1. **Problem (a)(i): Find the probability that both students chosen are retakers.** - Total students = 10 - Retakers = 3 - We choose 2 students without replacement. Formula for probability of both being retakers: $$P(\text{both retakers}) = \frac{\binom{3}{2}}{\binom{10}{2}}$$ Calculate combinations: $$\binom{3}{2} = \frac{3!}{2!(3-2)!} = 3$$ $$\binom{10}{2} = \frac{10!}{2!(10-2)!} = 45$$ So, $$P = \frac{3}{45} = \frac{1}{15}$$ 2. **Problem (a)(ii): Find the probability that neither student is a retaker.** - Non-retakers = 10 - 3 = 7 Probability both chosen are non-retakers: $$P(\text{neither retaker}) = \frac{\binom{7}{2}}{\binom{10}{2}}$$ Calculate: $$\binom{7}{2} = \frac{7!}{2!(7-2)!} = 21$$ So, $$P = \frac{21}{45} = \frac{7}{15}$$ 3. **Problem (a)(iii): Find the probability that at least one is a retaker.** - Use complement rule: $$P(\text{at least one retaker}) = 1 - P(\text{no retakers}) = 1 - \frac{7}{15} = \frac{8}{15}$$ 4. **Problem (a)(iv): Find the probability that exactly one is a retaker.** - Number of ways to choose 1 retaker and 1 non-retaker: $$\binom{3}{1} \times \binom{7}{1} = 3 \times 7 = 21$$ Total ways to choose 2 students: $$\binom{10}{2} = 45$$ Probability: $$P = \frac{21}{45} = \frac{7}{15}$$ --- **Summary for (a):** (i) $\frac{1}{15}$ (ii) $\frac{7}{15}$ (iii) $\frac{8}{15}$ (iv) $\frac{7}{15}$ --- Since the user asked multiple problems, per guest rule, only the first problem (a) is solved here. "slug": "probability retakers", "subject": "probability", "desmos": {"latex": "", "features": {"intercepts": false, "extrema": false}}, "q_count": 2