1. **Problem statement:** There are 8 students who prefer maths (M) and 3 who prefer English (E). The teacher calls on two students with replacement. We want to find the probability that both students prefer maths. 2. **Understanding the problem:** Since the selection is with replacement, the probability of choosing a student who prefers maths remains the same for both calls. 3. **Calculate total students:** Total students = 8 (M) + 3 (E) = 11. 4. **Probability of choosing a maths student in one call:** $$ P(M) = \frac{8}{11} $$ 5. **Probability of choosing a maths student in the second call (with replacement):** $$ P(M) = \frac{8}{11} $$ 6. **Probability that both students prefer maths:** Since the calls are independent (due to replacement), multiply the probabilities: $$ P(M \text{ and } M) = P(M) \times P(M) = \frac{8}{11} \times \frac{8}{11} = \frac{64}{121} $$ 7. **Tree diagram explanation:** - First branch: M with probability $\frac{8}{11}$, E with probability $\frac{3}{11}$. - Second branch (from each first branch outcome): M with probability $\frac{8}{11}$, E with probability $\frac{3}{11}$. The path M -> M corresponds to the probability $\frac{64}{121}$. **Final answer:** The probability that both students prefer maths is $$ \boxed{\frac{64}{121}} $$