Question

(10 pts) Suppose an electronic system comprises five components, and let Xi denote the time in hours until component j fails to operate (j = 1, 2, 3, 4, 5). Suppose that X1, X2, X3, X4 and X5 are i.i.d. random variables, each of which has the following p.d.f. f(x) = ( 2e −2x for x > 0, 0 otherwise. Suppose that the system will operate as long as at least one of the five components operate. Determine the probability that the system operates at least for 4 hours.

Answer 1

Answer:

0.00335

Step-by-step explanation:

Probability that one of the components works for x hours before failing is given by

f(x) = 2e⁻²ˣ

We can then find the probability that one of the components works for x = 4 hours before failing

f(x) = 2e⁻⁸ = 0.0006709253

Probability that the system works for 4 hours = probability that at least 1 of the components of the system works for 4 hours =

1 - (Probability that none of the components of the system works for 4 hours)

Probability that one of the components doesn't work for 4 hours = 1 - 0.000670925 = 0.9993290747

Probability that none of the 5 components work for 4 hours = (0.9993290747)⁵ = 0.9966498721

Probability that the system works for 4 hours = probability that at least 1 of the components of the system works for 4 hours

= 1 - (Probability that none of the components of the system works for 4 hours)

= 1 - 0.9966498721 = 0.0033501279

Hope this Helps!!!