To solve part (b), we need to find the probability distribution for the number of times a 1 or a 6 is rolled in 3 throws of a die. The probability of rolling a 1 or a 6 in a single throw is \(\frac{1}{3}\), and the probability of not rolling a 1 or a 6 is \(\frac{2}{3}\).

Calculating each probability:

\(P(0) = \left( \frac{2}{3} \right)^3 = \frac{8}{27}\)

\(P(1) = \binom{3}{1} \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^2 = 3 \times \frac{1}{3} \times \frac{4}{9} = \frac{12}{27}\)

\(P(2) = \binom{3}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^1 = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27}\)

\(P(3) = \left( \frac{1}{3} \right)^3 = \frac{1}{27}\)

For part (c), the expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{8}{27} + 1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}\)

\(E(X) = \frac{0}{27} + \frac{12}{27} + \frac{12}{27} + \frac{3}{27}\)

\(E(X) = \frac{27}{27} = 1\)