = 0.1915\)

Summing these probabilities gives:

\(P(X < 3) = 0.01341 + 0.07661 + 0.1915 = 0.28152\)

Rounding to three decimal places, the probability is:

\(\boxed{0.282}\)