A system is composed of 5 components, each of which is either working or failed.

Question

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector (x,, x2, X3, x4,xs), where xi is equal to 1 if component i is working and is equal to 0 if component i is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if 1. components 3 and 4 are both working, or if components 1, 3, and 5 are all working. Let W be the event that the system will work. Specify all the outcomes in W.

Explanation / Answer

Here we have a system of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector:
(x_1, x_2, x_3, x_4, x_5)
where x_i (note, the subscript is the letter " i ", not 1) is equal to 1 if component i is working, and is equal to 0 if component i is failed.

There are 32 outcomes. Each of the five components can either be working or failed (i.e. there are two outcomes per component in the system). Since the machines are independent, then one of them will never influence the state of the other four. Thus, the total number of outcomes for the five components is 2 x 2 x 2 x 2 x 2 = 25 = 32

Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components 1,3 & 5 are all working. Let W be the event that the system will work. Specify all the outcomes in W. Again, give an explanation to support your answer.

The best way to do this is to list all of the outcomes and identify which will be in W.
x_1 x_2 x_3 x_4 x_5 x_1 & x_2 x_3 & x_4 x_1 & x_3 & x_5 W

0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 1 1 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 1 0 1 0 0 0 0
0 0 1 1 0 0 1 0 1
0 0 1 1 1 0 1 0 1
0 1 0 0 0 0 0 0 0
0 1 0 0 1 0 0 0 0
0 1 0 1 0 0 0 0 0
0 1 0 1 1 0 0 0 0
0 1 1 0 0 0 0 0 0
0 1 1 0 1 0 0 0 0
0 1 1 1 0 0 1 0 1
0 1 1 1 1 0 1 0 1
1 0 0 0 0 0 0 0 0
1 0 0 0 1 0 0 0 0
1 0 0 1 0 0 0 0 0
1 0 0 1 1 0 0 0 0
1 0 1 0 0 0 0 0 0
1 0 1 0 1 0 0 1 1
1 0 1 1 0 0 1 0 1
1 0 1 1 1 0 1 1 1
1 1 0 0 0 1 0 0 1
1 1 0 0 1 1 0 0 1
1 1 0 1 0 1 0 0 1
1 1 0 1 1 1 0 0 1
1 1 1 0 0 1 0 0 1
1 1 1 0 1 1 0 1 1
1 1 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1

There are 15 outcomes in W. W = (x_1 & x_2) | (x_3 & x_4) | (x_1 & x_3 & x_5). The system will work if one of three criteria is met. The above truth table will illustrate how to solve. Columns 1 through 5 are all 32 of the possibilities. Column six deals with the first criterion. This column contains a “1†only when the first two columns are both
“1â€s. There are eight such outcomes. Column seven is for the second criterion. Similar to the first, a “1†is entered when columns three and four have a “1â€. Eight more for W. The eighth column is for the third criterion. It becomes a “1†when columns 1, 3, and 5 are “1â€. There are four more. So, we should have 8 + 8 + 4 = 20 in W, right? No. We have double-counted outcomes that meet two of the three criteria. Column nine represents W, and thus the OR of the three previous columns. Fifteen of the outcomes meet at least one way
of having the system work.

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