Find the distributions of $T$ and $N$. 
A flashlight needs $2$ batteries to be operational. You start with $4$ batteries numbered $1,2,3,4$. Whenever a battery fails, it is replaces by the lowest numbered working battery. Let each battery life be $Exponential(0.01)$. Let $T$ be the time at which there is only one working battery left and let $N$ be the number of the one battery that is still good. Then find the distributions of $T$ and $N$.

I am not getting head or tail of this problem. It is not specified which batteries are the initial ones.
 A: (a) By the memorylessness of the Exponential, when each replacement occurs, the next two working batteries each have probability $1/2$ of outlasting the other. Battery $1$ must outlast $3$ others. In all other cases, Battery $i$ must outlast $5-i$ others. Therefore,
$$P(N=i) = \begin{cases}
\left(\dfrac{1}{2}\right)^{3} & \text{if $i=1$} \\
\left(\dfrac{1}{2}\right)^{5-i} & \text{if $i\geq 2$.}
\end{cases}$$
(b) Let $X_i,\; i=1,2,3,4\;$ be the lifetime of Battery $i$, and let $Y_i,\; i=1,2,3\;$ be the time between the $(i-1)^{th}$ and $i^{th}$ replacement.
So $Y_1=\min\{X_1,X_2\}$ and $Y_1,Y_2,Y_3$ are iid due to memorylessness and the $X_i$s being iid. The cdf of $Y_1$ is
\begin{align}
F_Y(y) &= P(\min\{X_1,X_2\}\lt Y) \\
&= 1-P(\min\{X_1,X_2\}\gt y) \\
&= 1-P(X_1\gt y)P(X_2\gt y) \\
&= 1-(1-F_X(y))^2 \\
&= 1-(e^{-0.01y})^2 \\
&= 1-e^{-0.02y}.
\end{align}
So $T\sim Exp(0.02)$. Now, $\;T=Y_1+Y_2+Y_3\;$ so $T\sim Gamma(0.02,3)$ since the sum of $k$ iid $Exp(\lambda)$ r.v.s has distribution $Gamma(\lambda,k)$.
