This question is regarding Conditional Probability + Bayes. Circuit with the operational probability of the elements is attached.
I got the answer for probability that the electricity will flow from the circuit A to B as: = 0.788 (Hope this is correct)
Part of the question ask following.
Is it possible to save wire that connects the elements without affecting functionality of the network? Explain which part of wiring can be removed.
How could we answer the above question. Is that the wire which directly connects e1 to B? But the working probability of e1 to B (0.5) is higher than e1 to e4 to B (0.5 * 0.8 = 0.4)
Please provide a clue on answering this part.
In the attached image, starting point is A. Circuit is consisting with 3 elements (e1, e2, e3) attached parallelly. And to that, e4 is attached. e1 has a direct wiring to the end-B as well. Operational Probabilities of each element. P(e1) = 0.5; P(e2) = 0.2; P(e3) = 0.3; P(e4) = 0.8;
Your calculation as shown in the comments is wrong because in the outermost multiplication you apply the multiplication rule, which holds for independent events, to events that are dependent because $e_1$ appears in both of them. This gives you an indication which wire can be removed: It’s the one that causes this dependence, that involves $e_1$ in the first factor even though the probability that the system conducts because of $e_1$ is already fully accounted for in the second factor.