4. On the average, a computer experiences breakdowns every 5 months. The time until the first breakdown and the times between any two consecutive breakdowns are independent Exponential random variables. After the third breakdown, a computer requires a special maintenance. (a) Compute the probability that a special maintenance is required within the next 9 months. (b) Given that a special maintenance was not required during the first 12 months, what is the probability that it will not be required within the next 4 months

a) P=0.16

b) Accumulated breakdowns: 0 --> P(no special maintenance) = 0.962

Accumulated breakdowns: 1 --> P(no special maintenance) = 0.856

Accumulated breakdowns: 2 --> P(no special maintenance) = 0.641

Step-by-step explanation:

As the time between breakdowns is independent and exponential random variables, we can model the amount of breakdowns per unit of time as a Poisson process.

The rate of breakdowns is

a) For a special maintenance, three consecutive breakdowns have to happen within the next five months, so we have:

The probability of requiring a special maintenance within the next 9 months is P=0.16.

b) The probability of requiring a special maintenance only depends on the amounts of breakdowns since the last special maintenance. It does not depend on the amount of time without special maintenance.

In this case, we don't know if the computer has had any breakdown since the last special maintenance.

In the case the amount of breakdowns is 0 since the last special maintenance, we have to have 3 breakdowns to have a special maintenance in the next 4 months:

In the case the amount of breakdowns is 1 since the last special maintenance, we have to have 2 breakdowns to have a special maintenance in the next 4 months:

In the case the amount of breakdowns is 1 since the last special maintenance, we have to have 2 breakdowns to have a special maintenance in the next 4 months:

8

step-by-step explanation:

yes.. it's proven to be 36.