This Exam P sample reference tests Beta Distribution. The beta normalizing constant is 182. Integrating its polynomial density from 0.9 to 1 gives 0.415371, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is essentially half the correct tail probability and results from using k=91 rather than the normalized value 182.
BThis is close to 0.9¹⁴=0.2288, the probability that all 14 Bernoulli indicators fall below the threshold in the order-statistic check; that single configuration is not the requested beta tail.
DThis rounds 0.9¹³ and 0.9¹⁴ to 0.25 and 0.23 before subtracting nearly equal terms. The resulting 1-14(0.25)+13(0.23)=0.49 is nearest to 0.479; the powers must retain full precision.
EThis substitutes the beta mean 13/15 for the stated cutoff. The tail above 13/15 is 0.57463, which rounds to 0.575, but the required cutoff is 0.9.
Original practice · fully worked
Original variant: archived quality-score order statistic
Four independent quality scores are each uniformly distributed between 0 and 1. The third smallest score is archived as M. Calculate the probability that M exceeds 0.75.
A 0.0039
B 0.0508
C 0.2109
D 0.2617
E 0.7383
Variant answer in brief
The third smallest score exceeds 0.75 exactly when at most two of the four scores are at or below 0.75. The binomial sum is 0.26171875, so choice D is correct.
Setup
Setup
Count how many of the four scores do not exceed the threshold.
K=#{i:Ui≤0.75}∼Binomial(4,0.75)
Model
Model
For the third ordered value to be above 0.75, no more than two observations may lie at or below 0.75.
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