(d) Let $ a$ white ball be transferred from the first bag to the second.

The Probability of selecting a white ball from the first bag $ = \frac{5}{9}.$

Now the second bag has $8$ white and $9$ black.

The probability of selecting white ball from the second bag $ = \frac{8}{{17}}$.

Hence required probability $ = \frac{5}{9} \times \frac{8}{{17}} = \frac{{40}}{{153}}$

If a black ball be transferred from the first bag to the second, then the probability $ = \frac{4}{9} \times \frac{7}{{17}} = \frac{{28}}{{153}}$

Therefore required probability $ = \frac{{40}}{{153}} + \frac{{28}}{{153}} = \frac{4}{9}.$