Question

(12 points) Consider the system comprised of three components as shown below. Suppose The lifetime of Component 1 is exponentially-distributed with parameter 11 = 1/10. • The lifetime of Component 2 is exponentially-distributed with parameter 12 = 1/20. • The lifetime of Component 3 is exponentially-distributed with parameter 13 = 1/15. The system is working if both (A) Component 1 is working, and (B) Component 2 or/and Component 3 is working. Compute the probability that the system is still working at time t = 8.

Answer 1

Solution:

Given data:

let p1 ,p2 and p3 be the probability that they are working at t = 8

then

required answer = p1 * ( 1 - (1 -p2)*(1-p3))

that is A should work and (either 2 or 3 work)

2 or 3 work = 1 - 2 and 3 both does not work

now

for $\lambda$ ,

the component work at time t is

P(X >t) = e^(-$\lambda$*t) as cdf of exponential is 1 - e^(-$\lambda$*t)

p1 = e^(-1/10*8) =0.4493289

p2 = e^(-1/20*8) = 0.670320

p3 = e^(-1/15*8) = 0.586646

hence answer is

  p1 * ( 1 - (1 -p2)*(1-p3))

= 0.4493289 * ( 1 - (1 - 0.670320)*(1 - 0.586646)))

= 0.3880968

please give me thumbup....