1. **State the problem:** We have 11 animals: 7 bats and 4 mice. One animal leaves at random, then a second animal leaves at random without replacement. We want to complete the tree diagram and find the probability that a mouse leaves first and then a bat. 2. **Understand the tree diagram:** - First animal: Bat with probability $\frac{7}{11}$, Mouse with probability $\frac{4}{11}$. - Second animal after first is Bat: probabilities given as Bat $\frac{6}{10}$ and Mouse $\frac{4}{10}$. - Second animal after first is Mouse: probabilities missing, need to find. 3. **Calculate missing probabilities:** If the first animal is a mouse, then 4 mice become 3 mice left, and bats remain 7. Total animals left: $11 - 1 = 10$. - Probability second animal is Bat given first is Mouse: $$P(\text{Bat second} | \text{Mouse first}) = \frac{7}{10}$$ - Probability second animal is Mouse given first is Mouse: $$P(\text{Mouse second} | \text{Mouse first}) = \frac{3}{10}$$ 4. **Calculate the probability that a mouse leaves first and then a bat:** $$P(\text{Mouse first and Bat second}) = P(\text{Mouse first}) \times P(\text{Bat second} | \text{Mouse first}) = \frac{4}{11} \times \frac{7}{10} = \frac{28}{110} = \frac{14}{55}$$ 5. **Summary:** - Missing probabilities are $\frac{7}{10}$ for Bat and $\frac{3}{10}$ for Mouse after a Mouse leaves first. - Probability mouse leaves first then bat leaves second is $\frac{14}{55}$. Final answer: $$\boxed{\frac{14}{55}}$$