The School Raffle Probability
A school sells 500 raffle tickets for a fundraiser. Priya buys 8 tickets. There are three prizes: first prize (a tablet), second prize (a gift card), and third prize (a book). Winners are drawn without replacement - each ticket can win at most one prize. Priya's friend claims, 'With 8 tickets out of 500, you have about a 1.6% chance of winning something, so don't count on it.' The organizer clarifies that prizes are drawn one at a time from the remaining tickets.
1. Compute the probability that Priya wins at least one prize, and compare it with the friend's 1.6% claim.
Read the worked answer
P(no prize) = (492/500) × (491/499) × (490/498). Compute: 492/500 = 0.984; 491/499 ≈ 0.98397; 490/498 ≈ 0.98394. Product ≈ 0.984 × 0.98397 × 0.98394 ≈ 0.9524. So P(at least one) ≈ 1 − 0.9524 = 0.0476, about 4.8%. The friend's 1.6% (8/500) is the chance for a single draw only; with three draws her chance is roughly triple, about 4.8%.
2. What is the probability Priya wins exactly one prize?
Read the worked answer
Exactly one win = win one of the three draws and lose the other two. By symmetry, P(win first only) = (8/500)(492/499)(491/498) ≈ 0.016 × 0.986 × 0.986 ≈ 0.01556. Three positions → ≈ 3 × 0.01556 ≈ 0.0467, about 4.7%. Since winning two or three prizes is very unlikely (≈ 0.0009), this is consistent with P(at least one) ≈ 4.8%.
3. The friend says the draws are 'independent, so you just add 8/500 three times.' Identify the error and state the correct reasoning.
Read the worked answer
The draws are not independent because tickets are removed without replacement; the probability changes slightly each draw (8/500, then 8/499 or 7/499 depending on outcome). Adding 8/500 three times gives 24/500 = 4.8%, which happens to be very close to the exact answer (≈4.76%) because the population is large relative to the draws, but the reasoning is flawed: probabilities of overlapping events cannot simply be summed unless they are mutually exclusive. The correct method uses the complement (1 − P(no prize)) or careful case analysis.