$$\begin{gathered} \mathrm{Given}, \mathrm{P}\left(\mathrm{A}\right) = \frac{3}{2 + 3} = \frac{3}{5}, \mathrm{P}\left(\mathrm{B}\right) = \frac{2}{2 + 1} = \frac{2}{3} \\ \mathrm{and} \mathrm{P}\left(\overline{\mathrm{A}}\right) = 1 - \frac{3}{5} = \frac{2}{5}, \mathrm{P}\left(\overline{\mathrm{B}}\right)= 1 -\frac{2}{3} = \frac{1}{3} \\ \therefore \mathrm{Required} \mathrm{probability} \\ = \mathrm{P}\left(\mathrm{A} \cap \overline{\mathrm{B}}\right) + \mathrm{P}\left(\overline{\mathrm{A}} \cap \mathrm{B}\right) + \mathrm{P}\left(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\right) \\ = \mathrm{P}\left(\mathrm{A}\right) . \mathrm{P}\left(\overline{\mathrm{B}}\right) + \mathrm{P}\left(\overline{\mathrm{A}}\right) . \mathrm{P}\left(\mathrm{B}\right) + \mathrm{P}\left(\overline{\mathrm{A}}\right) . \mathrm{P}\left(\overline{\mathrm{B}}\right) \\ = \frac{3}{5} . \frac{1}{3} +\hspace{0.17em}\frac{2}{5} . \frac{2}{3} + \frac{2}{5} . \frac{1}{3} \\ = \frac{1}{5} +\hspace{0.17em}\frac{4}{15} + \frac{2}{15} \\ =\frac{3 + 4 + 2 }{15} = \frac{9}{15} \\ = \frac{3}{5} \end{gathered}$$