This Exam P sample reference tests Order Statistics. This is the CDF of a sample maximum from three independent uniform observations. All three observations must fall below the threshold, giving 0.7 cubed or 0.343 and selecting choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.027 is 0.3 cubed, the probability that all three observations exceed seven.
BThe value 0.081 is three times 0.3 cubed. Multiplying the all-above event by three double-counts a single outcome pattern.
CThe value 0.189 is 3(0.7)(0.3)², the probability that exactly one of the three observations is below seven.
EThe value 0.441 is 3(0.7)²(0.3), the probability that exactly two observations are below seven.
Original practice · fully worked
Original variant: median priority in a sample without replacement
Seven maintenance tickets carry distinct priority scores 1 through 7. Three tickets are selected uniformly without replacement, and M is the median of their three scores. Calculate Pr(M≥5).
A 1/35
B 12/35
C 13/35
D 135/343
E 22/35
Variant answer in brief
The sample median is at least five exactly when at least two selected tickets come from scores 5, 6, and 7. Hypergeometric counting gives (12+1)/35=13/35, so choice C.
Setup
Setup
Partition the tickets into three high scores, 5 through 7, and four lower scores. Every three-ticket subset is equally likely.
#{all samples}=(37)=35
Model
Model
For the middle selected score to be at least five, the sample must contain exactly two or exactly three high-score tickets.
#{exactly two high}=(23)(14)=12
#{exactly three high}=(33)(04)=1
Compute
Compute
Add the two disjoint favorable counts and divide by all samples.
Pr(M≥5)=3512+1=3513=0.371428…
Answer
Answer
The probability that the sample median is at least five is 13/35.
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