15.7 - A Gamma Example

Example 15-4 Section

Engineers designing the next generation of space shuttles plan to include two fuel pumps —one active, the other in reserve. If the primary pump malfunctions, the second is automatically brought on line. Suppose a typical mission is expected to require that fuel be pumped for at most 50 hours. According to the manufacturer's specifications, pumps are expected to fail once every 100 hours. What are the chances that such a fuel pump system would not remain functioning for the full 50 hours?


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

\(f_Y(y)=\dfrac{1}{100^2 \Gamma(2)}e^{-y/100} y^{2-1}=\dfrac{1}{10000}ye^{-y/100} \)

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

\(P(Y<50)=\int^{50}_0 \dfrac{1}{10000}ye^{-y/100} dy\)

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 \(v=-100e^{-y/100} \)

Now, for the integration: