This Exam P sample reference tests Marginal Distributions. Summing the joint mass over the second coordinate gives marginal probabilities 0.1, 0.3, and 0.6 at the positive values of X. The resulting first and second moments are 2.5 and 6.7, so the variance is 0.45 and choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThis is the raw second moment 6.70 with the decimal point shifted; a variance must subtract the squared mean.
CThis candidate fails the direct mean-centered sum and is too large once the mass concentrated near X = 3 is centered at 2.5.
DThis remains on the scale of the raw second moment because it subtracts only a small correction instead of the full squared mean 6.25.
EThis is E[X²] = 6.70, not Var(X).
Original practice · fully worked
Original variant: inspection tags and manual reviews
A quality team records the number A of priority tags on a production lot and whether a manual review B is requested. Their joint probabilities are P(A=0,B=0)=0.10, P(A=0,B=1)=0.05, P(A=1,B=0)=0.20, P(A=1,B=1)=0.25, P(A=2,B=0)=0.15, and P(A=2,B=1)=0.25. Calculate the variance of A.
A 0.4000
B 0.4500
C 0.4875
D 1.2500
E 2.0500
Variant answer in brief
Marginalizing over the review indicator gives probabilities 0.15, 0.45, and 0.40 for A = 0, 1, and 2. The first two raw moments are 1.25 and 2.05, so the variance is 0.4875 and choice C.
Setup
Setup
Add the two joint cells in each row to obtain the distribution of the tag count.
Pr(A=0)=0.10+0.05=0.15
Pr(A=1)=0.20+0.25=0.45
Pr(A=2)=0.15+0.25=0.40
Model
Model
Use the first and second raw moments of this three-point marginal distribution.
Var(A)=E[A2]−E[A]2
Compute
Compute
Evaluate the moments and subtract the squared mean.
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