To solve part (ii), we need to find the probability distribution for the number of new pens in a sample of two pens.

Let \(X\) be the random variable representing the number of new pens in the sample. \(X\) can take values 0, 1, or 2.

1. Probability of 0 new pens: \(P(X = 0) = \frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\).

2. Probability of 1 new pen: \(P(X = 1) = \left( \frac{3}{10} \times \frac{7}{9} \right) + \left( \frac{7}{10} \times \frac{3}{9} \right) = \frac{7}{15}\).

3. Probability of 2 new pens: \(P(X = 2) = \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}\).

For part (iii), calculate the expected value \(E(X)\):

\(E(X) = 0 \times \frac{7}{15} + 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}\).