Manufacturing problem-- bayesian probability

[DDM]

Please help:

Plant A produces 40% of the company's pin volume and 67% of the company's defective pins
Plant B produces 60% of the company's total volume and 33% of the total defective pins
What is the probability a defective pin comes from plant A? and probability it comes from plant B?

Thanks very much

 

[Dr.Peterson]

Plant A produces 40% of the company's pin volume and 67% of the company's defective pins
Plant B produces 60% of the company's total volume and 33% of the total defective pins
What is the probability a defective pin comes from plant A? and probability it comes from plant B?

Please check that you copied this correctly. Or might it be meant as a trick question?

As written, it seems to answer itself: if 67% of the defective pins are from plant A, then that is the probability that a pin comes from plant A, given that it is defective. You don't need the rest of the information.

 

[DDM]

Please check that you copied this correctly. Or might it be meant as a trick question?

As written, it seems to answer itself: if 67% of the defective pins are from plant A, then that is the probability that a pin comes from plant A, given that it is defective. You don't need the rest of the information.

Thank you for the quick reply. I have the answer to the problem but I don't know how it is calculated: The probability that Plant A produced the defective pin is 57.2% and Plant B 42.8%. It's a conditional probability playing off of the two pieces of information--contribution to total volume and total defects. Hopefully this helps you to help me. Thanks

 

[Dr.Peterson]

Thank you for the quick reply. I have the answer to the problem but I don't know how it is calculated: The probability that Plant A produced the defective pin is 57.2% and Plant B 42.8%. It's a conditional probability playing off of the two pieces of information--contribution to total volume and total defects. Hopefully this helps you to help me. Thanks

Did you copy the exact wording of the problem? Please check, and do so if you didn't. The wording of probability questions is highly sensitive.

I can get numbers close to the supposed answers if I change the problem to something that would have to be worded differently.

Do you understand what I said last time about why the problem doesn't make sense?

 

[DDM]

Did you copy the exact wording of the problem? Please check, and do so if you didn't. The wording of probability questions is highly sensitive.

I can get numbers close to the supposed answers if I change the problem to something that would have to be worded differently.

Do you understand what I said last time about why the problem doesn't make sense?

Here is the exact language. Thanks again

The company has two factories, the older of which produces 40% of the total output. This means that a pin picked up at random has a 40% probability of coming from the old factory, whether it is defective or perfect; this is the prior probability. We find that the older factory's defective rate is twice that found in the newer factory. If a customer calls and complains about finding a defective pin, which of the two factories should the manager call?

The prior probability would suggest that the defective pin was most likely to have come from the new plant, which produces 60% of the total. On the other hand, that plant produces only one-third of the company's total of defective pins. When we revise the priors to reflect this additional information, the probability that the new plant made the defective pin turns out to be only 42.8%; there is a 57 .2% probability that the older plant is the culprit. This new estimate becomes the posterior probability.

 

[Dr.Peterson]

Plant A produces 40% of the company's pin volume and 67% of the company's defective pins
Plant B produces 60% of the company's total volume and 33% of the total defective pins
What is the probability a defective pin comes from plant A? and probability it comes from plant B?

Here is the exact language. Thanks again

The company has two factories, the older of which produces 40% of the total output. This means that a pin picked up at random has a 40% probability of coming from the old factory, whether it is defective or perfect; this is the prior probability. We find that the older factory's defective rate is twice that found in the newer factory. If a customer calls and complains about finding a defective pin, which of the two factories should the manager call?

The prior probability would suggest that the defective pin was most likely to have come from the new plant, which produces 60% of the total. On the other hand, that plant produces only one-third of the company's total of defective pins. When we revise the priors to reflect this additional information, the probability that the new plant made the defective pin turns out to be only 42.8%; there is a 57 .2% probability that the older plant is the culprit. This new estimate becomes the posterior probability.

Do you see that your paraphrase means something entirely different from the original?

You were right that plant A produces 40% of the pins and plant B produces 60%; but the first paragraph does not say that plant A produces 67% of the total number of defective pins. It says that the rate of defective pins from each factory are in the ratio 2:1, but that does not make them 2/3 and 1/3 (much less the rounded values 0.67 and 0.33!), because they are not complements.

If the second paragraph here is from the book, then the book is wrong. Plant B does not produce 1/3 of the company's defective pins! (In fact, their conclusion means that 42.8% of the defective pins come from plant B, as I pointed out initially.) Who wrote this?

Rather, suppose that the defective rate for plant B is r, and for plant A is 2r. That is, P(defective | B) = r, and P(defective | A) = 2r.

Put that in the Bayes formula as given to you by pka, and see what you get. (The variable r will drop out.)

 

[pka]

Plant A produces 40% of the company's pin volume and 67% of the company's defective pins
Plant B produces 60% of the company's total volume and 33% of the total defective pins
What is the probability a defective pin comes from plant A? and probability it comes from plant B?

Here is the exact language.
The company has two factories, the older of which produces 40% of the total output. This means that a pin picked up at random has a 40% probability of coming from the old factory, whether it is defective or perfect; this is the prior probability. We find that the older factory's defective rate is twice that found in the newer factory. If a customer calls and complains about finding a defective pin, which of the two factories should the manager call?

For the good of the order, please please take note that had you posted the exact wording of the question to begin with, a lot of confusion could have been avoided. After first posting the exact wording, you can then post your reading of the question. At that point we can give help on the salient points.

 

[DDM]

For the good of the order, please please take note that had you posted the exact wording of the question to begin with, a lot of confusion could have been avoided. After first posting the exact wording, you can then post your reading of the question. At that point we can give help on the salient points.

Yes, will do. Appreciate the admonishment-- being my first post is my only excuse. It is also unfortunate that as Dr. Peterson pointed, the book was wrong! Thanks again

 

[DDM]

Do you see that your paraphrase means something entirely different from the original?

You were right that plant A produces 40% of the pins and plant B produces 60%; but the first paragraph does not say that plant A produces 67% of the total number of defective pins. It says that the rate of defective pins from each factory are in the ratio 2:1, but that does not make them 2/3 and 1/3 (much less the rounded values 0.67 and 0.33!), because they are not complements.

If the second paragraph here is from the book, then the book is wrong. Plant B does not produce 1/3 of the company's defective pins! (In fact, their conclusion means that 42.8% of the defective pins come from plant B, as I pointed out initially.) Who wrote this?

Rather, suppose that the defective rate for plant B is r, and for plant A is 2r. That is, P(defective | B) = r, and P(defective | A) = 2r.

Put that in the Bayes formula as given to you by pka, and see what you get. (The variable r will drop out.)

Dr. Peterson, your clever suggestion worked--thank you. Paragraph 2 is directly from the book and as you point out and I now see: it is wrong! "Against the Gods--The Remarkable Story of Risk" by Peter Bernstein, 1996

 

[Dr.Peterson]

Dr. Peterson, your clever suggestion worked--thank you. Paragraph 2 is directly from the book and as you point out and I now see: it is wrong! "Against the Gods--The Remarkable Story of Risk" by Peter Bernstein, 1996

Clearly he isn't a mathematician; he got the answer more or less right, but he misstated the explanation.

At least it provided a good lesson for those who are reading along here: there's a reason for the request in our submission guidelines, to "Post the exercise or your question completely and accurately". Whether the error is in your thinking, or in the original source, we can more easily see what is going on when we see everything.

 

[BCF00]

I am reading the same book and have the same question as to how the author arrived at his conclusion and if perhaps there is an error. Please forgive the repetition but this is driving me bonkers trying to figure this out and I didn’t understand the initial post. Reference: Against the Gods the Remarkable Story of Risk, Peter L. Bernstein 1998. The author makes mention of a pin- manufacturing company that on average has 12 defective pins per 100,000 pins produced, then goes on to explain the life and contributions of Thomas Bayes. At the end of the chapter the author presents a question to illustrate Bayes’ theorem, again referencing the pin-manufacturing company. I am quoting the book from here. “The company has two factories, the older of which produces 40% of the total output. This means that a pin picked up at random has a 40% probability of coming from the old factory, whether it is defective or perfect; this is the prior probability. We find that the old factory’s defective rate is twice that found in the newer factory. If a customer calls and complains about finding a defective pin, which of the two factories should the manager call? The prior probability would suggest that the defective pin was most likely to have come from the new plant, which produces 60% of the total. On the other hand, that plant produces only one-third of the company’s total defective pins. When we revive the priors to reflect this additional information, the probability that the new plant made the defective pins turns out to be only 42.8%; there is a 57.2% probability that the older plant is the culprit. This new estimate becomes the posterior probability.” End of quoting from the book. Please help me with two questions, first is the author’s answer correct? Second, I will explain my thinking, what am I doing wrong?

If a customer received a shipment of 100,000 pins about 12 should be defective and the old factory producing twice the errors as the new factory should be responsible for 8 errors and the new factory responsible for 4 errors. The probability that the defective pin came from the old factory is 8/12, .6666 and the probability that the defective pin came from the new factory is 4/12, .3333. Thanks for any help you can provide.

 

[Dr.Peterson]

I am reading the same book and have the same question as to how the author arrived at his conclusion and if perhaps there is an error. Please forgive the repetition but this is driving me bonkers trying to figure this out and I didn’t understand the initial post.

Have you read this thread, particularly my post #7?? YES, there is an error in the book. And the initial post is even worse.

Please help me with two questions, first is the author’s answer correct?

See above. The book's answer is (approximately) correct; the explanation is not.

Second, I will explain my thinking, what am I doing wrong?

If a customer received a shipment of 100,000 pins about 12 should be defective and the old factory producing twice the errors as the new factory should be responsible for 8 errors and the new factory responsible for 4 errors. The probability that the defective pin came from the old factory is 8/12, .6666 and the probability that the defective pin came from the new factory is 4/12, .3333.

No, this is wrong. Where did you get your number 12? And what statement in the problem (as opposed to the author's incorrect reasoning) says that the new factory makes 2/3 of the total errors? You are misreading it, as the OP did, and the author also did.

Again, (re)read #7, carefully, do what it says, and write back.

 

[BCF00]

Have you read this thread, particularly my post #7?? YES, there is an error in the book. And the initial post is even worse.


See above. The book's answer is (approximately) correct; the explanation is not.


No, this is wrong. Where did you get your number 12? And what statement in the problem (as opposed to the author's incorrect reasoning) says that the new factory makes 2/3 of the total errors? You are misreading it, as the OP did, and the author also did.

Again, (re)read #7, carefully, do what it says, and write back.

1.) I got the number 12 from the author of the book, he first referenced a pin-manufacturing company that has on average 12 defects for every 100,000 pins produced. Later the author states that the company is made up of two factories, an old factory that produces 40% of total output and also that the old factory's defective rate is twice that of the new factory.
2.) Nowhere does it make that statement other that the authors incorrect reasoning.
3.) I think I might have it... the 2/3 defects coming from the old factory have to be weighted by the 40% total company output..? Using the formula: Probability) that a pin came from the old factory (given) there is a defective pin. = in the numerator (2/3 x .40) divided by in the denominator (2/3 x .40) + (1/3 x .60) = .266667/ .466667 = .5714 that the pin is from the old factory?

 

[Dr.Peterson]

1.) I got the number 12 from the author of the book, he first referenced a pin-manufacturing company that has on average 12 defects for every 100,000 pins produced. Later the author states that the company is made up of two factories, an old factory that produces 40% of total output and also that the old factory's defective rate is twice that of the new factory.

If this problem is meant to continue from that one, then you could suppose that the old factory has an error rate of 12/100,000, and the new one has a rate of only 6/100,000. My suggestion in #6 was to call the rates 2r and r.

3.) I think I might have it... the 2/3 defects coming from the old factory have to be weighted by the 40% total company output..? Using the formula: Probability) that a pin came from the old factory (given) there is a defective pin. = in the numerator (2/3 x .40) divided by in the denominator (2/3 x .40) + (1/3 x .60) = .266667/ .466667 = .5714 that the pin is from the old factory?

Your answer is correct; but I wouldn't use the numbers 2/3 and 1/3. (It is definitely not true that "2/3 of defects come from the old factory"; in fact, the distinction between the proportion of total defects and the individual rates of defects should be one of the main points in discussing Bayes.)

If you do the work with r, you'll find that it cancels out, so that using the values you did has no effect.

 

[BCF00]

If this problem is meant to continue from that one, then you could suppose that the old factory has an error rate of 12/100,000, and the new one has a rate of only 6/100,000. My suggestion in #6 was to call the rates 2r and r.


Your answer is correct; but I wouldn't use the numbers 2/3 and 1/3. (It is definitely not true that "2/3 of defects come from the old factory"; in fact, the distinction between the proportion of total defects and the individual rates of defects should be one of the main points in discussing Bayes.)

If you do the work with r, you'll find that it cancels out, so that using the values you did has no effect.

(.4 x 2 ) / (.4 x 2) + (.6 x ) = .5714 I see... Thank you...