Let box A, box B, and box C be the three boxes, respectively. The given event is a Compound Event which can be written as: \[ [(1G \text{ from box A}) \text{ and } (1R \text{ from box B}) \text{ and } (1R \text{ from box C})] \] or \[ [(1R \text{ from box A}) \text{ and } (1G \text{ from box B}) \text{ and } (1R \text{ from box C})] \] or \[ [(1R \text{ from box A}) \text{ and } (1R \text{ from box B}) \text{ and } (1G \text{ from box C})] \] Thus, the probability is: \[ P = \left(\frac{{{}^1C_1}}{{{}^4C_1}} \times \frac{{{}^2C_1}}{{{}^4C_1}} \times \frac{{{}^1C_1}}{{{}^4C_1}} \right) \]\[+ \left(\frac{{{}^3C_1}}{{{}^4C_1}} \times \frac{{{}^2C_1}}{{{}^pC_1}} \times \frac{{{}^1C_1}}{{{}^pC_1}} \right) + \left(\frac{{{}^3C_1}}{{{}^4C_1}} \times \frac{{{}^2C_1}}{{{}^4C_1}} \times \frac{{{}^3C_1}}{{{}^4C_1}} \right) \] \[ P = \left(\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4} \right) + \left(\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4} \right) + \left(\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} \right) \] \[ P = \frac{2}{64} + \frac{6}{64} + \frac{18}{64} \] \[ P = \frac{26}{64} \] \[ P = \frac{13}{32}. \]