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07 Feb 2020, 07:41
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
78% (02:34) correct
23%
(02:21)
wrong
based on 80
sessions
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Machine–A produces 40% of the total output and Machine-B produces 60% of the total output. An average of nine units out of a thousand goods manufactured by Machine-A and one unit of 250 units produced by Machine-B prove to be defective. What is the probability that a unit chosen at random from the total daily output of the factory is defective?
A. 0.6
B. 0.06
C. 0.006
D. 0.0006
E. 0.00006
A. 0.6
B. 0.06
C. 0.006
D. 0.0006
E. 0.00006
07 Feb 2020, 21:37
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Bunuel
Probability that a unit from machine-A is defective = 40/100*9/1000 = 0.0036
& Probability that a unit from machine-B is defective = 60/100*1/250 = 6/10*4/1000 = 0.0024
So, total probability that a unit chosen is defective = 0.0036 + 0.0024 = 0.0060
Option C
Posted from my mobile device
07 Feb 2020, 21:55
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Let us assume that total output is 2500 units per day
Machine A produces 40% ==> 1000 units ( with 9 units defective units)
MachineB produces 60% ==> 1500 units ( with 6 units defective 1 unit per 250 units)
When One unit is selected random from day's output:
The total possible outcomes = 2500
Desired outcomes( defective ) = 9+6 = 15
The Probability of Defective unit = Desired Outcomes / Total possible outcomes = 15/ 2500 = 3/500 = 0.006
Ans - C
Machine A produces 40% ==> 1000 units ( with 9 units defective units)
MachineB produces 60% ==> 1500 units ( with 6 units defective 1 unit per 250 units)
When One unit is selected random from day's output:
The total possible outcomes = 2500
Desired outcomes( defective ) = 9+6 = 15
The Probability of Defective unit = Desired Outcomes / Total possible outcomes = 15/ 2500 = 3/500 = 0.006
Ans - C
