Let \(W\), \(C\), and \(B\) represent the events that Anya walks, cycles, and takes the bus, respectively. Let \(L\) represent the event that Anya is late.

The probability that Anya cycles and is late is \(P(C \cap L) = P(C) \times P(L \mid C) = 0.65 \times 0.1 = 0.065\).

The total probability that Anya is late, \(P(L)\), is given by:

\(P(L) = P(W \cap L) + P(C \cap L) + P(B \cap L)\)

\(= (0.3 \times 0.15) + (0.65 \times 0.1) + (0.05 \times 0.6)\)

\(= 0.045 + 0.065 + 0.03 = 0.14\)

The probability that Anya cycles given that she is late is:

\(P(C \mid L) = \frac{P(C \cap L)}{P(L)} = \frac{0.065}{0.14} = 0.464\)

Thus, the probability that Anya cycles given that she is late is \(0.464\) or \(\frac{13}{28}\).