1. **State the problem:** We have 75 learners playing at most two games among football (F), volleyball (V), and netball (N). Given data: - Only football: 9 - Only volleyball: 12 - Only netball: 5 - Volleyball and netball overlap: 10 - Football or volleyball but not netball: 32 - Number of learners playing football equals those playing volleyball. We need to find: (a)(i) Number of learners who play none of the three games. (a)(ii) Probability of selecting a learner who plays none. (b) Whether the teacher's target of more than 60 learners playing football or volleyball is reached. 2. **Define variables and sets:** - Let $F$ be the set of football players. - Let $V$ be the set of volleyball players. - Let $N$ be the set of netball players. 3. **Given:** - $|F_{only}|=9$ - $|V_{only}|=12$ - $|N_{only}|=5$ - $|V \cap N|=10$ - $|F \cup V \text{ but not } N|=32$ - Total learners $=75$ - $|F|=|V|$ 4. **Analyze the overlap $F \cup V$ but not $N$:** This includes: - Only football: 9 - Only volleyball: 12 - Football and volleyball but not netball: $x$ So, $$9 + 12 + x = 32$$ $$x = 32 - 21 = 11$$ 5. **Calculate total football and volleyball players:** - Football players $|F| = |F_{only}| + |F \cap V \text{ but not } N| = 9 + 11 = 20$ - Volleyball players $|V| = |V_{only}| + |F \cap V \text{ but not } N| + |V \cap N| = 12 + 11 + 10 = 33$ But given $|F|=|V|$, so our calculation must be consistent. 6. **Adjust for equality of $|F|$ and $|V|$:** Since $|F|=|V|$, let $|F \cap V \text{ but not } N| = y$. Then: $$|F| = 9 + y$$ $$|V| = 12 + y + 10 = 22 + y$$ Set equal: $$9 + y = 22 + y$$ This is impossible unless $9=22$, so the assumption that $|V \cap N|=10$ is fully separate from $|F \cap V \text{ but not } N|$ is correct but the problem states 32 learners play football or volleyball but not netball. 7. **Re-examine the problem statement:** "32 play football or volleyball but not netball" means: $$|F \cup V| - |N| = 32$$ But since players play at most two games, and the only overlaps are between two games, the sets are disjoint except for overlaps. 8. **Calculate total football and volleyball players:** $$|F \cup V| = |F| + |V| - |F \cap V|$$ We know $|F|=|V|=k$ (say). Also, $|F \cap V| = y$ (football and volleyball but not netball). Given: $$|F \cup V| - |N| = 32$$ But $|N|$ includes only netball players and overlaps with volleyball (10) and only netball (5). 9. **Sum of all players:** Total players = 75 = sum of all disjoint groups + those who play none. Groups: - Only football: 9 - Only volleyball: 12 - Only netball: 5 - Volleyball and netball: 10 - Football and volleyball (not netball): $y$ - None: $z$ Sum: $$9 + 12 + 5 + 10 + y + z = 75$$ $$36 + y + z = 75$$ $$y + z = 39$$ 10. **From football and volleyball equality:** $$|F| = 9 + y$$ $$|V| = 12 + y + 10 = 22 + y$$ Set equal: $$9 + y = 22 + y$$ Again impossible, so the problem likely means the number of learners who play football equals those who play volleyball including overlaps. 11. **Reconsider the problem statement:** "The number of learners who play football is equal to those who play volleyball." So: $$|F| = |V|$$ Where: $$|F| = 9 + y$$ $$|V| = 12 + y + 10$$ Set equal: $$9 + y = 22 + y$$ No solution unless $9=22$ which is false. 12. **Possible interpretation:** The 32 learners who play football or volleyball but not netball includes only football, only volleyball, and football-volleyball overlap. Given only football = 9, only volleyball = 12, so overlap football-volleyball but not netball = $$32 - (9 + 12) = 11$$ Then total football players: $$9 + 11 = 20$$ Total volleyball players: $$12 + 11 + 10 = 33$$ Not equal. 13. **Assuming the problem means the number of learners who play football equals those who play volleyball including all overlaps, but the 10 volleyball-netball players are not football players, so football players = 20, volleyball players = 33, which contradicts the problem. 14. **Therefore, the only way to satisfy $|F|=|V|$ is if the 10 volleyball-netball players also play football, but the problem states at most two games per learner, so no triple overlaps. 15. **Hence, the problem's data is inconsistent or the 10 volleyball-netball players are not counted in volleyball players for this equality. 16. **Proceed to find number of learners who play none:** Sum of all players playing at least one game: $$9 + 12 + 5 + 10 + 11 = 47$$ Total learners = 75 So, $$z = 75 - 47 = 28$$ 17. **Probability of selecting a learner who plays none:** $$P = \frac{28}{75} \approx 0.3733$$ 18. **Teacher's target:** Target is more than 60 learners who play football or volleyball. Currently, $$|F \cup V| = |F| + |V| - |F \cap V| = 20 + 33 - 11 = 42$$ So target not reached. **Final answers:** (a)(i) Number of learners who play none = 28 (a)(ii) Probability = $\frac{28}{75} \approx 0.3733$ (b) Teacher's target is not reached because only 42 learners play football or volleyball, less than 60.