Light bulbs with exponential life Five light bulbs are controlled simultaneously. Each has an exponential lifespan of parameters $λ_i$ with i = 1 .. 5, independent.
I have to find:
1) expected time of a failure and distribution
2) expected time of two failures and distribution
3) expected time and breakdown of all light bulbs
These are my partial answers:
1) $ Y = \min (X_1 , X_2 , X_3, X_4, X_5) \sim exp(\sum_{i = 1}^{5}\lambda_i)$
$E[Y] = \frac{1}{\sum_{i = 1}^{5}\lambda_i}$ 
$F_Y(x) = {\begin{cases}1-e^{-(\sum_{i = 1}^{5}\lambda_i) \cdot x}&x\geq 0,\\0&x<0\end{cases}}$
2) Y = ? 
$E[Y] = \frac{\lambda_1}{\lambda_1 + ... + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 2}^{5}\lambda_i}\right) + \frac{\lambda_2}{\lambda_1 + ... + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1_{i \ne 2}}^{5}\lambda_i}\right) 
+ \frac{\lambda_3}{\lambda_1 + ... + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1_{i \ne 3}}^{5}\lambda_i}\right)
+ \frac{\lambda_4}{\lambda_1 + ... + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1_{i \ne 4}}^{5}\lambda_i}\right)+ \frac{\lambda_5}{\lambda_1 + ... + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1}^{4}\lambda_i}\right)$
(Using the suggestion of the answer of @Misha Lavrov)
$F_Y(x) = $ ?
3) $ Y = \max (X_1 , X_2 , X_3, X_4, X_5)$
$E[Y] =$ ?
$F_Y(x) = {\begin{cases}P(Y \le x) = P(X_1 \le x, \dots, X_5 \le x) \\
= \prod_{i=1}^5 P(X_i \le x) = \prod_{i=1}^5 (1 - e^{-\lambda_i x})&x\geq 0,\\0&x<0\end{cases}}$ 
(Using the suggestion of the answer of @BruceET)
 A: Although the distributions of the time-to-two-failures and time-to-all-failures are complicated things, their expected values are easy to find just from what we know about exponential distributions.
Consider a simpler case with only two lightbulbs, which have independent failure rates $\lambda_1, \lambda_2$. We know that the expected time until the first fails is $\frac1{\lambda_1 + \lambda_2}$. Moreover, we know that the lightbulb which fails is the first lightbulb with probability $\frac{\lambda_1}{\lambda_1 + \lambda_2}$ and the second with probability $\frac{\lambda_2}{\lambda_1 + \lambda_2}$. Finally, the exponential is memoryless, so after the first failure, the remaining exponential variable "starts over".
If the first lightbulb fails, then the second, the expected waiting time is
$
    \frac1{\lambda_1 + \lambda_2} + \frac1{\lambda_2}.
$
If the second lightbulb fails, then the first, the expected waiting time is
$
    \frac1{\lambda_1 + \lambda_2} + \frac1{\lambda_1}.
$
So the overall expected waiting time is the weighted average of these:
$$
    \frac{\lambda_1}{\lambda_1 + \lambda_2}\left(\frac1{\lambda_1 + \lambda_2} + \frac1{\lambda_2}\right) + \frac{\lambda_2}{\lambda_1 + \lambda_2} \left(\frac1{\lambda_1 + \lambda_2} + \frac1{\lambda_1}\right)
$$
which simplifies to $\frac1{\lambda_1} + \frac1{\lambda_2} - \frac1{\lambda_1 + \lambda_2}$.
A different argument which gives this answer is that we can add $\frac1{\lambda_1}$ and $\frac1{\lambda_2}$ together for the expected time until both lightbulbs fail, but this double-counts the time before either bulb fails. The expected time before the first failure is $\frac1{\lambda_1 + \lambda_2}$, so we subtract it.
You can generalize either argument to the five-lightbulb case.
A: You already have an excellent hint from @MishaLavrov.(+1) Here are two additional suggestions toward a solution.
First, let the five rates all be $\lambda$ and $W$ be the waiting time
for the last bulb to die. Then, by the 'no-memory' property of exponential
random variables:
 $$E(W) = \frac{1}{5\lambda} + \frac{1}{4\lambda} + \frac{1}{3\lambda} + \frac{1}{2\lambda} + \frac{1}{\lambda}.$$ 
Also, the CDF of $W$ is $F_W(t) = (1 - e^{-\lambda t})^5.$ 
These solutions account for results of the following simulation in R statistical software:
set.seed(1222); m = 10^6; lam = 2
w = replicate(m, max(rexp(5, lam)))
mean(w); 2*sd(w)/sqrt(m)
## 1.140875          # aprx E(W)
## 0.001206653       # 95% margin of simulation error
(1/lam)*sum(1/(1:5))
## 1.141667          # exact E(W)

mean(w <= 1)
## 0.483739          # aprx P(W < 1)
(1-exp(-lam))^5
## 0.4833244         # exact P(W < 1)

Second, generalizing the method above for $E(W)$ can be done by extending @MishaLavrov's Answer, but you seem to be finding that approach difficult for five bulbs.
(Maybe not so difficult after you've thought about the equal-rate case.)
Anyhow, getting the CDF $F_W(t) = \prod_{i=1}^5 (1 - e^{-\lambda_i t}),\, t > 0$ is not difficult. Getting $E(W)$ from $F_W$ may be a bit tedious, but
that approach uses straightforward algebra and calculus.
Note: If one of the five $\lambda_i$ is very much smaller than the others
(one light bulb is very much more durable), then $E(W)$ depends heavily on
that one failure rate.
Note: In response to question in Comment.
Let $W$ be the waiting time until the fifth bulb fails. Thus
$W = \max\{X_1, \dots X_5\}$ and
$$F_W(t) = P(W \le t) = P(X_1 \le t, \dots, X_5 \le t) \\
= \prod_{i=1}^5 P(X_i \le t) = \prod_{i=1}^5 (1 - e^{-\lambda_i t}),$$
where independence is used to get from the first line to the second.
To finish the general case, multiply the five terms (collecting terms to minimize the mess),
differentiate to get the PDF of $W$ and then integrate as needed to get
$E(W).$ The last integration may require some integration by parts. 
