MAS-I · Probability Models · Free Lesson

Calculate system reliability for series, parallel, and bridge configurations of independent components, and analyze Markov chain transition matrices.

Free CAS MAS-I (Modern Actuarial Statistics I) lesson in Probability Models. 20 min read, ~2,926 words.

A reinsurer rates a satellite with five redundant subsystems and needs the probability the whole craft survives a year. Same math, smaller stakes: a Markov model for policy lapse asks what fraction of a cohort stays in force after ten renewals. Reliability and Markov chains share one engine, the multiplication of independent probabilities arranged into a structure.

Series systems. Components are in series when every one must work for the system to work. With independent components and individual reliabilities RiR_i,

Rs=∏i=1nRiR_s = \prod_{i=1}^{n} R_i

Ten components each at 0.99 give 0.9910=0.90440.99^{10} = 0.9044. Adding cheap parts to a series chain is expensive in reliability.

KEY: In series, multiply reliabilities. In parallel, multiply unreliabilities. The two operations are duals.

Parallel systems. Components are in parallel when the system works if at least one component works. With independent unreliabilities qi=1−Riq_i = 1 - R_i,

Rp=1−∏i=1n(1−Ri)R_p = 1 - \prod_{i=1}^{n}(1 - R_i)

Three components each at 0.80 give 1−0.23=0.9921 - 0.2^3 = 0.992. Redundancy buys reliability cheaply.

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For a quick check on a series chain, approximate ∏Ri≈1−∑(1−Ri)\prod R_i \approx 1 - \sum (1 - R_i) when the sum of component unreliabilities ∑i(1−Ri)\sum_i (1 - R_i) is much smaller than 1, so products of unreliabilities are negligible. For nn identical components with unreliability qq, use nq≪1nq \ll 1; small individual unreliabilities alone are insufficient.

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