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
Answer and calculation
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.
{M>0.75}={K≤2} Compute
Compute
Add the three eligible binomial probabilities.
Pr(K≤2)=k=0∑2(k4)(0.75)k(0.25)4−k Pr(K≤2)=0.00390625+0.046875+0.2109375 Pr(K≤2)=0.26171875 Answer
Answer
The requested probability rounds to 0.2617.
0.2617(D)