First, find the probability of throwing a 4. Since the probabilities for scores 5 and 6 are 0.2 each, the total probability for scores 1 to 4 is:

\(1 - 0.4 = 0.6\)

Since scores 1 to 4 have equal probabilities, the probability of throwing a 4 is:

\(\frac{0.6}{4} = 0.15\)

Let \(p = 0.15\) be the probability of throwing a 4, and \(q = 1 - p = 0.85\) be the probability of not throwing a 4.

We need to find the probability of getting a score of 4 on not more than 1 of the 3 throws. This is the sum of the probabilities of getting a score of 4 on exactly 0 or 1 throw:

\(P(0) = q^3 = (0.85)^3\)

\(P(1) = \binom{3}{1} p q^2 = 3 \times 0.15 \times (0.85)^2\)

Thus, the total probability is:

\(P(0) + P(1) = (0.85)^3 + 3 \times 0.15 \times (0.85)^2\)

Calculating these values:

\((0.85)^3 = 0.614125\)

\(3 \times 0.15 \times (0.85)^2 = 0.325875\)

Adding these gives:

\(0.614125 + 0.325875 = 0.939\)