What is my lambda here The life of automobile voltage regulators has an exponential distribution with a mean life of six years. You purchase a six-year-old automobile, with a working voltage regulator and plan to own it for six years.
What is the probability that the voltage regulator fails during your ownership?
Is my $\lambda = 2$ because after 12 total years you expect 2 failures? That's what my gut tells me, as setting the $\lambda=1$ would suggest that the next 6 years during ownership are as statistically healthy as a brand new regulator. It's the fact that you're looking at only the last 6 years that's making me question myself.
 A: The probability that the lifetime $T$ of a voltage regulator with mean lifetime $\mu$ exceeds six years from the date of manufacture ($t=0$)  is
$$\Pr(T>6) = \exp{\frac{-6}{\mu}}$$
The probability that its lifetime exceeds twelve years given that it's already lasted six years is
$$\Pr(T>12 \mid T>6)$$
Re-write the conditional probability as the ratio of a joint to a  marginal probability, & note that the joint probability simplifies (if a lifetime's over twelve years it's necessarily over six):
$$\Pr(T>12 \mid T>6) = \frac{\Pr(T>12, T>6)}{\Pr(T>6)}\\
=\frac{\Pr(T>12)}{\Pr(T>6)}$$
Compare the answer you get to this with the first probability you calculated, then read up on the memoryless property of the exponential distribution. Does it seem like a reasonable assumption for the distribution of voltage regulator lifetimes?
A: The exponential distribution has a mean that is somewhat "before" the 50%-tile of its associated survival curve. Because of the memoryless property, the probability of survival to 12 years is  exactly one-half the probability of survival to 6 years squared. The probability of failure to any time is 1-Pr_survival.
If you are only talking about one item then the number of expected failures cannot exceed 1, unless of course, you replace it with an identical item. The rate, lambda is not the inverse of the expected number of failures but it is the inverse of the mean lifetime. So lambda is 1/6 and :
> exp(-1)  # the expected survival of one unit at 6 years: exp(-t/lambda)
[1] 0.3678794

If these are "dieing" at this rate, then the number of needed regulators to time T is distributed as a Poisson variate.
