Question

Lifetime of certain device is exponentially distributed random variable T( λ ). Probability that T > 10 is e-3: P{ T> 10} = e-3 The system consists of 6 devices of such type and is in working condition if all of its components are in working condition. a) Find expectation μx and standard deviation σx for given distribution. b) Find the probability that the system will not fail I) in the next 3 years? II) in the next 8 years? III) in the next 8 years given that it didn’t fail in the first 3years? IV)If you have 6 such system, what is probability that exactly two will not fail in the next 3 years

Answer 1

The lifetime of the devide is exponentially distributed random variable T.

We are given, T ~ Exponential (λ).

The probability density function (p.d.f.) of T is given by,

,

And the cumulative distribution functon (c.d.f.) of T is given by,

,   

Now, we are given,

i.e.   

i.e.

i.e. [using the definition of c.d.f., given t = 10]

i.e.

which implies [since λ > 0]

Therefore,   

a) Since, T ~ Exponential (λ)

The expectation and standard deviation of T are given by,

  

and

Answer: Therefore, the expectation of the given distribution is and standard deviation is .

b) We have a system that consists of 6 of the given devices and is in working condition if all of its components are in working condition.

I) Therefore, the probability that the system will not fail in the next 3 years

= The probability that all it's component devices are in working condition for at least next 3 years

= The probability that the lifetime of each of it's components will be greater than or equal to 3 years

   [where, Ti is the lifetime of the ith devide]

[since the lifetime of one component is independent of another, the probability is the product of all individual probability]

   [all the individual devices are exponentially distributed with same parameter]

  

  

  

  = 0.00452 (approx.)

Answer: Therefore the probability that the system will not fail in the next 3 years is 0.00452 (approx).

II) Again, the probability that the system will not fail in the next 8 years

= The probability that all it's component devices are in working condition for at least next 8 years

= The probability that the lifetime of each of it's components will be greater than or equal to 8 years

   [where, Ti is the lifetime of the ith devide]

[since the lifetime of one component is independent of another, the probability is the product of all individual probability]

   [all the individual devices are exponentially distributed with same parameter]

  

  

  

  = 0.00000 (approx.)

Answer: Therefore the probability that the system will not fail in the next 8 years is 0 (rounded upto 5 decimal places).

III)  The probability that the system will not fail in the next 8 years given that it didn’t fail in the first 3 years

= P (the system will not fail in the next 8 years | it didn’t fail in the first 3 years)

= P (the system will not fail in the next 8 years AND it didn’t fail in the first 3 years)/P (the system did not fail in the first 3 years)

= P (the system will not fail in the next 8 years)/P (the system did not fail in the first 3 years) [since, the intersection of the system will not fail in the next 8 years AND it didn’t fail in the first 3 years is the system will not fail in the next 8 years]

  

= 0 (approx) [since the numerator is 0]

Answer: The probability that the system will not fail in the next 8 years given that it didn’t fail in the first 3 years is 0 (rounded upto 5 decimal places).

IV) Let, X denotes one of the system which will not fail in the next 3 years

Then the probability of success for X is 0.00452 (from a)II) of this problem) the value of X lies betwen 1,2,3,4,5 and 6.

Then we have, X ~ binominal (n, p) where n = 6 and p = 0.00452

The probability mass function of X is given by,

   

Therefore the probability that out of 6 systems exactly 2 will not fail in the next 3 years is

  

= 15 X 0.0000204304 X 0.98204221

= 0.0003 (approx)

  Therefore the probability that out of 6 systems exactly 2 will not fail in the next 3 years is 0.0003 (approx.).