# $k$-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?

-
Unfortunately there is no "nice" answer when the probabilities are different. But you could try the inclusion-exclusion principle on the $6$ events $W_iW_j$ of probability $p_ip_j$ since the event "system works" is just the union of the $6$ events $W_1W_2, W_1W_3, W_1W_4, W_2W_3, W_2W_4, W_3W_4$ which will give a "nice" symmetric and systematic form for the result. – Dilip Sarwate Feb 9 '12 at 22:44

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.

-
"in terms of elementary symmetric polynomials"? +1 for observaton. Please elaborate. Thank u. – hhh Feb 19 at 14:39

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.

-
$A-B+C=3$ when $p_i=1$ for each $i$. Perhaps A-B-C? – Ben Derrett Feb 9 '12 at 23:17
@Ben Derrett: Thanks. Out of time, so deleted the part with the mistake. – André Nicolas Feb 9 '12 at 23:33

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 $$1 - \prod_{i=1}^4(1-p_i)\left(1+\sum_{j=1}^4\frac{p_j}{1-p_j}\right)$$

-