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

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 $\binom{50}{3} = 19{,}600$ minimal path sets. The 48-out-of-50 system has $\binom{50}{48} = \binom{50}{2} = 1{,}225$ 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 $19{,}600 \times 1{,}225 = 24{,}010{,}000$.
Choice A is incorrect because 24,010,000 is not fewer than 20,000.
Choice B is incorrect because 24,010,000 is not fewer than 100,000.
Choice C is incorrect because 24,010,000 is not fewer than 2,000,000.
Choice D is incorrect because 24,010,000 is not fewer than 20,000,000.

A memoryless distribution is exponential. by the memoryless property, given that a loss exceeds the deductible of 500, the expected payment is the mean of the distribution, which is 1,000. So the average insurance payment is 1,000.
Choice A is incorrect because it understates the payment; the memoryless property means the expected excess is the full mean.
Choice B is incorrect because 500-700 is too low; the memoryless property gives exactly 1,000.
Choice C is incorrect because 700-900 is too low; the expected excess loss equals the mean.
Choice E is incorrect because the expected payment is exactly 1,000, not 1,100 or more.

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

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

Substituting gives the same result as the inclusion-exclusion formula. Choice ((C)) is the correct expression.


Choice B is incorrect because it uses an improper integrand $(1-t)$ beyond $t=1$.
Choice
(A) is incorrect because it represents $E[\min(T_1,T_2)]$ rather than $E[\max(T_1,T_2)]$.


Choice D is incorrect because the last integral should have upper limit 1, not $\infty$.


Choice C is incorrect because the last integral extends $(1-t)$ beyond its support $[0,1)$.