Here are four fractions labelled A, B, C and D.
1 answer:
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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
0.000298 = 0.0298% probability that the system will collapse