$k$-out-of-$n$ system probabilities 
An engineering system consisting of $n$ components is said to be a $k$-out-of-$n$ system ($k \le n$) when the system functions if and only if at least $k$ out of the $n$ components function. Suppose that all components function independently of each other.
  If the $i^{th}$ component functions with probability $p_i$, $i = 1, 2, 3, 4$, compute the probability that a 2-out-of-4 system functions.

This problem in itself does not seem very difficult to solve, but I suspect I am not doing it the way it was intended to be done, because the formulas that come out are very ugly. I calculated the probability by conditioning on whether or not the $1^{st}$ and $2^{nd}$ components worked, and it came out to be
$$
p_3 p_4 + p_2 (p_3 + p_4 - 2 p_3 p_4) +  p_1 (p_3 + p_4 - 2 p_3 p_4 + p_2 (1 - 2 p_3 - 2 p_4 + 3 p_3 p_4))
$$
Even if this is right, there's no way it's what the answer is supposed to look like. Can someone give me a push in the right direction?
 A: We derive an expression which is also somewhat complicated, but is based on simple fail-safe ideas, and can be easily imitated in similar problems. It is very convenient to let $q_i=1-p_i$. So $q_i$ is the probability that component $i$ fails.
0 bad:  Maybe they all work. The probability of this is 
$$p_1p_2p_3p_4.$$
1 bad: Maybe exactly one goes bad. The probability of this is
$$q_1p_2p_3p_4+p_1q_2p_3p_4+p_1p_2q_3p_4+p_1p_2p_3q_4$$
(one $q$ and three $p$'s, in all possible ways).
2 bad: This is a little more complicated, there are $6$ terms. The probability is
$$q_1q_2p_3p_4+q_1p_2q_3p_4+ q_1p_2p_3q_4+p_1q_2q_3p_4+p_1q_2p_3q_4+p_1p_2q_3q_4$$
(two $q$'s and two $p$'s, in all possible ways).
To find the desired probability, add together the probabilities of the $3$ cases.
Remark:   In this  particular problem, it is easier to find the probability of failure, 
since there are fewer cases to examine. The system fails if $3$ or more components fail. 
The probability that exactly $3$ fail is 
$$p_1q_2q_3q_4+ q_1p_2q_3q_4+q_1q_2p_3q_4+q_1q_2q_3p_4,$$
and the probability they all fail is 
$$q_1q_2q_3q_4.$$
Add, and subtract the result from $1$ to get the probability the system works.
A: Well, a 2-out-of-4 system functions as long as there are not just zero or one functioning components. So,
$$\begin{align}\mathbb{P}[\text{system functioning}]&=1-\prod_{i=1}^4(1-p_i)-\sum_{i=1}^4p_i\prod_{j\neq i}(1-p_j)
\end{align}
$$
This is a symmetric polynomial, so if you like you could express it in terms of elementary symmetric polynomials.
A: If you want something even prettier. I think this is right,
$$\sum_{i=1}^3 \sum_{j=i+1}^4 \frac{(1-p_i)(1-p_j)}{p_ip_j}p_1p_2p_3p_4$$
A: If you want something "prettier" you could take all the possible cases and write them using products and sums, such as $$\prod_{i=1}^4 p_i \left(1+\sum_{j=1}^4\frac{1-p_j}{p_j} + \sum_{k=1}^3 \sum_{l=k+1}^4 \frac{(1-p_k)(1-p_l)}{p_k \; p_l} \right)$$
or you could work out the probability that one or none work and subtract that from $1$ to get $$1 - \prod_{i=1}^4(1-p_i)\left(1+\sum_{j=1}^4\frac{p_j}{1-p_j}\right)$$ 
