1. **State the problem:** We have two bags with marbles: - Bag 1: 1 green (G), 1 red (R), 2 blue (B) - Bag 2: 1 blue (B), 1 green (G), 2 red (R) We randomly remove one marble from each bag. We want the probability of removing one green and one blue marble in total. 2. **Identify possible favorable outcomes:** - Green from Bag 1 and Blue from Bag 2 - Blue from Bag 1 and Green from Bag 2 3. **Calculate probabilities for each bag:** - Total marbles in Bag 1: $1+1+2=4$ - Total marbles in Bag 2: $1+1+2=4$ 4. **Calculate probability of each favorable event:** - $P(\text{Green from Bag 1}) = \frac{1}{4}$ - $P(\text{Blue from Bag 2}) = \frac{1}{4}$ - $P(\text{Blue from Bag 1}) = \frac{2}{4} = \frac{1}{2}$ - $P(\text{Green from Bag 2}) = \frac{1}{4}$ 5. **Calculate combined probabilities (independent events):** - $P(\text{Green Bag 1 and Blue Bag 2}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ - $P(\text{Blue Bag 1 and Green Bag 2}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ 6. **Add probabilities for either event:** $$ P(\text{Green and Blue}) = \frac{1}{16} + \frac{1}{8} = \frac{1}{16} + \frac{2}{16} = \frac{3}{16} $$ **Final answer:** The probability of removing one green and one blue marble is $\boxed{\frac{3}{16}}$.