Let \(D\) be the event that the dog chases ducks, and \(G\) be the event that the dog chases geese. Let \(A\) be the event that the dog is attacked, and \(NA\) be the event that the dog is not attacked.

We need to find \(P(G \mid NA)\), which is given by:

\(P(G \mid NA) = \frac{P(G \cap NA)}{P(NA)}\)

First, calculate \(P(G \cap NA)\):

\(P(G \cap NA) = P(G) \times P(NA \mid G) = \frac{2}{5} \times \left(1 - \frac{3}{4}\right) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)

Next, calculate \(P(NA)\):

\(P(NA) = P(D \cap NA) + P(G \cap NA)\)

\(P(D \cap NA) = P(D) \times P(NA \mid D) = \frac{3}{5} \times \left(1 - \frac{1}{10}\right) = \frac{3}{5} \times \frac{9}{10} = \frac{27}{50}\)

Thus,

\(P(NA) = \frac{27}{50} + \frac{1}{10} = \frac{27}{50} + \frac{5}{50} = \frac{32}{50}\)

Finally, substitute back to find \(P(G \mid NA)\):

\(P(G \mid NA) = \frac{\frac{1}{10}}{\frac{32}{50}} = \frac{1}{10} \times \frac{50}{32} = \frac{5}{32}\)