Here is a question for which I am not able to figure out the approach to solving it.
Problem statement:
Suppose that $n$ light bulbs in a room are switched on at the same instant. The life time of each bulb is exponentially distributed with parameter $\mu = 1$, and are independent.
Suppose you walk into the room and find $m$ bulbs glowing. Starting from the instant of your walking in, what is the distribution of the time it takes until you see a bulb blow out?
My approach: If the lifetime of the bulbs is taken to have the CDF $1-e^{-t}$, and if the random variables are denoted by $X_i$ where $i$ ranges from $1$ to $n$, then we need to first find the distribution of $X_i>t$ conditioned on the fact that at time $T$, I walked in and found $m$ bulbs still glowing. Call the event that at time $T$, $m$ bulbs were glowing, as event $B$.
If the above assumption is right, then all we have to do is find the distribution of the minimum of the random variables $X_i \mid B~$ i.e the minimum of the random variables $X_i$, each of which is conditioned on $B$.
Is this approach correct? I got a weird answer through this approach. If $Y_i = X_i \mid B~$, then CDF of $\min(Y_i)$ turned out to be $$\left \{ \frac{(e^{-t})(n-m)}{n(1-e^{-T})} \right \}^n$$ Checking for the limit $T \to 0$ gives an answer that does not make sense but nevertheless, I would greatly appreciate it if you can please provide some feedback.
You mention "amnesia" in the title of the post. This is a hint to use the memorylessness property of the exponential distribution.
Solution sketch: Conditioned on the event $B$, the distribution of the remaining life time (after you walk in) of each bulb that is still lit when you walk in is also exponentially distributed with parameter $1$. Thus, the "distribution of the time it takes until you see a bulb blow out" is the minimum of $m$ independent Exponential($1$) random variables.