Worked 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 July 31, 2026

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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)}}

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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

Answer and calculation

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)}}

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How to solve this Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: infer a defect rate before counting

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability