# Thread: Probability of faulty part

1. ## Probability of faulty part

Need some guidance on how to solve such a problem. A company has 3 assembly lines(A,B,C). 5% of A's items needs rework, 8% of B's and 10% of C's. Supposing that 50% of all components are made on A, 30% on B and 20% on C. Question is, if a part is randomly selected what is the chance it came from line A? B? C?

So, each line produces the following percents of faulty parts.
A) .05*.5=.025(2.5%)
B) .08*.3=.024(2.4%)
C) .10*.2=.02(2.0%)

So would the answers be those exact percentages as well?

Let me know if im on the right track here possibly?

2. Originally Posted by netring
Need some guidance on how to solve such a problem. A company has 3 assembly lines(A,B,C). 5% of A's items needs rework, 8% of B's and 10% of C's. Supposing that 50% of all components are made on A, 30% on B and 20% on C. Question is, if a part is randomly selected what is the chance it came from line A? B? C?

So, each line produces the following percents of faulty parts.
A) .05*.5=.025(2.5%)
B) .08*.3=.024(2.4%)
C) .10*.2=.02(2.0%)

So would the answers be those exact percentages as well?

Let me know if im on the right track here possibly?
There is a mistake in the wording of this question, or the answer is 50, 30 and 20%.

CB

3. Yeah I'm sorry you're right, it's supposed to be chance the part that needs rework came from a,b,c

4. Originally Posted by netring
Need some guidance on how to solve such a problem. A company has 3 assembly lines(A,B,C). 5% of A's items needs rework, 8% of B's and 10% of C's. Supposing that 50% of all components are made on A, 30% on B and 20% on C. Question is, if a faulty part is randomly selected what is the chance it came from line A? B? C?

So, each line produces the following percents of faulty parts.
A) .05*.5=.025(2.5%)
B) .08*.3=.024(2.4%)
C) .10*.2=.02(2.0%)

So would the answers be those exact percentages as well?

Let me know if im on the right track here possibly?
Those percentages are the percentage faulty parts from each conveyor out of the total parts.

Given that a part is faulty,
the probability that it came from a particular conveyor is different.

The probability it came from A is $\displaystyle \frac{0.05(0.5)}{0.05(0.5)+0.08(0.3)+0.1(0.2)}=\fr ac{0.025}{0.069}$

The probability it came from B is $\displaystyle \frac{0.08(0.03)}{0.05(0.5)+0.08(0.3)+0.1(0.2)}=\f rac{0.024}{0.069}$

the probability it came from C is $\displaystyle \frac{0.1(0.2)}{0.05(0.5)+0.08(0.2)+0.1(0.2)}=\fra c{0.02}{0.069}$