SOA Exam P Sample Question #662. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #662

This is a discrete convolution for two independent copies of the hourly count. Adding the six qualifying ordered-pair probabilities gives 57/400, so the official answer is choice E.

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Discrete Random Variables question

Answer in brief

This is a discrete convolution for two independent copies of the hourly count. Adding the six qualifying ordered-pair probabilities gives 57/400, so the official answer is choice E.

Setup

Let X and Y be the two technicians' counts. They are independent and share the stated probability mass function.

\Pr(X=k)=\Pr(Y=k)=\frac{(1,2,4,8,3,2)_k}{20},\qquad k=0,1,\ldots,5

Model

A total of at least eight is possible only for six ordered pairs. Independence turns each joint probability into a product of marginal masses.

\{(x,y):x+y\ge 8\}=\{(3,5),(4,4),(5,3),(4,5),(5,4),(5,5)\}

\Pr(X=x,Y=y)=\Pr(X=x)\Pr(Y=y)

Compute

Use the distribution weights directly and retain both orders whenever the counts differ.

\Pr(X+Y\ge 8)=\frac{8\cdot2+3\cdot3+2\cdot8+3\cdot2+2\cdot3+2\cdot2}{20^2}

\Pr(X+Y\ge 8)=\frac{57}{400}=0.1425

Answer

The required probability is 57/400.

\boxed{\frac{57}{400}\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis counts only outcomes in which both technicians resolve at least four problems. It omits the qualifying ordered pairs (3,5) and (5,3).
BThis total counts each unequal qualifying pair in only one order. The technicians are distinguishable, so both (x,y) and (y,x) must be retained.
CThis is the probability of exactly eight resolved problems, obtained from (3,5), (4,4), and (5,3), rather than the probability of at least eight.
DThis counts qualifying outcomes having at least one count equal to five, but it leaves out the additional qualifying outcome (4,4).

Original practice · fully worked

Original variant: two-drone parcel sortie

Two rescue drones independently complete a parcel sortie. The number delivered by either drone has probabilities 0.10, 0.10, 0.20, 0.40, and 0.20 for counts 0, 1, 2, 3, and 4, respectively. Calculate the probability that the two drones deliver at least six parcels in total.

A
24%
B
28%
C
32%
D
36%
E
44%

Variant answer in brief

Enumerating the ordered count pairs with total at least six gives probability 0.44. The result selects choice E.

Setup

Let R and S be the parcel counts for the first and second drone. Their common probability weights for counts zero through four are 1, 1, 2, 4, and 2 out of 10.

\Pr(R=k)=\Pr(S=k)=\frac{(1,1,2,4,2)_k}{10},\qquad k=0,1,\ldots,4

Model

The qualifying ordered pairs are those with a sum of six, seven, or eight.

\{(r,s):r+s\ge6\}=\{(2,4),(3,3),(4,2),(3,4),(4,3),(4,4)\}

Compute

Multiply the independent weights for each ordered pair and add.

\Pr(R+S\ge6)=\frac{2\cdot2+4\cdot4+2\cdot2+4\cdot2+2\cdot4+2\cdot2}{10^2}

\Pr(R+S\ge6)=\frac{44}{100}=0.44

Answer

The chance of delivering at least six parcels is 0.44.

\boxed{0.44\quad\text{(E)}}

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ActuaryProof. “SOA Exam P Sample Question #662 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.