A power system will collapse if a ground fault occurs on any two of lines A, B, and C simultaneously. Assume that ground faults on lines A, B, and C are independent and have identical probability of 0.01. Hint: parts (a) and (b), use Bernoulli Trials; part (c), use Bayes Rule. a) What is the probability that the system will collapse

Answer:

0.000298 = 0.0298% probability that the system will collapse

Step-by-step explanation:

Number of possible ways the system can fail:

2 lines from a set of 3 fail, or all three fail.

Two from a set of 3:

Can be A and B, A and C, or B and C. Three possible outcomes then.

The probability of each is:

0.01*0.01*0.99

Two failling, each with 0.01 probability, and 1 working, with 0.99 probability.

All three:

Each with 0.01 probability.

a) What is the probability that the system will collapse

Both cases, so

[tex]p = 3*(0.01)^2*0.99 + (0.01)^3 = 0.000298[/tex]

0.000298 = 0.0298% probability that the system will collapse

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х
0
-2
-4
-6
-8
у
8
12
16
20
24
y = -2x + 8
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