Question
After a delivery vehicle was jolted, a warehouse finds that exactly eight of its twelve sealed parcels have damaged outer seals; the other four seals are intact. An inspector chooses six parcels uniformly at random from the twelve, opens them, and does not put any chosen parcel back into the selection pool.
Let \(X\) be the number of damaged-seal parcels in the six-parcel sample. What values of \(X\) are actually possible, and what is the exact probability that exactly four selected parcels have damaged seals? Every six-parcel subset is equally likely.
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Hint 1
There are only four intact parcels in the entire delivery. Can a six-parcel sample contain fewer than two damaged seals?
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Solution
The sample contains six parcels in total. Because at most four of the twelve parcels have intact seals, at least \(6-4=2\) of those inspected must have damaged seals. No more than six sampled parcels can be damaged, and the lot has eight such parcels available. Thus the feasible values are
\[ \max(0,6-4)=2\le X\le\min(6,8)=6, \]
with every integer in between achievable.
A six-parcel subset is uniformly selected from the \(\binom{12}{6}=924\) subsets. To get \(X=4\), choose four from the eight damaged-seal parcels and two from the four intact ones. The favourable count is \(\binom84\binom42=70\cdot6=420\). Hence
\[ P(X=4)=\frac{\binom84\binom42}{\binom{12}{6}}=\frac{420}{924}=\boxed{\frac5{11}}. \]
The full favourable subset counts for \(X=2,3,4,5,6\) are respectively \(28,224,420,224,28\), summing to 924 and confirming that the distribution is normalized. This is a hypergeometric count, not six independent Bernoulli trials: removing one parcel changes the lot available for later selections.
Assumptions
There are exactly eight marked parcels and four unmarked parcels in the finite lot.
Six distinct parcels are sampled as a uniformly random subset without replacement.
\(X\) counts damaged outer seals in the sample, not additional hidden damage inside parcels.
Final Answer
The support is \(X\in\{2,3,4,5,6\}\). The probability \(P(X=4)=5/11\approx45.45\%\).