The Raffle With Replacement Confusion
A school fundraiser sells 200 raffle tickets. There are 5 prize draws. After each draw, the winning ticket is placed back in the drum before the next draw (with replacement). A student buys 10 tickets. The student calculates the probability of winning at least one prize as 5 × (10/200) = 0.25, reasoning that each draw gives a 10/200 chance and there are 5 draws. A teacher says this is wrong because the draws are independent and the ticket can win more than once. The student insists the answer must be 0.25 because 'you just add the chances.' The committee also considers switching to drawing without replacement, which would change the probabilities.
1. Identify the error in the student's 0.25 calculation and compute the correct probability of winning at least one prize with replacement.
Read the worked answer
The student added probabilities of non-mutually-exclusive events. Winning on draw 1 and winning on draw 2 can both happen, so P(A or B) ≠ P(A) + P(B) in general.
Correct approach: P(at least one win) = 1 − P(no wins on any draw).
With replacement, each draw is independent. P(no win on one draw) = 1 − 10/200 = 190/200 = 0.95.
P(no win in 5 draws) = 0.95^5 ≈ 0.7738.
P(at least one win) = 1 − 0.7738 ≈ 0.2262, about 22.6%.
The student's 0.25 overcounts because it double-counts scenarios where the same ticket wins multiple prizes. The correct value is slightly less than 0.25, which makes sense because the events overlap.
2. If the committee switches to drawing without replacement, would the probability of the student winning at least one prize increase or decrease? Justify without computing the exact value.
Read the worked answer
Without replacement, a winning ticket is removed, so the student's remaining tickets have slightly higher chance per remaining draw, but the student also cannot win the same ticket twice. The net effect on 'at least one win' is an increase.
Reasoning: 'At least one win' with replacement includes cases where the same ticket wins twice, which 'wastes' a win. Without replacement, each win removes a ticket, so the student's tickets are drawn from a shrinking pool. The complement event (no wins) becomes less likely because the number of non-student tickets decreases faster relative to the student's tickets.
Formally, P(no win without replacement) = (190/200)(189/199)(188/198)(187/197)(186/196), which is less than 0.95^5. So P(at least one win) is greater. The probability increases, though only slightly because the pool is large relative to 5 draws.