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Machine–A produces 40% of the total output and Machine-B produces 60%

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Bunuel
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Machine–A produces 40% of the total output and Machine-B produces 60% [#permalink] New post   07 Feb 2020, 07:41
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74% (02:34) correct 26% (02:21) wrong based on 68 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
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C
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Re: Machine–A produces 40% of the total output and Machine-B produces 60% [#permalink] New post   07 Feb 2020, 21:37
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Bunuel wrote:
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


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

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KalyanChamarthi
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Re: Machine–A produces 40% of the total output and Machine-B produces 60% [#permalink] New post   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
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Re: Machine–A produces 40% of the total output and Machine-B produces 60% [#permalink] New post   20 Apr 2020, 12:09
Bunuel wrote:
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


solved using weighted average
9/1000=0.009
1/250=4/1000=0.004
a=2/5, b=3/5
2/3=0.004-w/0.009-w,
0.002-18m=0.012-3m,
m=0.006

Ans (C)
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Re: Machine–A produces 40% of the total output and Machine-B produces 60% [#permalink] New post   20 Apr 2020, 13:33
Hint : Consider the total unit as 10,000

so 36+24 / 10,000 = 0.006
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Re: Machine–A produces 40% of the total output and Machine-B produces 60% [#permalink]
20 Apr 2020, 13:33
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