This Exam P sample reference tests Conditional Distributions. Restricting the joint distribution to the stated event gives conditional masses 0.125, 0.625, and 0.250 at values 0, 25, and 50, respectively. The resulting conditional variance is 224.609375 in millions of squared monetary units, which rounds to choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BIgnoring the conditioning event gives the unconditional variance of Y as 243.75 million, which rounds to 244.
CConditioning on the single value X=25,000 instead of the full event X at most 25,000 gives a variance of exactly 250 million.
DInterchanging the table axes and conditioning the wrong variable on the analogous two-level event gives 288.8889 million, which rounds to 289.
EComputing the unconditional variance of X rather than the requested conditional variance of Y gives 306.25 million, which rounds to 306.
Original practice · fully worked
Original variant: submersible diagnostic flag
An underwater drone's emergency repair burden C, in hundreds of dollars, is 0, 20, or 50 with probabilities 0.50, 0.30, and 0.20. A diagnostic flag appears with conditional probabilities 0.20, 2/3, and 1.00 at those respective burden levels. Given that the flag appears, calculate the conditional variance of C.
A 28
B 364
C 376
D 784
E 1160
Variant answer in brief
Bayes weighting gives conditional probabilities 0.20, 0.40, and 0.40 for burden levels 0, 20, and 50 after the flag appears. Their conditional variance is 376, so choice C is correct.
Setup
Setup
Multiply each prior burden probability by its corresponding flag likelihood.
Pr(C=0,F)=0.50(0.20)=0.10
Pr(C=20,F)=0.30(2/3)=0.20
Pr(C=50,F)=0.20(1.00)=0.20
Model
Model
Add the joint flag masses and normalize them to obtain the posterior burden distribution.
Pr(F)=0.10+0.20+0.20=0.50
Pr(C=0∣F)=0.20,Pr(C=20∣F)=0.40,Pr(C=50∣F)=0.40
Compute
Compute
Calculate the first two moments of the posterior distribution and form the variance.
E[C∣F]=20(0.40)+50(0.40)=28
E[C2∣F]=202(0.40)+502(0.40)=1160
Var(C∣F)=1160−282=376
Answer
Answer
The flag-conditioned variance is 376 squared hundreds of dollars.
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