The Raffle Totals
Two Year 10 classes sell raffle tickets for a school trip. Class 10A has 30 students; their sales have a mean of 12 tickets and a median of 10. Class 10B has 28 students; their sales have a mean of 15 tickets and a median of 16.
One student in 10B sold 90 tickets to a family business. No other student in either class sold more than 30. In total, 780 tickets were sold, and three prizes will be drawn at random from all the tickets sold, without replacement.
At assembly the principal announces that '10B clearly worked harder than 10A'. A parent then complains that 10B students will win most of the prizes 'because they sold more tickets'.
1. Verify the total of 780 tickets and state which class sold more, and by how much.
Read the worked answer
Total = (mean × number of students) for each class.
- 10A: 12 × 30 = 360 tickets
- 10B: 15 × 28 = 420 tickets
- Total: 360 + 420 = 780 ✓ (matches the figure given, so the means are consistent)
10B sold more, by 420 − 360 = 60 tickets.
Note that the class sizes differ, so totals alone do not tell us about effort per student - that is what the mean and median describe.
2. Evaluate the parent's and the principal's claims using the mean, the median and the effect of the 90-ticket sale.
Read the worked answer
Effect of the 90-ticket sale: Remove it from 10B: 420 − 90 = 330 tickets over 27 students → 330 ÷ 27 ≈ 12.2 tickets per student. So without that single sale, 10B's total (330) would be below 10A's (360), and its mean would drop from 15 to about 12.2.
But the median is 16 for 10B and 10 for 10A. The median ignores the size of the biggest sale, so the typical 10B student still sold about 6 more tickets than the typical 10A student.
Judgement: the principal is partly wrong - the gap in totals and in the mean is created almost entirely by one exceptional sale, so '10B worked harder' is not supported by the mean. However, the median shows 10B really was stronger across the class.
The parent's prize claim is different: prizes are drawn from tickets, not from students, so a class that holds more tickets does have a higher chance of winning - that is fair, not a flaw.
3. Find the probability that the first prize goes to a 10A student, and the probability that at least one of the three prizes goes to a 10A student.
Read the worked answer
First prize: 10A holds 360 of the 780 tickets. P = 360/780 = 6/13 ≈ 0.462 (46.2%)
At least one of three prizes: it is easier to find the complement - no 10A ticket wins any prize.
- P(1st not 10A) = 420/780
- After one 10B ticket is removed: P(2nd not 10A) = 419/779
- P(3rd not 10A) = 418/778
P(none from 10A) = (420/780) × (419/779) × (418/778) ≈ 0.5385 × 0.5379 × 0.5373 ≈ 0.1556
P(at least one from 10A) = 1 − 0.1556 ≈ 0.844 (about 84%)
The denominators shrink because the draw is without replacement. A quick sanity check: with 46% of tickets, 10A failing to win any of three prizes should indeed be uncommon, but not negligible.