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I am stuck with a problem in probability. It says that the system consists of 16 components distributed in corners and intersections as the following:

  s---s---s---s 
  |   |   |   | 
  s---s---s---s 
  |   |   |   | 
  s---s---s---s 
  |   |   |   | 
  s---s---s---s 

Failure of two adjacent components will lead to system failure. The probability of failure for each component is 0.0005.

I tried different ways but doesn't work, and I was thinking to do it as a simulation but I didn't have any idea

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  • $\begingroup$ This question is unclear to me. $\endgroup$ – Dzoooks Apr 16 '19 at 22:57
  • $\begingroup$ I revised the figure so it will display the way I think you intended. If I did not understand your intent correctly, feel free to change it back. $\endgroup$ – awkward Apr 17 '19 at 13:03
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It's easy to see that the number of pairs of two adjacent components is 8*3=24 (each large edge has 3 pairs and there are 8 edges). Because the probability of failure is so small, we can ignore the case where three or more fail simultaneously. For failure, two have to fail simultaneously and the two have to be adjacent. The probability becomes -

$$p = {16 \choose 2}(0.0005)^2 (1-0.0005)^{14} \frac{24}{{16 \choose 2}}$$

This is a lower bound since we didn't consider 3 failing, 4 failing and so on. But those probabilities are extremely low.

If you want to include 3 failing as well your estimate becomes (24*14 because two adjacent need to fail and choose any of the remaining 14 for the third component that failed):

$$p = {16 \choose 2}(0.0005)^2 (1-0.0005)^{14} \frac{24}{{16 \choose 2}} + {16 \choose 3}(0.0005)^3 (1-0.0005)^{13} \frac{24 \times 14}{{16 \choose 3}}$$

4 onwards becomes complicated with negligible addition to the answer.

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  • $\begingroup$ Thank you, but actually the 8 * 3 how you got this? if each line has 4 components $\endgroup$ – Abdula Apr 17 '19 at 0:00
  • $\begingroup$ Each line has 4 components, but 3 consecutive pairs. $\endgroup$ – Rohit Pandey Apr 17 '19 at 0:31
  • $\begingroup$ sorry for this again, but the system is as shown, only one layer of 16 components, I cannot understand the edges number $\endgroup$ – Abdula Apr 17 '19 at 21:54
  • $\begingroup$ The system is one large square grid with four components on each edge of the square. Take any given edge. How many pairs of components does the edge have? It has four components in a line (1-2-3-4), so three pairs that are connected to each other (1-2, 2-3 and 3-4) $\endgroup$ – Rohit Pandey Apr 21 '19 at 5:48
  • $\begingroup$ Thank you very much $\endgroup$ – Abdula Apr 22 '19 at 0:00
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Another answer that uses simulation. I will provide the python code. Note that if you number the components from top-left to bottom right, two components are adjacent if the difference in the numbers they got is either 1 or 4. Using this fact:

import numpy as np
sys_fails = 0
for n in range(10000):
  failed = []
  for comp_no in range(16):
    if np.random.uniform() < .0005:
        failed.append(comp_no)
  sys_fails += did_sys_fail(failed)
prob_system_failed = sys_fails/10000


def did_sys_fail(failed):
  if len(failed)>1:
     for i1 in range(len(failed)):
       for j1 in range(i1+1,len(failed)):
          if abs(failed[i1]-failed[j1])==1 or abs(failed[i1]-failed[j1])==4:
            return 1
  return 0

Because the probability of failure is so low, you will have high variance in the result. One way to overcome this is to use importance sampling where you increase the failure probability (to say, 50%), but then adjust each simulation.

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