Two fuel pumps on a space shuttle one is active and one rese

Two fuel pumps on a space shuttle, one is active and one reserved. If the active fails then the reserve automatically starts. If we have a typical mission, the fuel is pumped at most 50 hours Pumps are expected to fail once every 100 hours. What are the chances/probability the pump system will not remain functioning for the full 50 hours?

? = 100 ?= 2

Write P(Y<50) as an integral and use integration by parts to solve.

Solution

. We are given that ?, the average number of failures in a 100-hour interval is 1. Therefore, ?, the mean waiting time until the first failure is 1/?, or 100 hours. Knowing that, let\'s now let Y denote the time elapsed until the ? = 2nd pump breaks down. Assuming the failures follow a Poisson process, the probability density function of Y is:

fY(y)=11002?(2)e?y/100y2?1=110000ye?y/100

for y > 0. Therefore, the probability that the system fails to last for 50 hours is:

P(Y<50)=?500110000ye?y/100dy

Integrating that baby is going to require integration by parts. Let\'s let:

u=y  and dv=e?y/100

So that:

du=dy and v=?100e?y/100