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Sagot :
Answer:
0.2941 = 29.41% probability that it was manufactured during the first shift.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Defective
Event B: Manufactured during the first shift.
Probability of a defective item:
1% of 50%(first shift)
2% of 30%(second shift)
3% of 20%(third shift).
So
[tex]P(A) = 0.01*0.5 + 0.02*0.3 + 0.03*0.2 = 0.017[/tex]
Probability of a defective item being produced on the first shift:
1% of 50%. So
[tex]P(A \cap B) = 0.01*0.5 = 0.005[/tex]
What is the probability that it was manufactured during the first shift?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.005}{0.017} = 0.2941[/tex]
0.2941 = 29.41% probability that it was manufactured during the first shift.
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