(a) The probability of not getting a 4 in a single throw is \(\frac{5}{6}\). The probability of getting a 4 for the first time on the 6th throw is \(\left( \frac{5}{6} \right)^5 \times \frac{1}{6}\). Therefore, the probability of obtaining a 4 in fewer than 6 throws is:

\(1 - \left( \frac{5}{6} \right)^5\)

Calculating this gives:

\(1 - \left( \frac{5}{6} \right)^5 = 1 - \frac{3125}{7776} = \frac{4651}{7776}\)

(b) The probability of getting a 4 in a single throw is \(\frac{1}{6}\). We use the binomial distribution to find the probability of getting at least 3 fours in 10 throws:

\(1 - P(0, 1, 2) = 1 - \left( \binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} + \binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^9 + \binom{10}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^8 \right)\)

Calculating each term:

\(\binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} = 0.1615056\)

\(\binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^9 = 0.3230111\)

\(\binom{10}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^8 = 0.290710\)

Summing these gives:

\(0.1615056 + 0.3230111 + 0.290710 = 0.7752267\)

Thus, the probability of getting at least 3 fours is:

\(1 - 0.7752267 = 0.2247733\)

Rounding to three decimal places gives \(0.225\).