Verified solution · Exam P
This Exam P sample reference tests Joint Probability Mass Functions. This is a conditional expectation obtained by restricting a joint probability table. Within the at-most-one-filling columns, the masses for zero, one, and two root canals are 0.66, 0.07, and 0.02; dividing their weighted sum 0.11 by total mass 0.75 gives 11/75≈0.15 and choice B.
Independent solution
Setup
Restrict the table to the two filling columns allowed by the conditioning event and sum within each root-canal row.
Model
Add the restricted masses to obtain the conditioning probability.
Compute
Divide the restricted first-moment numerator by the conditioning probability.
Answer
The conditional expected number rounds to 0.15.
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Original practice · fully worked
A paired calibration reading (X,Y) has joint probability density f(x,y)=x+y for 0<x<1 and 0<y<1, and zero elsewhere. A report states that Y is greater than X. Calculate the conditional expected value of Y.
Symmetry gives probability one half for Y>X. Integrating y(x+y) over 0<x<y<1 gives the restricted first moment 3/8, so the conditional mean is (3/8)/(1/2)=3/4 and choice D.
Setup
The density is symmetric in x and y, so the two strict orderings have equal probability.
Model
Integrate Y times the density over the triangular conditioning region.
Compute
Evaluate the restricted moment and divide by the conditioning probability.
Answer
The conditional expected upper coordinate is 0.750.
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ActuaryProof. “SOA Exam P Sample Question #338 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
