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Answer:
The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.
Step-by-step explanation:
The three different assembly lines are: A₁, A₂ and A₃.
Denote R as the event that a component needs rework.
It is given that:
[tex]P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20[/tex]
Compute the probability that a randomly selected component needs rework as follows:
[tex]P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069[/tex]
Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:
[tex]P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623[/tex]
Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.
If a randomly selected component needs rework, the probability that it came from line A1 is 36.23%.
Given that a company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component, and of those manufactured by line A1, 5% need rework to remedy a defect, whereas 8% of A2's components need rework and 10% of A3's need rework, supposing that 50% of all components are produced by Line A1, 30% are produced by line A2, and 20% come from line A3, if a randomly selected component needs rework, to determine what is the probability that it came from line A1 the following calculation must be performed:
- 0.5 x 0.05 = 0.025
- 0.025 + (0.3 x 0.08) + (0.2 x 0.1) = 0.025 + 0.024 + 0.02 = 0.069
- 69 = 100
- 25 = X
- 25 x 100/69 = X
- 36.23 = X
Therefore, if a randomly selected component needs rework, the probability that it came from line A1 is 36.23%.
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