Worked solution · Exam P

SOA Exam P Sample Question #457 Solution

This Exam P sample reference tests Mixture Distributions. This is a mixture probability over four latent count states. Weighting each conditional zero-visit probability gives 0.389948, so the probability of one or more visits is 0.610052 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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 Mixture Distributions question

Setup

Let X be the latent illness count and V the resulting visit count. Work through the complement event V=0.

\Pr(V\ge1)=1-\Pr(V=0)

Model

The zero state produces no visits with certainty; each positive state produces a Poisson zero probability exp(-k).

\Pr(V=0)=0.28+0.12e^{-1}+0.42e^{-2}+0.18e^{-3}

Compute

Evaluate the weighted mixture and take its complement.

\Pr(V=0)=0.3899480242

\Pr(V\ge1)=1-0.3899480242=0.6100519758

Answer

The probability of at least one visit rounds to 0.61.

\boxed{0.61\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThe value 0.18 is only the probability of the highest latent illness state and does not aggregate visits across the other states.
BThe value 0.39 is the probability of zero visits, 0.389948 rounded, so it is the complement of the requested event.
DThe value 0.72 is Pr(X>0). It assumes every positive illness count guarantees a visit, ignoring the positive Poisson probability of zero.
EOmitting the structural zero-state term gives 1-[0.12 exp(−1)+0.42 exp(−2)+0.18 exp(−3)]=0.890052, near 0.89.

Original practice · fully worked

Original variant: passes from a random scanner panel

A control panel activates 1, 2, or 4 scanners with probabilities 0.25, 0.50, and 0.25, respectively. Conditional on the number activated, scanners pass a diagnostic independently with probability 0.80. Calculate the probability that at least two scanners pass.

A
0.2432
B
0.3200
C
0.5632
D
0.6400
E
0.9728

Answer and calculation

The one-scanner state contributes zero. The two- and four-scanner states contribute 0.50(0.64)=0.32 and 0.25(0.9728)=0.2432, totaling 0.5632 and choice C.

Setup

Let N be the active-scanner count and K the number that pass. Conditional on N, K is binomial.

K\mid N=n\sim\operatorname{Binomial}(n,0.80)

Model

Evaluate the target under each possible panel size.

\Pr(K\ge2\mid N=1)=0

\Pr(K\ge2\mid N=2)=(0.80)^2=0.64

\Pr(K\ge2\mid N=4)=1-(0.20)^4-4(0.80)(0.20)^3=0.9728

Compute

Weight the three conditional probabilities by the panel-size probabilities.

\Pr(K\ge2)=0.25(0)+0.50(0.64)+0.25(0.9728)

\Pr(K\ge2)=0.3200+0.2432=0.5632

Answer

The probability that at least two scanners pass is 0.5632.

\boxed{0.5632\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Mixture Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: passes from a random scanner panel

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables