This Exam P sample reference tests Conditional Probability. Conditioning on exactly three affected employees makes every three-person subset equally likely. Only 20 of the 120 subsets lie entirely in department A, so the complement is 5/6=0.833333 and choice E is correct.
How to solve this Conditional Probability question
Setup
Setup
All employees have the same independent event probability. Once the total count is fixed at three, the common Bernoulli factors cancel across candidate subsets.
Pr(S∣∣S∣=3)=(310)1
Model
Model
Use the complement: no department-B employee is included precisely when all three affected employees come from the six-person department A.
Pr(none from B∣∣S∣=3)=(310)(36)
Compute
Compute
Complement the all-A subset probability.
Pr(at least one from B∣∣S∣=3)=1−12020
Pr(target)=65=0.8333333333
Answer
Answer
The conditional probability rounds to 0.833.
0.833(E)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is approximately Pr(exactly three events)=C(10,3)(0.2)³(0.8)⁷=0.2013. It repeats the event already given instead of conditioning on it.
BThis assigns an unsupported one-half split between departments. The relevant three-person subsets are not divided evenly by whether they include department B.
CThis is 1-(0.8)⁴=0.5904, the unconditional probability that at least one of four department-B employees has an event; it ignores the fixed total of three.
DThis is 1-(6/10)³=0.784. It samples department labels with replacement, allowing the same employee to be selected repeatedly.
Original practice · fully worked
Original variant: sculpture-heavy inspection
A museum shipment contains eight framed prints and six sculptures. An inspector selects five of the fourteen objects uniformly without replacement. Calculate the probability that the inspection contains at least three sculptures.
A 0.0030
B 0.0599
C 0.2797
D 0.3427
E 0.6573
Variant answer in brief
The favorable samples contain three, four, or five sculptures. Their 686 combinations out of 2002 five-object samples give 0.342657, so choice D is correct.
Setup
Setup
All five-object samples are equally likely.
Nall=(514)=2002
Model
Model
Sum the hypergeometric counts for three, four, and five sculptures.
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.
