Worked solution · Exam P

SOA Exam P Sample Question #448 Solution

This Exam P sample reference tests Order Statistics. A payment is at most t exactly when its underlying loss is at most 5+t. One payment therefore has probability (5+t)/10 of being at most t, and independence squares this for the larger payment, giving ((5+t)/10)² and choice D.

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 Order Statistics question

Setup

Let U and V be the two payments obtained by applying the same positive-part threshold to independent losses X and Y.

U=(X-5)_+,\qquad V=(Y-5)_+

Model

For a threshold t in the stated range, translate one payment event back to its loss variable.

U\le t\quad\Longleftrightarrow\quad X\le5+t

\Pr(U\le t)=\frac{5+t}{10}

Compute

The maximum is at most t only when both independent payments are at most t.

\Pr(\max(U,V)\le t)=\Pr(U\le t,V\le t)

\Pr(\max(U,V)\le t)=\left(\frac{5+t}{10}\right)^2

Answer

The required distribution function is the square of the one-payment CDF.

\boxed{\left(\frac{5+t}{10}\right)^2\quad\text{(D)}}

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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 expression (t/5)² treats each payment as continuous Uniform(0,5). A deductible payment also has probability 1/2 at zero, so its CDF is not t/5.
BThe expression (t/10)² ignores the five-unit threshold shift and uses the payment level as though it were the original loss level.
CThe expression (5+t)/10 is the CDF of one payment. The maximum of two independent payments requires both events, so this probability must be squared.
EThis subtracts ((5-t)/10)² from one, reversing the threshold shift and complementing the wrong event. The correct one-payment bound is X≤5+t.

Original practice · fully worked

Original variant: survival until the second component failure

Four independent components have exponentially distributed lifetimes with mean 10 hours. A device shuts down when the second component fails. Calculate the probability that the device is still operating after 5 hours.

A
0.1353
B
0.3935
C
0.4865
D
0.8282
E
0.9760

Answer and calculation

At five hours, one component survives with probability exp(−0.5). The device is operating exactly when zero or one of the four components has failed; summing those two binomial probabilities gives 0.4865, choice C.

Setup

Compute the five-hour survival and failure probabilities for one exponential component.

s=\Pr(T>5)=e^{-5/10}=e^{-0.5}

q=1-s=1-e^{-0.5}

Model

The second failure has not occurred by five hours exactly when the four-component failure count is at most one.

F\sim\operatorname{Binomial}(4,q)

\Pr(\text{device operating at 5})=\Pr(F=0)+\Pr(F=1)

Compute

Evaluate the zero- and one-failure masses.

\Pr(F\le1)=s^4+4qs^3

\Pr(F\le1)=e^{-2}+4(1-e^{-0.5})e^{-1.5}

\Pr(F\le1)=0.4865147909\ldots

Answer

The device remains operating with probability approximately 0.4865.

\boxed{0.4865\quad\text{(C)}}

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How to solve this Order Statistics question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: survival until the second component failure

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables