Question

A system is made up of two subsystems, A and B, connected in-series (i.e., the system fails if either subsystem fails). Subsystem A is made up of components 1 and 2, connected in series. Component 1 fails according to a Poisson process with rate 0.2 per month, whereas component 2 fails as a Poisson process with rate 0.08 per month. Subsystem B is made up of a single component 3, which is subject to repeated disruptions. The disruptions arrive as a Poisson. process with rate 1.2 per month but only 20% of them cause the failure of component 3.

a) Find the probability that the system will fail during a 3-month period.
b) Find the probability that there will be no more than two failures of the system during a 6-month period.
c) Find the probability that the next system failure will occur within a year.

Answer 1

Answer: a) The probability that the system will fail during a 3-month period can be calculated as the sum of the probabilities of subsystem A and subsystem B failing, since the system fails if either one fails. The probability of subsystem A failing in 3 months is given by P(A) = 1 - e^(-0.2 * 3) * (1 + 0.2 * 3 + (0.2 * 3)^2 / 2!) = 0.0664. The probability of subsystem B failing in 3 months is given by P(B) = 1 - e^(-1.2 * 3) * (1 + 0.2 * 3 + (0.2 * 3)^2 / 2!) = 0.36. Therefore, the probability that the system will fail in 3 months is P(system) = P(A) + P(B) = 0.0664 + 0.36 = 0.4264.

b) To find the probability that there will be no more than two failures of the system during a 6-month period, we need to use the cumulative distribution function (CDF) of the Poisson distribution. Let X be the number of failures of the system in 6 months. The CDF is given by F(x) = P(X <= x) = 1 - P(X > x), where P(X > x) is the probability that there will be more than x failures. Therefore, the probability that there will be no more than two failures of the system during a 6-month period is given by F(2) = 1 - P(X > 2). We can use the formula for the Poisson distribution to find P(X > 2), which is P(X > 2) = e^(-6 * (P(A) + P(B))) * ((6 * (P(A) + P(B)))^3 / 3! + (6 * (P(A) + P(B)))^2 / 2! + 6 * (P(A) + P(B)) / 1! + 1). Plugging in the values for P(A) and P(B) from part (a), we get F(2) = 1 - e^(-6 * 0.4264) * ((6 * 0.4264)^3 / 3! + (6 * 0.4264)^2 / 2! + 6 * 0.4264 / 1! + 1) = 1 - 0.2079 = 0.7921.

c) To find the probability that the next system failure will occur within a year, we need to find the cumulative distribution function (CDF) of the time between failures, which is the sum of the time between failures of subsystem A and subsystem B. Let X be the time between failures of subsystem A, which follows an exponential distribution with rate 0.2 per month. Let Y be the time between failures of subsystem B, which also follows an exponential distribution with rate 1.2 * 0.2 = 0.24 per month. Then the CDF of the time between failures is given by F(t) = P(X + Y <= t) = 1 - P(X + Y > t0.2 * 12) * e^(-0.24 * 12) = 1 - 0.17 = 0.83. Therefore, the probability that the next system failure will occur within a year is 0.83.

Explanation: