This Exam P sample reference tests Conditional Probability. Conditioning on a total of three affected individuals makes every three-person subset equally likely. Counting subsets with two or three members of the eight-person group gives 112/120=0.933333, so choice E is correct.
How to solve this Conditional Probability question
Setup
Setup
Given that exactly three members of the ten-person population are selected by identical independent event mechanisms, the selected set is uniform over all three-person subsets.
#{possible selected sets}=(310)=120
Model
Model
There are eight members in the distinguished group and two outside it. The favorable set contains either exactly two or exactly three distinguished members.
K∣(T=3)∼Hypergeometric(N=10,K0=8,n=3)
Pr(K≥2∣T=3)=(310)(28)(12)+(38)(02)
Compute
Compute
Count the two disjoint favorable cases and divide by the total number of possible selected sets.
(28)(12)=28(2)=56,(38)(02)=56
Pr(K≥2∣T=3)=12056+56=1514=0.933333…
Answer
Answer
The conditional probability rounds to 0.933.
0.933(E)
Continue without hunting through PDFs
All 718 Exam P sample solutions in syllabus order
The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.384 is the with-replacement binomial probability C(3,2)(0.8)²(0.2) of exactly two distinguished members. It ignores both sampling without replacement and the all-three case.
BThe value 56/120=0.466667 counts exactly two distinguished members under the correct without-replacement model, but it omits the 56 subsets containing three distinguished members.
CCounting all 56 all-distinguished subsets but only C(8,2)=28 of the exactly-two subsets gives (56+28)/120=0.700. The omitted factor C(2,1)=2 chooses which outside member is present.
DTreating the three selected people as independent draws with distinguished probability 0.8 gives C(3,2)(0.8)²(0.2)+(0.8)³=0.896. A fixed finite population requires hypergeometric sampling instead.
Original practice · fully worked
Original variant: three-category laboratory kit
A storeroom contains five calibration cartridges, four assay cartridges, and three control cartridges. A technician chooses four of the twelve cartridges uniformly without replacement to build a kit. Calculate the probability that the kit contains at least one cartridge of every type.
A 0.121212
B 0.181818
C 0.242424
D 0.454545
E 0.545455
Variant answer in brief
A four-cartridge kit representing all three types must have category counts 2,1,1. Summing the three possible doubled categories gives 270 favorable kits out of 495, or 6/11, so choice E is correct.
Setup
Setup
All four-cartridge subsets are equally likely. First count the complete sample space.
#{kits}=(412)=495
Model
Model
With four selections and three represented types, one type appears twice and each other type appears once.
(nC,nA,nR)∈{(2,1,1),(1,2,1),(1,1,2)}
Compute
Compute
Count the kits for each possible doubled type and add the disjoint cases.
(25)(14)(13)=120
(15)(24)(13)=90
(15)(14)(23)=60
Pr(all types)=495120+90+60=116=0.545454…
Answer
Answer
The probability that every cartridge type is represented is approximately 0.545455.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
