An engineering system consisting of n components is said to be a k-out-of-n system (k≤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 and are not identical. If the ith component functions with probability pi, i=1,2,3,4, compute the probability that a 2-out-of-4 system functions.

Option 1: $$ \Bbb ℙ(2,3,4) $$

$$=P_1 [1−(1−P_2)(1−P_3)(1−P_4)]+ $$ $$P_2 [1−(1−P_1)(1−P_3)(1−P_4)]+ $$ $$P_3 [1−(1−P_1)(1−P_2)(1−P_4)]+ $$ $$P_4 [1−(1−P_1)(1−P_2)(1−P_3)] $$

Option 2:

Would it be the same as 4-choose-2? $$ \Bbb ℙ(2,3,4) $$ $$ \Bbb =P_1 P_2 P_3 P_4 $$ $$ - P_1P_2 - P_1P_3 - P_1P_4 - P_2P_3 - P_2P_4 - P_3P_4 - P_1P_2P_3P_4 $$

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    $\begingroup$ What's the question? Are you looking for confirmation of your answer, or for a simpler form, or to understand the given answer? $\endgroup$ – ConMan May 9 '17 at 2:06
  • $\begingroup$ i dont think my answer is right. looking for the answer for, 2-out-of-4 components functioning for the entire system to be functioning. $\endgroup$ – Chootar Laal May 9 '17 at 2:22

Almost ... no where near what you need..

$$\quad{p_1\big(1-(1-p_2)(1-p_3)(1-p_4)\big)}\\ + {p_2\big(1-(1-p_1)(1-p_3)(1-p_4)\big)}\\ + {p_3\big(1-(1-p_1)(1-p_2)(1-p_4)\big)}\\ + {p_4\big(1-(1-p_1)(1-p_2)(1-p_3)\big)}$$

What you have is the probability that the first component functions and not all three of the others fail, or the second does and not all three of the others do, or... so forth.   On closer inspection, those are not disjoint events.

You require the probability that two, three, or four components function.   I suggest using the principle of inclusion and exclusion.


  • $\begingroup$ thx for the input Graham. so what you added at the end is the part for adding in probabilities that the components fail? confused. $\endgroup$ – Chootar Laal May 9 '17 at 2:29
  • $\begingroup$ the case where Prob (success) for each of the 4 components are identical and independent is the binomial case, so that's easy enough but when its independent and NOT identical is more complicated and that's the case i am after here. $\endgroup$ – Chootar Laal May 9 '17 at 2:34
  • $\begingroup$ Are you saying that's the 4-choose-2 case? i'll take a gander. $\endgroup$ – Chootar Laal May 9 '17 at 13:12
  • $\begingroup$ i dont understand your example Graham. $\endgroup$ – Chootar Laal May 9 '17 at 13:23
  • $\begingroup$ I remember using a graphical example with circles grayed out or leaved empty when I was in undergraduate, maybe that would help him understand ? $\endgroup$ – Revolucion for Monica May 10 '17 at 9:40

This is what I have come up with. But I am unable to come up with the generalized equation, so there may be some errors in here.

$$ P_1P_2P_3P_4 $$ $$ +(1-P_1)P_2P_3P_4 $$ $$ +P_1(1-P_2)P_3P_4 $$ $$ +P_1P_2(1-P_3)P_4 $$ $$ +P_1P_2P_3(1-P_4) $$ $$ +(1-P_1)(1-P_2)P_3P_4 $$ $$ +(1-P_1)(1-P_2)(1-P_3)P_4 $$ $$ +(1-P_1)(1-P_2)(1-P_3)(1-P_4) $$ $$ +P_1(1-P_2)(1-P_3)P_4 $$ $$ +P_1(1-P_2)(1-P_3)(1-P_4) $$ $$ +P_1P_2(1-P_3)(1-P_4) $$

  • $\begingroup$ Yes? No? Maybe? heck No? $\endgroup$ – Chootar Laal May 18 '17 at 13:11

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