ExponentialDistribution
ProductDistribution
Probability
SurvivalFunction
CDF
BooleanCountingFunction
Binomial

Probability & Statistics
Mathematics

Expected Lifetime of a Multi-Component System

Use Boolean logic to represent and evaluate conditions involving specific numbers of failures

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A system consists of four identical components operating independently, each with an exponentially distributed lifetime. Determine the probability that the system experiences no failures within a specified time period, and the probability that exactly one component fails within a given interval.

Each component lifetime follows an exponential distribution with failure rate 𝜆 =

-3

per hour:

In[1]:=

dist=ExponentialDistribution1

1000

;

All four components must survive beyond 500 hours:

In[2]:=

Probabilityt1>

500

&&t2>

500

&&t3>

500

&&t4>

500

,{t1,t2,t3,t4}ProductDistribution[{dist,4}]

Out[2]=



We can also obtain the same result using the SurvivalFunction:

In[3]:=

SurvivalFunctiondist,

500

^4

Out[3]=



Use two different methods to compute the probability that exactly one component will fail in the first 1200 hours.

Compute the probability that the first component will fail times the number of ways of choosing the "first" component:

In[4]:=

prob1=Binomial[4,1]Probabilityt1≤

1200

&&t2>

1200

&&t3>

1200

&&t4>

1200

,{t1,t2,t3,t4}ProductDistribution[{dist,4}]

Out[4]=

4(-1+

6/5



)

24/5



Alternatively, by using

BooleanCountingFunction

we can define the logical condition that exactly

events are

True

In[5]:=

ExactlyK[k_,v_]:=BooleanCountingFunction[{k},Length[v]]@@v

Find the logical expression for the case of exactly one

True

In[6]:=

ExactlyK[1,{a,b,c,d}]//BooleanConvert

Out[6]=

(a&&!b&&!c&&!d)||(!a&&b&&!c&&!d)||(!a&&!b&&c&&!d)||(!a&&!b&&!c&&d)

Compute the probability that exactly one of the four exponentially distributed component lifetimes is ≤ 1200 hours (i.e., exactly one failure within 1200 hours):

In[7]:=

prob2=ProbabilityExactlyK1,t1≤

1200

,t2≤

1200

,t3≤

1200

,t4≤

1200

,{t1,t2,t3,t4}ProductDistribution[{dist,4}]

Out[7]=

4(-1+

6/5



)

24/5



In[8]:=

N[%]

Out[8]=

0.0763759

Observe that the two methods give equivalent results:

In[9]:=

prob1prob2

Out[9]=

True

Related Symbols

ExponentialDistribution
ProductDistribution
Probability
SurvivalFunction
CDF
BooleanCountingFunction
Binomial

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Contributed by: Wolfram Staff

Wolfram Language Example Repository

Expected Lifetime of a Multi-Component System

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