This Exam P sample reference tests Counting Methods. A qualifying four-item subset must have category counts 2, 1, and 1. Counting the three possible doubled categories gives 105 favorable subsets out of 210, so the probability is 1/2 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 3/100 is already smaller than the single valid 2-1-1 case with the first category doubled, whose probability is 60/210=2/7, so it severely undercounts favorable subsets.
BThe value 1/24 is also smaller than one complete valid category-count case and cannot include the three disjoint ways to choose the doubled category.
CA probability of 1/4 would correspond to 52.5 of the 210 equally likely subsets, so it cannot arise from a valid subset count and omits substantial favorable mass.
DThe value 9/25=0.36 is obtained by treating the four selections as independent draws with replacement from category probabilities 0.5, 0.3, and 0.2; the actual sampling is without replacement.
Original practice · fully worked
Original variant: seed-vault viability panel
A seed vault has 5 alpine accessions, 4 coastal accessions, and 3 desert accessions ready for a viability panel. A botanist chooses 5 distinct accessions uniformly at random. Calculate the probability that the panel contains exactly two desert accessions and at least one accession from each of the other two habitats.
A 0.1136
B 0.1515
C 0.1953
D 0.2652
E 0.3182
Variant answer in brief
After choosing two of the three desert accessions, the remaining three positions split between alpine and coastal as 1+2 or 2+1. These cases give 210 of 792 panels, or 0.2652 and choice D.
Setup
Setup
Count the equally likely panels of five distinct accessions.
Nall=(512)=792
Model
Model
Choose two desert accessions. The remaining three selections must consist of one alpine and two coastal accessions or two alpine and one coastal accession.
Ngood=(23)[(15)(24)+(25)(14)]
Compute
Compute
Evaluate the two valid alpine-coastal splits and divide by the total panel count.
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