(a) The probabilities for the tree diagram are:

• From Box A to Box B (Red to Red): \(\frac{x+1}{x+10}\)
• From Box A to Box B (Red to Blue): \(\frac{9}{x+10}\)
• From Box A to Box B (Blue to Red): \(\frac{x}{x+10}\)
• From Box A to Box B (Blue to Blue): \(\frac{10}{x+10}\)

(b) The probability that both balls chosen are blue is:

\(\frac{4}{10} \times \frac{10}{x+10} = \frac{4}{x+10}\)

(c) Given \(\frac{4}{x+10} = \frac{1}{6}\), solve for \(x\):

\(\frac{4}{x+10} = \frac{1}{6}\)

\(4 \times 6 = x + 10\)

\(x + 10 = 24\)

\(x = 14\)

Now, find the probability that the ball chosen from box A is red given that the ball chosen from box B is red:

\(P(A \text{ Red } | B \text{ Red }) = \frac{P(A \text{ Red } \cap B \text{ Red })}{P(B \text{ Red })}\)

\(P(A \text{ Red } \cap B \text{ Red }) = \frac{6}{10} \times \frac{x+1}{x+10} = \frac{6}{10} \times \frac{15}{24} = \frac{3}{8}\)

\(P(B \text{ Red }) = \frac{6}{10} \times \frac{15}{24} + \frac{4}{10} \times \frac{14}{24} = \frac{8}{73}\)

\(P(A \text{ Red } | B \text{ Red }) = \frac{3/8}{8/73} = \frac{45}{73} \approx 0.616\)