1. **State the problem:** Tina has two bags, A and B, with red and blue counters. We want to complete the probability tree diagram showing the probabilities of drawing red or blue counters from each bag. 2. **Given probabilities:** - Bag A: $P(\text{Red})=\frac{5}{8}$, $P(\text{Blue})=\frac{3}{8}$ - Bag B: $P(\text{Red})=\frac{4}{9}$, $P(\text{Blue})=\frac{5}{9}$ 3. **Probability tree structure:** - First split: Bag A's outcomes (Red or Blue) - Second split: For each Bag A outcome, Bag B's outcomes (Red or Blue) 4. **Calculate joint probabilities:** - $P(\text{Red from A and Red from B}) = P(\text{Red from A}) \times P(\text{Red from B}) = \frac{5}{8} \times \frac{4}{9} = \frac{20}{72} = \frac{5}{18}$ - $P(\text{Red from A and Blue from B}) = \frac{5}{8} \times \frac{5}{9} = \frac{25}{72}$ - $P(\text{Blue from A and Red from B}) = \frac{3}{8} \times \frac{4}{9} = \frac{12}{72} = \frac{1}{6}$ - $P(\text{Blue from A and Blue from B}) = \frac{3}{8} \times \frac{5}{9} = \frac{15}{72} = \frac{5}{24}$ 5. **Interpretation:** Each branch in the tree diagram is labeled with these probabilities. The first branches from Bag A are $\frac{5}{8}$ (Red) and $\frac{3}{8}$ (Blue). From each of these, branches split to Bag B's Red ($\frac{4}{9}$) and Blue ($\frac{5}{9}$) with the joint probabilities calculated above. 6. **Summary:** The completed tree diagram shows: - Bag A Red branch: $\frac{5}{8}$ - Bag B Red: $\frac{5}{18}$ - Bag B Blue: $\frac{25}{72}$ - Bag A Blue branch: $\frac{3}{8}$ - Bag B Red: $\frac{1}{6}$ - Bag B Blue: $\frac{5}{24}$ This completes the probability tree with all probabilities labeled clearly.