1. **State the problem:** We have a bag with 10 gum balls, 7 candy bars, and 3 toffees, totaling $10 + 7 + 3 = 20$ candies. Two candies are drawn without replacement. 2. **Part a) Probability of keeping a gum ball:** - Total candies: 20 - Number of gum balls: 10 - Probability first candy is a gum ball: $\frac{10}{20} = \frac{1}{2}$ - After drawing one gum ball, remaining gum balls: 9, total candies left: 19 - Probability second candy is a gum ball: $\frac{9}{19}$ - Probability both candies are gum balls: $\frac{10}{20} \times \frac{9}{19} = \frac{1}{2} \times \frac{9}{19} = \frac{9}{38}$ 3. **Part b) Probability of keeping any candy:** - Total candies: 20 - Number of candies (gum balls + candy bars + toffees): 20 (all candies) - Probability first candy is any candy: $\frac{20}{20} = 1$ - Probability second candy is any candy: $\frac{19}{19} = 1$ - Probability both candies are any candy: $1 \times 1 = 1$ 4. **Part c) Probability of not keeping any candy:** - Since all items are candies, the probability of not keeping any candy is 0. **Final answers:** - a) $\frac{9}{38}$ - b) $1$ - c) $0$