First, define $ E_1 $, $ E_2 $, and $ E_3 $ as the events of drawing a white, red, and blue ball from bag $ P $, respectively. The probabilities are calculated as follows:

$ P(E_1) = \frac{3}{10} $ (probability of drawing a white ball from $ P $)

$ P(E_2) = \frac{2}{10} $ (probability of drawing a red ball from $ P $)

$ P(E_3) = \frac{5}{10} $ (probability of drawing a blue ball from $ P $)

Now, consider the composition of bag $ Q $ after transferring one ball from bag $ P $:

If $ E_1 $ occurs (a white ball is transferred), bag $ Q $ will have 3 white balls, 3 red balls, and 5 blue balls. The probability of drawing a red ball from $ Q $ becomes $ \frac{3}{11} $.

If $ E_2 $ occurs (a red ball is transferred), bag $ Q $ will have 2 white balls, 4 red balls, and 5 blue balls. The probability of drawing a red ball from $ Q $ becomes $ \frac{4}{11} $.

If $ E_3 $ occurs (a blue ball is transferred), bag $ Q $ will have 2 white balls, 3 red balls, and 6 blue balls. The probability of drawing a red ball from $ Q $ becomes $ \frac{3}{11} $.

The overall probability $ P(E) $ of drawing a red ball from bag $ Q $ is computed using the law of total probability:

$ P(E) = P(E_1) \cdot \frac{3}{11} + P(E_2) \cdot \frac{4}{11} + P(E_3) \cdot \frac{3}{11} $

Substitute the values:

$ P(E) = \left(\frac{3}{10} \times \frac{3}{11}\right) + \left(\frac{2}{10} \times \frac{4}{11}\right) + \left(\frac{5}{10} \times \frac{3}{11}\right) $

$ P(E) = \frac{9}{110} + \frac{8}{110} + \frac{15}{110} = \frac{32}{110} = \frac{16}{55} $

The probability that a ball chosen from bag $ Q $ is red after transfer is $ \frac{16}{55} $.

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