Verified solution · Exam P

SOA Exam P Sample Question #486 Solution

This Exam P sample reference tests Independence. Each selected amount has probability 0.80 of staying at or below the threshold. Independence makes the probability that all three stay below equal to 0.80³, so the complementary probability is 0.488 and choice B is correct.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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

Read the solution Open official source

Independent solution

How to solve this Independence question

Setup

Let E_i denote the event that selected amount i crosses the stated threshold. The given one-selection probability is 0.20.

\Pr(E_i)=0.20,\qquad \Pr(E_i^c)=0.80

Model

It is simpler to complement the event that at least one crossing occurs. The complement is that none of the three independent selections crosses.

\Pr\!\left(\bigcap_{i=1}^{3}E_i^c\right)=\prod_{i=1}^{3}\Pr(E_i^c)=(0.80)^3

Compute

Subtract the no-crossing probability from one.

\Pr\!\left(\bigcup_{i=1}^{3}E_i\right)=1-(0.80)^3

1-0.512=0.488

Answer

The probability of at least one threshold crossing is 0.488.

\boxed{0.488\quad\text{(B)}}

Continue without hunting through PDFs

All 718 Exam P sample solutions in syllabus order

The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.

See inside the Exam P study manual PDF

Error diagnosis

Why the other letters tempt

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

AThis signs the all-three term incorrectly when summing exact counts: 3(0.2)(0.8)²+3(0.2)²(0.8)-(0.2)³=0.472. The exactly-three outcome must be added, not subtracted.
CThis is (0.80)³=0.512, the probability that none of the three selections crosses the threshold. It is the complement of the requested event.
DThis is 1-0.472=0.528, the complement of the already mis-signed calculation associated with choice A.
EThis adds the three individual probabilities, 3(0.20)=0.600, without correcting for overlaps in which two or three crossings occur.

Original practice · fully worked

Original variant: infer a defect rate before counting

Independent units from a production line share the same defect probability. The probability that all three units in a calibration group are defect-free is 0.729. For a new group of five units, calculate the probability that exactly one is defective.

A
0.10000
B
0.32805
C
0.40951
D
0.59049
E
0.72900

Variant answer in brief

The calibration probability gives a defect-free probability of the cube root of 0.729, or 0.90. Thus the defect probability is 0.10, and the five-unit exactly-one probability is 5(0.10)(0.90)⁴=0.32805, choice B.

Setup

Let q be the common probability that one unit is defect-free. Independence translates the calibration statement into a cube.

q^3=0.729

Model

Recover the one-unit defect probability, then model the number of defects in the new five-unit group as binomial.

q=0.90,\qquad p=1-q=0.10

D\sim\operatorname{Binomial}(5,0.10)

Compute

Choose the one defective position and require the other four units to be defect-free.

\Pr(D=1)=\binom51(0.10)(0.90)^4

\Pr(D=1)=5(0.10)(0.6561)=0.32805

Answer

The probability of exactly one defect is 0.32805.

\boxed{0.32805\quad\text{(B)}}

Keep practicing

Turn this solution into a complete study sequence

The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.

See the Exam P study manual PDF — $49 More Exam P General Probability practice problems Free Exam P formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #486 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Independence question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: infer a defect rate before counting

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution