This Exam P sample reference tests Hypergeometric Distribution. After conditioning, both selections come from the seven non-spade cards, two of which are diamonds. A hypergeometric variance calculation gives 50/147=0.3401, choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AA value near 0.24 results from halving the conditioned diamond proportion to 1/7 because two cards are selected, then using 2(1/7)(6/7)=0.245. The per-draw diamond proportion is 2/7.
BThe value 0.28 is the unconditioned hypergeometric variance 2(2/10)(8/10)(8/9). It leaves the spades in the population despite the conditioning event.
CThe value 0.32 is the unconditioned binomial variance 2(2/10)(8/10). It ignores both the conditioning and sampling without replacement.
EThe value 0.41 is 2(2/7)(5/7), the binomial variance after conditioning. It omits the finite-population correction 5/6.
Original practice · fully worked
Original variant: conditional covariance in a sample without replacement
A container holds four copper tokens, three graphite tokens, and five ceramic tokens. Three tokens are sampled without replacement. Given that exactly one sampled token is graphite, calculate the conditional covariance between the numbers of copper and ceramic tokens sampled.
A -0.4938
B -0.4321
C 0
D 0.4321
E 0.4938
Variant answer in brief
Conditioning leaves two draws from nine non-graphite tokens. The copper count has hypergeometric variance 35/81, and the ceramic count equals two minus the copper count, so their covariance is -35/81=-0.4321, choice B.
Setup
Setup
After fixing one graphite token in the sample, the other two tokens come from the nine non-graphite tokens.
C∼Hypergeometric(N=9,K=4,n=2)
W=2−C
Model
Model
Find the variance of the copper count with the without-replacement correction.
Var(C)=2(94)(95)(87)=8135
Compute
Compute
Use the deterministic relation between the two non-graphite counts.
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