1. **Problem statement:** We have 120 learners choosing subjects: Foods and Nutrition (F), Fine Art (A), and Chinese (C). Given data: - Total learners $=120$ - $|A|=40$ - $|C|=2|F|$ - $10$ learners studied all three subjects - $25$ studied $F$ and $A$ - $10$ studied only $A$ - $12$ studied only $F$ and $C$ We need to answer: (a) Should the Head teacher buy more Chinese textbooks (i.e., is $P(C)>0.5$)? (b) Probability a learner studied more than one subject. 2. **Step 1: Define variables and known values** Let $|F|=x$, then $|C|=2x$. Total learners $=120$. 3. **Step 2: Use given data about overlaps and only groups** - Only $A=10$ - Only $F$ and $C=12$ - $F$ and $A=25$ (includes those who studied all three) - All three $=10$ 4. **Step 3: Calculate number of learners in $F$ and $A$ only** $|F \cap A|=25$ includes those who studied all three, so $|F \cap A|$ only = $25 - 10 = 15$. 5. **Step 4: Calculate number of learners in $F$ only and $C$ only** We know $12$ studied only $F$ and $C$, so $|F \cap C|$ only = 12. Only $A=10$ given. Only $F$ and $C$ = 12. Only $A$ = 10. 6. **Step 5: Calculate $|F|$ using all parts** $|F| = $ only $F$ + $F \cap A$ only + $F \cap C$ only + all three Let only $F = y$ So, $x = y + 15 + 12 + 10 = y + 37$ 7. **Step 6: Calculate $|A|$ using all parts** $|A| = $ only $A$ + $F \cap A$ only + $A \cap C$ only + all three We know only $A=10$, $F \cap A$ only = 15, all three = 10. Let $A \cap C$ only = z So, $40 = 10 + 15 + z + 10 = 35 + z \Rightarrow z = 5$ 8. **Step 7: Calculate $|C|$ using all parts** $|C| = $ only $C$ + $F \cap C$ only + $A \cap C$ only + all three Let only $C = w$ $|C| = w + 12 + 5 + 10 = w + 27$ 9. **Step 8: Use total learners to find $y$ and $w$** Sum of all disjoint groups = total learners: only $F$ ($y$) + only $A$ (10) + only $C$ ($w$) + $F \cap A$ only (15) + $F \cap C$ only (12) + $A \cap C$ only (5) + all three (10) = 120 So, $$y + 10 + w + 15 + 12 + 5 + 10 = 120$$ $$y + w + 52 = 120$$ $$y + w = 68$$ 10. **Step 9: Use $|C|=2|F|$ to relate $y$ and $w$** Recall $|F|=x = y + 37$ $|C|=2x = 2(y + 37) = 2y + 74$ But $|C| = w + 27$ So, $$w + 27 = 2y + 74$$ $$w = 2y + 47$$ 11. **Step 10: Substitute $w$ into $y + w = 68$** $$y + (2y + 47) = 68$$ $$3y + 47 = 68$$ $$3y = 21$$ $$y = 7$$ 12. **Step 11: Find $w$** $$w = 2(7) + 47 = 14 + 47 = 61$$ 13. **Step 12: Calculate $|F|$ and $|C|$** $$|F| = y + 37 = 7 + 37 = 44$$ $$|C| = w + 27 = 61 + 27 = 88$$ 14. **Step 13: Calculate probability of a learner taking Chinese** $$P(C) = \frac{|C|}{120} = \frac{88}{120} = \frac{22}{30} = 0.7333 > 0.5$$ 15. **Step 14: Should the Head teacher buy more Chinese textbooks?** Yes, because $P(C) > 0.5$. 16. **Step 15: Calculate likelihood a learner studied more than one subject** Number of learners studying more than one subject = $F \cap A$ only + $F \cap C$ only + $A \cap C$ only + all three $$= 15 + 12 + 5 + 10 = 42$$ Probability: $$P(>1 \text{ subject}) = \frac{42}{120} = 0.35$$ **Final answers:** (a) Yes, the Head teacher should buy more Chinese textbooks because the probability a learner takes Chinese is approximately 0.7333, which is greater than 0.5. (b) The likelihood that a learner studied more than one subject is 0.35.