Question
A fundraising raffle has 40 printed tickets, exactly ten of which are marked as winners. The tickets are shuffled. Five are sold to a morning group and, without returning any of them to the pile, eight more are sold to an afternoon group. The buyer groups are disjoint, and each sale batch is a random sample of the remaining tickets.
Afterward an organiser checks the records but learns only that four winning tickets were sold across the two groups combined. Nobody recorded how those four winners were divided between morning and afternoon.
Given this information, what is the probability that exactly three of the afternoon group's eight tickets were winners? Explain whether the ten winners in the original forty-ticket stock still matter once the combined total of four is known.
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Hint 1
Condition first on the composition of the 13 sold tickets. With exactly four known winners among them, every choice of four winning positions within those 13 is equally likely.
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Solution
Let \(K\) be the afternoon winning count and \(T\) the total winning count in the 13 sold tickets. Conditional on \(T=4\), the positions of the four winners are exchangeable within those 13 slots: any four-slot arrangement has the same probability. The 8 afternoon positions and 5 morning positions have unequal sizes, so the conditional law is hypergeometric on this sold-ticket population:
\[ P(K=k\mid T=4)=\frac{\binom8k\binom5{4-k}}{\binom{13}{4}} \]
for feasible \(k\in\{0,1,2,3,4\}\). The requested event \(K=3\) puts one winner in the five morning positions. Therefore
\[ P(K=3\mid T=4)=\frac{\binom83\binom51}{\binom{13}{4}} =\frac{56\cdot5}{715}=\boxed{\frac{56}{143}}. \]
To cross-check all possibilities, the favourable position counts for \(K=0,1,2,3,4\) are \(5,80,280,280,70\), which total 715. A further joint-stock check follows by choosing the ten true winning positions among all 40 tickets. For any fixed configuration with \(T=4\) and \(K=k\), there are \(\binom{27}{6}\) ways to assign the other six winning tickets among the unsold 27. This factor is common to all feasible \(k\) and cancels in the conditional ratio. Consequently the original total of ten winners among 40 matters to the probability of observing \(T=4\), but not to the conditional split of those four between the two sold batches.
Assumptions
All forty tickets are exchangeable before the two disjoint, without-replacement sale batches.
Exactly ten tickets in the original population are winners; conditioning on four among the 13 sold has positive probability.
Morning sells five tickets and afternoon eight; only the combined winner count is reported.
Final Answer
\(P(\text{three afternoon winners}\mid\text{four winners among all 13 sold})=56/143\approx39.16\%\).