One of four generators supplies electricity to a remote research station in the Arctic. When the first generator breaks down, the resident research team has to start another of the remaining three generators. After two generators break down, the team has to call their homebase to place an order for replacement generators to be sent.
There are two brands of generators at the research station. The lifetime of brand A is known to follow an exponential distribution with mean 5 months, while the lifetime of brand B is known to follow an exponential distribution with mean 4 months. Unfortunately, the brand labels have been removed from the generators, so the research team only knows that they have 2 generators of brand A and 2 generators of brand B.
The team switched on a generator at random when they arrived at the research station, and will switch on a second generator at random when the first generator breaks down.
Assuming that the lifetimes of the individual generators are independent of each other.
(b) Find the expected time until the research team has to call their homebase to order new generators.
(c) Find the variance of the time until the research team has to call their homebase to order new generators.
No idea how to approach (b) and (c).
Is (b) $E_x[\widetilde {\mathrm{Var}} (Y \mid X)]$ where $Y$ is the time and $X$ is both generators being $A$, both being $B$ and being different? If so, how is the variance calculated when the the two generators are different?
Clearly, it doesn't matter which order the generators are chosen (e.g. if A is chosen first and B second, the random total lifetime has the same distribution as if B were chosen first and A second). In fact, we can model the random total lifetime simply by counting the number of Brand A generators chosen: call this variable $X$, hence $$X \sim \operatorname{Hypergeometric}(N = 4, n = 2, m = 2), \\ \Pr[X = x] = \frac{\binom{2}{x}\binom{2}{2-x}}{\binom{4}{2}} = \frac{1}{6}\binom{2}{x}^2, \quad x \in \{0,1, 2\}.$$
Then the total lifetime random variable given $X$ is $$T \mid X = X T_A + (2-X)T_B,$$ where $T_A \sim \operatorname{Exponential}(1/5)$ and $T_B \sim \operatorname{Exponential}(1/4)$. Now that the notation is set up, all you need to do is calculate
$$\operatorname{E}[T] = \operatorname{E}[\operatorname{E}[T \mid X]]$$
$$\operatorname{Var}[T] = \operatorname{Var}[\operatorname{E}[T \mid X]] + \operatorname{E}[\operatorname{Var}[T \mid X]].$$