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14 Aug 2018, 02:32
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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A machine is made up of two components, A and B. Each component either works or fails. The failure or non-failure of one component is independent of the failure or non-failure of the other component. The machine works if at least one of the components works.
If the probability that each component works is \(\frac{2}{3}\), what is the probability that the machine works?
A. \(\frac{1}{9}\)
B. \(\frac{4}{9}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{8}{9}\)
If the probability that each component works is \(\frac{2}{3}\), what is the probability that the machine works?
A. \(\frac{1}{9}\)
B. \(\frac{4}{9}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{8}{9}\)
14 Aug 2018, 04:23
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Bunuel
Method 1:
Machine works if atleast one component works i.e. we get the following cases for machine to work
Case 1: First Component works and Second Fails = Probability = (2/3)*(1/3) = 2/9
Case 2: First Component Fails and Second Works = Probability = (1/3)*(2/3) = 2/9
Case 3: First Component works and Second Works = Probability = (2/3)*(2/3) = 4/9
Total Probability = (2/9)+(2/9)+(4/9) = 8/9
Answer: option E
Method 2:
Probability of machine to work = 1 - Probability of machine to fail to wrk
i.e. Probability of machine to work = 1 - Probability of Both components failing = 1- (1/3)*(1/3) = 1- (1/9) = 8/9
Answer: Option E
Hope this helps!!!
General Discussion
14 Aug 2018, 02:37
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Bunuel
A works but B doesn't = 2/3*1/3=2/9
B works but A doesn't = 2/9
both A and B work = 2/3*2/3=4/9
p(machine works) = sum of above = 2/9 + 2/9 + 4/9 = 8/9
