k out of n probability

Question

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? $$ \Bbb ℙ(2,3,4) $$ $$ \Bbb =P_1 P_2 P_3 P_4 $$ $$ - P_1P_2 - P_1P_3 - P_1P_4 - P_2P_3 - P_2P_4 - P_3P_4 - P_1P_2P_3P_4 $$

Answer 1

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.

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

$${p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4}\\{-p_1p_2p_3-p_1p_2p_4-p_1p_3p_4-p_2p_3p_4}\\{+p_1p_2p_3p_4}$$

Answer 2

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.

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

Answer 3

The system functions when 2 or more than 2 components work. Thus it should be $\binom{4}{2}+ \binom{4}{3}+ \binom{4}{4} $ combination of individual component probabilities. Thus $E $ is the event when system functions, $$ \begin {align*} P(E) = & p_1 p_2(1-p_3)(1-p_4)+ p_1 p_3(1-p_2)(1-p_4)+ p_1 p_4(1-p_2)(1-p_3)\\ +& p_2 p_3(1-p_2)(1-p_1)+ p_2 p_4(1-p_1)(1-p_3)+ p_3 p_4(1-p_1)(1-p_2)\\ +& p_1 p_2 p_3 (1-p_4)+ p_1 p_3 p_4(1-p_2)+ p_1 p_2 p_4(1-p_3)+ p_2 p_3 p_4(1-p_1)\\ +& p_1 p_2 p_3 p_4 \end {align*} $$