(i) The probability of taking a white paper clip from box A is \(\frac{1}{6}\). After transferring this white paper clip to box B, box B will have 7 red and 3 white paper clips. The probability of then taking a red paper clip from box B is \(\frac{7}{10}\). Therefore, the probability of both events occurring is:

\(P(W, R) = \frac{1}{6} \times \frac{7}{10} = \frac{7}{60} \approx 0.117\)

(ii) The probability of taking a red paper clip from box A is \(\frac{5}{6}\). After transferring this red paper clip to box B, box B will have 8 red and 2 white paper clips. The probability of then taking a red paper clip from box B is \(\frac{8}{10}\). Therefore, the probability of both events occurring is:

\(P(R, R) = \frac{5}{6} \times \frac{8}{10} = \frac{40}{60}\)

The total probability of taking a red paper clip from box B is the sum of the probabilities of the two mutually exclusive events (taking a red from A and then a red from B, or taking a white from A and then a red from B):

\(P(\text{red}) = \frac{40}{60} + \frac{7}{60} = \frac{47}{60} \approx 0.783\)

(iii) The probability that the paper clip taken from box A was red, given that the paper clip from box B is red, is calculated using conditional probability:

\(P(R | R) = \frac{P(R \cap R)}{P(R)} = \frac{\frac{40}{60}}{\frac{47}{60}} = \frac{40}{47} \approx 0.851\)