The Fair Stall: Set Logic, Probability and a Pricing Decision
At a school fair a stall offers a prize to a student drawn at random. A poster put up by Form 5A (35 students) states: '20 of us study Mathematics, 18 study English, and 8 study neither.'
A poster from Form 5B (25 students) states: '14 study Mathematics, 12 study English, and 5 study both.'
Every student in both classes studies at least one subject, but the two posters count only Mathematics and English.
The organiser charges $2 for a draw. A prize is given only when the drawn student studies Mathematics, and the prize is worth $5. She expects 40 draws during the fair. In the fair programme she wants to print two claims: 'more than half of Form 5A studies both Mathematics and English', and 'the proportion studying Mathematics is the same in both classes'.
1. Using Form 5A's poster, find the number of students who study both Mathematics and English, and the number who study exactly one of the two.
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Start with the total 35 students, of whom 8 study neither subject, so the number studying at least one of the two is n(M ∪ E) = 35 − 8 = 27
Both subjects Using n(M ∪ E) = n(M) + n(E) − n(M ∩ E): 27 = 20 + 18 − n(M ∩ E) 27 = 38 − n(M ∩ E) n(M ∩ E) = 38 − 27 = 11
Check with the regions of a Venn diagram:
- Mathematics only = 20 − 11 = 9
- English only = 18 − 11 = 7
- Both = 11
- Neither = 8 Sum = 9 + 7 + 11 + 8 = 35 ✓
Exactly one subject = 9 + 7 = 16 students
A frequent error is to subtract the 'neither' group twice, or to write 20 + 18 = 38 and divide by the 35 total, which would count the 11 who study both twice.
2. A Form 5A student is chosen at random. Find the probability that the student studies exactly one of the two subjects, and the probability that a student who studies at least one of them studies Mathematics.
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Exactly one subject From the Venn regions, 16 of the 35 students study exactly one. P = 16/35 ≈ 0.457 (3 d.p.)
Conditional probability The sample space is now restricted to the 27 students who study at least one of Mathematics or English. Of those, 20 study Mathematics (this includes the 11 who study both - the condition does not exclude them). P(Mathematics | at least one) = 20/27 ≈ 0.741 (3 d.p.)
Common trap: dividing by 35 gives 20/35 = 0.571, which answers a different question ('what is the chance a random student studies Mathematics?'). Once the condition is given, both the numerator and the denominator change - here only the denominator does, because all 20 Mathematics students are inside the 27.
Sanity check: 0.741 must be at least 20/35, and it is, because restricting to students who study something removes the 8 who study neither.
3. Test the two claims in the programme and, using expected values, advise the organiser whether a $5 prize with a $2 charge over 40 draws is sound.
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Claim 1: 'more than half of Form 5A studies both' Both = 11 out of 35 → 11/35 = 0.314 = 31.4% Half of 35 is 17.5, so 'more than half' would need at least 18 students. The claim is false; fewer than one in three study both.
Claim 2: 'the Mathematics proportion is the same in both classes' Form 5A: 20/35 = 0.571 = 57.1% Form 5B: 14/25 = 0.560 = 56.0% The figures are close but not the same. Since the claim is exact, it is false as worded; it could honestly be printed as 'about 56–57% in both classes'. (For completeness, Form 5B: both = 5, so at least one = 14 + 12 − 5 = 21, neither = 4.)
Is the stall sound? P(Mathematics in 5A) = 20/35 = 4/7 Expected number of prizes in 40 draws = 40 × 4/7 = 22.86 Expected prize cost = 22.86 × $5 = $114.29 Expected income = 40 × $2 = $80.00 Expected loss ≈ $34.29
Advice: the stall loses money as planned. Break-even prize value = 80 ÷ 22.857 = $3.50, so either cut the prize to about $3.50, raise the charge to 5 × 4/7 = $2.86 (say $3), or award the prize only when the drawn student studies both subjects, where P = 11/35 and expected prizes fall to 40 × 11/35 = 12.57. Note that 22.86 is an expected count, not a guarantee - on the day the actual number will vary, so the organiser should not treat the $34 as a fixed loss.