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We have a flashlight that uses two batteries and we have a package of
four new batteries. Suppose that each battery has an independent
exponential($lambda$) lifetime. Compute the distribution of the
number of hours $T$ that we can use the flashlight.

Try

Call $X$ the lifetime of each battery which given is that $X$ is Exp($lambda$). Let $T$ be number of hours we can use the flashlight.

LEt $Y$ be the lifetime of two batteries taking as single unit. So $(T leq t) = (Y leq t$). Thus ,

$$ P(Y leq t) = boxed{1 - e^{-lambda^2 t } }$$

is this the correct distribution?

To get the most out of the batteries, we'll put two batteries in the flashlight. Once one of them expires, we'll replace it (and only it) with the third. And when another battery expires (be it the second or the third), we'll replace it with the fourth.

Let $X sim Exp(lambda)$ and $Y sim Exp(lambda)$ be the expiry times of the initial two batteries. Let's consider the time one of them runs out and we replace it, and and let $Delta$ be the remaining lifetime of second. Then $Delta = |X - Y|$.

What's the distribution of $Delta$? To compute it, we'll need to consider the two cases, $X < Y$ and $X > Y$ in order to remove the absolute value.

$$begin{align}
mathbb{P}[Delta < x] &= mathbb{P}[Delta < x | X < Y]mathbb{P}[X < Y] + mathbb{P}[Delta < x | X > Y]mathbb{P}[X > Y] = \
&= frac{1}{2}mathbb{P}[Y - X < x | X < Y] + frac{1}{2}mathbb{P}[X - Y < x | X > Y] \
&=mathbb{P}[Y-X < x | X < Y]
end{align}$$

The last equality follows from symmetry: $mathbb{P}[Y-X < x | X < Y] = mathbb{P}[X - Y < x | X > Y]$.

To simplify this further, we can use the memotylessness of the exponential distribution.

$$mathbb{P}[Y - X < x | X < Y]$ = $mathbb{P}[Y < X + x | Y > X] =
1 - mathbb{P}[Y > X + x | Y > X] = \=1 - mathbb{P}[Y > x] = mathbb{P}[Y < x]$$

This essentially shows, if $Y>X$, then $Y-X$ has the same distribution as $X$ - exponential with parameter $lambda$. Therefore, $Delta sim Exp(lambda)$.

How is this significant?

We insert two batteries with lifetimes $X sim Exp(lambda)$ and $Ysim Exp(lambda)$. One of them runs out and we replace it with the third battery. Its lifetime is $Z sim Exp(lambda)$. At the time of replacement, the other has lifetime $Delta sim Exp(lambda)$ - the same distribution it initially had! So we're left in the same situation, just with only three batteries!

In conclusion: The time between the expiry of any two batteries is exponential with parameter $lambda$. Once three batteries expire, the flashlight stops working, as we only have one left, and it needs two. Therefore, the working time is the sum of three $Exp(lambda)$ random variables, which is (as seen here).

$$Gamma(3, lambda)$$.