Language and Sport Survey with a Fundraising Game
A school of 200 students surveys participation in a language and a sport. The results show 120 students study a language, 90 play a sport, and 40 do both. The principal wants to use the survey to decide whether to schedule language clubs and sports fixtures on the same afternoon. A teacher claims that 'most language students also play a sport', and the sports coordinator claims that 'sport and language participation are independent, so the timetable change will not affect either group'. The student council also proposes a fundraiser: roll a fair six-sided die once; if it lands on 6, the player wins €5, otherwise the player pays €1. The council wants to know whether this game is fair to the player and what prize on a 6 would make the expected gain €0.20 per play.
1. Construct the two-way table and calculate P(language | sport) and P(sport | language). Assess the teacher's claim.
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The given totals give the table:
| Sport | No sport | Total | |
|---|---|---|---|
| Language | 40 | 80 | 120 |
| No language | 50 | 30 | 80 |
| Total | 90 | 110 | 200 |
P(language | sport) = 40/90 = 4/9 ≈ 0.444. P(sport | language) = 40/120 = 1/3 ≈ 0.333. The teacher says 'most language students also play a sport', which would require P(sport | language) > 0.5. It is only about 0.333, so the claim is false: only one third of language students play a sport.
2. Test whether language participation and sport participation are independent. What does the result mean for the coordinator's claim?
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If the events were independent, P(language and sport) would equal P(language) × P(sport). P(language) = 120/200 = 0.6. P(sport) = 90/200 = 0.45. Their product is 0.6 × 0.45 = 0.27. The actual joint probability is 40/200 = 0.20. Since 0.20 ≠ 0.27, the events are not independent. Equivalently, P(language | sport) = 0.444 is not equal to P(language) = 0.6. Sport participation is associated with a lower language participation rate in this survey. The coordinator's claim of independence is not supported, so the timetable change could affect the groups differently.
3. Calculate the expected gain per play for the die game with a €5 prize. Find the prize on a 6 that would give an expected gain of €0.20 per play.
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With a €5 prize, expected gain = (1/6)(+5) + (5/6)(−1) = 5/6 − 5/6 = 0. The game is fair to the player: neither side has an expected advantage. Let the prize be x euros. Then (1/6)x + (5/6)(−1) = 0.20. Multiply by 6: x − 5 = 1.2, so x = 6.20. A prize of €6.20 on a 6 gives expected gain €0.20 per play. Check: (6.20)/6 − 5/6 = 1.20/6 = 0.20.