1. **State the problem:** Jessica has a probability of $\frac{4}{5}$ of getting an A in Mathematics and a probability of $\frac{2}{5}$ of getting an A in English. We want to find the probability that Jessica gets exactly one A in the two subjects. 2. **Represent the problem with a tree diagram:** - First branch: Mathematics A with probability $\frac{4}{5}$, Mathematics not A with probability $\frac{1}{5}$. - Second branch (from each Mathematics outcome): English A with probability $\frac{2}{5}$, English not A with probability $\frac{3}{5}$. 3. **Calculate the probabilities of each combined outcome:** - Both A: $\frac{4}{5} \times \frac{2}{5} = \frac{8}{25}$ - Math A, English not A: $\frac{4}{5} \times \frac{3}{5} = \frac{12}{25}$ - Math not A, English A: $\frac{1}{5} \times \frac{2}{5} = \frac{2}{25}$ - Neither A: $\frac{1}{5} \times \frac{3}{5} = \frac{3}{25}$ 4. **Find the probability of exactly one A:** This happens if Jessica gets A in Mathematics but not in English, or not in Mathematics but A in English. $$P(\text{one A}) = \frac{12}{25} + \frac{2}{25} = \frac{14}{25}$$ 5. **Conclusion:** The probability that Jessica gets exactly one A in the two subjects is $\frac{14}{25}$ or 0.56.