Solution:

Let represent the event that item A produces, represent the event that item produces, and represent the event that the generated product is discovered to be defective.

Then

Also (item is defective given that it is produced by machine

Likewise, (thing is defective because it was made by a machine $left.B\right)=1 \%=1 / 100\mathrm{B}\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{2}\right)}=\frac{\frac{2}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{2}{100}+\frac{2}{5} \times \frac{1}{100}}=\frac{\frac{2}{5} \times \frac{1}{100}}{\frac{1}{500}(6+2)}=\frac{2}{8}\Rightarrow \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)=\frac{1}{4}$