Free CAS MAS-I (Modern Actuarial Statistics I) Probability Models Practice Questions

Probability models on CAS Exam MAS-I cover parametric loss distributions, severity and frequency models, mixture distributions, and maximum likelihood estimation applied to property and casualty insurance data (CAS).

179 Questions
99 Easy
37 Medium
43 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
A 3-out-of-50 system is placed in series with a 48-out-of-50 system.

Calculate the number of minimal path sets.
Solution
B is correct.

For a k-out-of-n system, a minimal path set consists of exactly k components. For the series combination of a 3-out-of-50 and 48-out-of-50 system, a minimal path set must contain a path set from each subsystem.

The 3-out-of-50 system has (503)=19,600\binom{50}{3} = 19{,}600 minimal path sets. The 48-out-of-50 system has (5048)=(502)=1,225\binom{50}{48} = \binom{50}{2} = 1{,}225 minimal path sets.

For the series system, each minimal path set is the union of one from each. However, components may overlap. But since these are separate systems with separate components (100 components total), the number of minimal path sets is 19,600×1,225=24,010,00019{,}600 \times 1{,}225 = 24{,}010{,}000.
Question 2 Medium
You are given:
- Constant force of mortality μ=0.04\mu = 0.04
- Force of interest δ=0.06\delta = 0.06

Calculate Aˉx\bar{A}_x, the actuarial present value of a continuous whole life insurance of 1 on life (x).
Solution
D is correct.

With constant force of mortality, the continuous whole life insurance APV is:
Aˉx=∫0∞e−δtμe−μt dt=μμ+δ=0.040.04+0.06=0.040.10=0.40\bar{A}_x = \int_0^{\infty} e^{-\delta t} \mu e^{-\mu t}\,dt = \frac{\mu}{\mu + \delta} = \frac{0.04}{0.04 + 0.06} = \frac{0.04}{0.10} = 0.40
Question 3 Hard
You are given the following information about a parallel system with two components:

- The first component has a lifetime that is uniform on (0,1)(0, 1)
- The second component has a lifetime that is exponential with mean of 2

Determine which of the following is an expression for the expected lifetime of the system.
Solution
D is correct.

For a parallel system, the system lifetime is T=max⁡(T1,T2)T = \max(T_1, T_2).

The expected lifetime is:
E[T]=∫0∞[1−FT(t)] dt=∫0∞[1−F1(t)F2(t)] dtE[T] = \int_0^{\infty} [1 - F_T(t)]\,dt = \int_0^{\infty} [1 - F_1(t)F_2(t)]\,dt
=∫0∞[1−F1(t)F2(t)] dt=∫0∞[S1(t)+S2(t)−S1(t)S2(t)] dt= \int_0^{\infty} [1 - F_1(t)F_2(t)]\,dt = \int_0^{\infty} [S_1(t) + S_2(t) - S_1(t)S_2(t)]\,dt
Here S1(t)=1−tS_1(t) = 1-t for 0≤t<10 \le t < 1 (0 otherwise), and S2(t)=e−t/2S_2(t) = e^{-t/2}.
E[T]=∫01(1−t) dt+∫0∞e−t/2 dt−∫01(1−t)e−t/2 dtE[T] = \int_0^1 (1-t)\,dt + \int_0^{\infty} e^{-t/2}\,dt - \int_0^1 (1-t)e^{-t/2}\,dt
But this matches choice D if the last integral's upper limit were 1. Looking at choice C:
∫01(1−t) dt+∫01t e−t/2 dt+∫1∞e−t/2 dt\int_0^1 (1-t)\,dt + \int_0^1 t\,e^{-t/2}\,dt + \int_1^{\infty} e^{-t/2}\,dt
Rewriting: ∫01(1−t)dt+∫01te−t/2dt+∫1∞e−t/2dt\int_0^1(1-t)dt + \int_0^1 te^{-t/2}dt + \int_1^\infty e^{-t/2}dt. Note ∫01te−t/2dt=∫01e−t/2dt−∫01(1−t)e−t/2dt\int_0^1 te^{-t/2}dt = \int_0^1 e^{-t/2}dt - \int_0^1(1-t)e^{-t/2}dt.

Substituting gives the same result as the inclusion-exclusion formula. Choice D is the correct expression.
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