How Do You Calculate the Probability of Valve Failure Within 3 Years?

[abba02]

[SOLVED] Failure probability

1. Homework Statement [/b
The performance of the valves in Q5 has been assessed in more detail under conditions
closer to those experienced in-service and the distribution functions of the random time
to failure have been quantified. The useful life period, prior to wear-out, occurs from
installtion to 5years. During this period, all of the distribution functions are modeled
using an exponential distribution function of the form:
FT (t) = 1 − exp[−_λit] where i=1,2,3,4,5
If _λ1 = λ_2 = _λ3 = 0.05; λ_4 = 0.267; λ_5 = 0.189 (all in years−1), calculate the probability
of a loss of flow from the manifold sometime in the period (0,3)years.
ANSWER[P[F]=0.08643]

Homework Equations

FT (t) = 1 − exp[−_λit] where i=1,2,3,4,5

The Attempt at a Solution

ATTEMPT
Have tried to substitute .05 for lambada and 3 for t in the given equation but my answer is still very different from the given answer of 0.08643

 

[abba02]

The attempt at a solution:

V1,V2 AND V3 ARE IN SERIES AND SS1 ARE IN PARALLEL TO V4 AND V5

(PV1 OR PV2 OR PV3) AND PV4 AND PV5

FT (t) = 1 − exp[−_λit] where i=1,2,3,4,5

If _λ1 = λ_2 = _λ3 = 0.05; λ_4 = 0.267; λ_5 = 0.189 (all in years−1),

For PSS FT (t) =1- exp [-λ1+λ2+λ3*3] where t=3 = .36237

For PV4, FT (t) =1- exp [-λ4*3]=.55112

For PV5,FT (t) =1- exp [-λ5*3]= .43278

Therefore the probability of loss of flow from the manifold at time 3 years is

PSS1 AND PV4 AND PV5= .36237*.55112*.43278= .08643