I am trying to solve the following application problem:
There are $n$ components with independent lifetimes which are such that component $i$ functions for an exponential time with rate $\lambda_i$. Suppose that all components are initially in use and remain so until they fail. Find the probability that component 1 is the second component to fail.
I know I need to condition on which component fails first, but I don't know how to set up this condition in order to compute the probability in question.
One approach :
First consider all of the possible orders in which component $1$ can be the second component to fail.
Event $A$: $2$ fails, then $1$, then another component
Event $B$: $3$ fails, then $1$, then another component
Event $C$: ... (continuation)
Event $D$: $n$ fails first, then $1$, then another component
Then computing the probability for each ordering beginning with event $A$, we have \begin{align*} P(A) = P(2 \text{ fails, } \text{then } 1\text{, then another}) &= P(2\text{ fails first})\cdot P(1 \text{ fails before each except 2}) \\ &= \frac{\lambda_{2}}{\sum_{i=1}^{n}\lambda_{i}}\cdot \frac{\lambda_{1}}{\sum_{i=2}^{n}\lambda_{i}} \end{align*}
You can find the probabilities of the other ordering events similarily, and then as they are all mutually exclusive you simply take their sum to acquire your desired probability.
Your result should be the sum $$\frac{\lambda_{2}}{\sum_{i=1}^{n}\lambda_{i}}\cdot \frac{\lambda_{1}}{\sum_{i=2}^{n}\lambda_{i}} + ... + \frac{\lambda_{n}}{\sum_{i=1}^{n}\lambda_{i}}\cdot \frac{\lambda_{1}}{\sum_{i=n}^{n}\lambda_{i}}$$