Conditional probability and defect rate

  • Thread starter Kinetica
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  • #1
Kinetica
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Homework Statement



Of the items produced daily by a factory, 40% come from line I and 60% from line II. Line I
has a defect rate of 8%, whereas line II has a defect rate of 10%. If an item is chosen at random
from the day’s production, find the probability that it will not be defective.

Homework Equations





The Attempt at a Solution



My answer is 0.82 because Line I with no defect is 0.32+Line II with no defect is 0.50.
OR
1-Both lines defected (0.18)=0.82.


Is my approach correct?
 

Answers and Replies

  • #2
utstatistics
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I'd like to know how your numbers came about. Try drawing a tree diagram and see if you can find the results then; it'll help simplify the process.
 
  • #3
Kinetica
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I drew a table:
Def No Def
I 0.08 0.32 0.40
II 0.10 0.50 0.60
0.18 0.82 1

It is a little bit off but you can see that elements add up to 1. and you can also see the number 0.82 - this is my answer.
 
  • #4
utstatistics
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No no, I'm requesting that you write out how you got your numbers. I'll give a hint; question if your table is correct. If you have a .4 chance of it being in line 1, and a .6 chance of the item being in line 2, would you say that the chance of it being in line 1 AND defect is .08? Or would it be something else?
 
  • #5
Kinetica
88
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I see what you mean. My mistake is that I did not consider to multiply the probabilities by corresponding values.

Then, my question is. Some problems that I solved earlier were easily solved by using tables like this. How do I distinguish which method to use?
 
  • #6
utstatistics
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That's the tricky part. A key component is how the question is asked, but nothing beats practice. Probability is one of those subjects where you just have to keep doing problems over and over to gain intuition. Best of luck!

-J
 
  • #7
Ray Vickson
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I see what you mean. My mistake is that I did not consider to multiply the probabilities by corresponding values.

Then, my question is. Some problems that I solved earlier were easily solved by using tables like this. How do I distinguish which method to use?

Table, trees, formulas, whatever---they all say the same thing, perhaps in disguised form. There is no single right way: do whatever makes you fee comfortable.

Let's compute P(D) = prob item is defective.
(1) Formula: P(D) = P(D|I)*P(I) + P(D|II)*P(II) = (.08)*(.60)+(.10)*(.40) = 0.0.088
(2) Table: Say we make 1,000,000 items.
No. produced in Line I = (0.60)(1,000,000) = 600,000
No. produced in Line II = (0.40)(1,000,000) = 400,000
Of the 600,000 produced on LI, the number defective = (.08)(600,000) = 48,000
Of the 400,000 produced on LII the number defective = (.10)(400,000) = 40,000
Putting these in a table we have:

Total Defect Non-defect
Line I items 600,000 48,000 552,000
Line II items 400,000 40 000 360,000
Total 1,000,000 88,000 912,000
Thus, P(D) = 88,000/1,000,000 = 88/1000 = 0.088.

You can also do it in a tree, but I can't easily draw a tree here.

RGV
 

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