I have 3 similar problems but I don't quite understand the differences between them and how to approach each.

First 2 Problems

Lets say you have 2 circuits where all relays function independently:

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I'll call the top 1 circuit 1 and the bottom 1 circuit 2. Each of them has a flow going from A to B, and has a probability of closing the $i$th relay with a probability of $P_i$ where $i=1,2,3,4...$

I was working on these problems and the answers the book gave me for the probability of the circuit flows being completed were that for circuit 1 the probability was $P_{1}P_{3} + P_{1}P_{4} + P_{2}P_{3} + P_{2}P_{4}$ and for circuit 2 it was $P_{5}(P_{1}P_{2} + P_{3}P_{4} - P_{1}P_{2}P_{3}P_{4})$

I don't under stand why in the first case you don't have to subtract out the product of all 4 probabilites of the relays like you do in the second circuit.

3rd Problem:

An engineering system consists of $n$ components is said to be a k out of n system (where k is less than n) if the system functions if and only if at least k of the n components function. Suppose that all components function independently of each other. If the $i$th component functions with probability $P_{i}$ where $i=1,2,3,4$ what is the probably that a 2 out of 4 system functions?

For this one then I'm confused as to whether I can just add up all the combinations of pairs of components and add them together like the first circuit, or I have to subtract out overlaps like the second circuit.

I guess I'm really not understanding how the inclusion-exclusion rule is working in these problems. Thanks for any help.

  • $\begingroup$ As you noted, the first solution is clearly wrong: for example, if $P_i=\frac12$ for every $i$, the sum is $1$, which is impossible. $\endgroup$
    – Did
    Apr 26, 2012 at 5:35
  • $\begingroup$ "The probability was $P_1P_3+P_1P_4+\cdots$" The probability of what? $\endgroup$ Apr 26, 2012 at 5:41
  • $\begingroup$ The flow being able to go from point a to point b. A relay closing means closing the gap. $\endgroup$
    – Matt
    Apr 26, 2012 at 5:45

1 Answer 1


For the 3rd problem, no, you can't just add up all pairs. The easiest way to see this can't work is to note (in line with Didier's comment) that for various values of the $P_i$ adding up all the pairs would give you a probability exceeding one. You have to add up the probabilities corresponding to pairs, subtract the ones for triples, add back in the one for all four.

  • $\begingroup$ This makes a lot of sense, thanks. The answer for the first problem must be wrong then. $\endgroup$
    – Matt
    Apr 26, 2012 at 5:53

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