# Thread: Exponential Random Variable Problem

1. ## Exponential Random Variable Problem

Problem: A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times are exponential with means 1 year, 1.5 years, and 3 years. What is the average length of time the boat can remain at sea?

Some basic facts:

Density for an exponential random variable: f(x)= λe-λx for x>=0
E(T)=1/λ if T is an exponential random variable

Maybe relevant: P(S<T)=λS / (λS + λT) for exponential random variables S and T, and this is similar for many exponential random variables.

Where I've gotten:

The boat can remain at sea until 2 parts break.
Let T be the time that the boat can remain at sea.

Since the sample space can be divided into the order in which the parts fail, we have:

E(T)= Sum where 1<=i,j,k<=3 of: E(T|Ti<Tj<Tk)P(Ti<Tj<Tk)

where E(T|Ti<Tj<Tk)=E(Tj|Ti<Tj<Tk) since the boat can remain at sea until two parts fail.

Now, this would be a similar method that we've used with discrete random variables in class. Unfortunately, I'm a little rusty with my continuous probability. Is there somewhere to go from here, or maybe an easier way to approach the problem?

Designate the navigational devices 1 and 2 and 3, with corresponding times to failure $T_{1}$ and $T_{2}$ and $T_{3}$ respectively. Suppose that devices 1 and 2 are connected in parallell and that the parallell coupling is connected in series to device 3. Then the navigational system is operational if (1 and 2 are operational) or if (2 and 3 are operational).
The system is operational at time $t$ if
$\min(T_{1},T_{3}) > t \text{ or } \min(T_{2},T_{3}) > t \ .$