This Exam P sample reference tests Joint Distributions. Normalization restricts the free probability to a nonnegative interval. The symmetric marginal has variance 2/5+2a, so its minimum occurs at a=0; the other marginal then has the same variance 2/5, giving choice A.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BSetting the unknown entry equal to the fixed corner value 1/15 gives a=1/15 and Var(Y)=2/5+2/15=8/15. The free entry is not determined by that corner.
CImposing the unsupported condition a=b on 4a+b=3/5 gives a=3/25 and Var(Y)=2/5+6/25=16/25.
DSetting a equal to the other fixed off-center value 2/15 gives Var(Y)=2/5+4/15=2/3. That equality is not a table constraint.
EChoosing the largest feasible a=3/20, where b=0, gives Var(Y)=2/5+3/10=7/10. This maximizes the increasing variance rather than minimizing it.
Original practice · fully worked
Original variant: largest variance in a calibrated score family
A diagnostic score R takes values 0, 2, and 5 with probabilities q, 1/2, and 1/2-q, respectively, where 0 is at most q and q is at most 1/2. Calculate the largest possible value of Var(R).
A 2.25
B 2.50
C 3.25
D 7.00
E 9.50
Variant answer in brief
The variance is 9/4+10q-25q squared, a concave quadratic. Completing the square shows its maximum is 13/4=3.25 at q=1/5, so choice C.
Setup
Setup
Calculate the first two raw moments as functions of the probability parameter.
E[R]=2(21)+5(21−q)=27−5q
E[R2]=4(21)+25(21−q)=229−25q
Model
Model
Subtract the square of the mean from the second moment.
Var(R)=229−25q−(27−5q)2
=49+10q−25q2
Compute
Compute
Complete the square to expose the feasible maximizer.
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