# Thread: Probability of faulty parts shipped. Tough question!

1. ## Probability of faulty parts shipped. Tough question!

Okay, im doing some revision questions for probability and this question has me stuck.. ive spent hours on this one problem and i cant get my head around it. I have the answers to it, but they dont show the working. So here goes..

Question: A supplier has shipped a lot of 6 parts to a company. The lot contained 3 defective parts. Suppose the customer decided to randomly select two parts and test them for defects. How large a sample space is the customer potentially working with? List the sample space. Using the sample space list, determine the probability that the customer will select a sample with exactly I defect.

What I've tried: I've tried finding a sample list of 15 pairs of parts - each part chosen with each other (ie. with replacement i think?). So ive got a list like: (PART1,PART2),(PART1,PART3),and so on for part1....,then (PART2,PART3),(PART2,PART4).. and etc for each part.. So I end up with 15 possible pairs that can be chosen from the 6, you know?
But this sample list doesnt contain information about whether the part is faulty. So how do i do this?
I also tried listing each pair of faulty parts, ie. let 0 = not faulty, and 1 = faulty. Then for each possible pair i have is 00, 01, 10, 11. Then i tried doing combinations of the 6 parts, ie. 111000, 011100, 001110, 000111, 101100, etc etc but the sample space is greater than 15?? (as you woould expect!).

Answer: The sample space is 15 (but the answer wont list the samples so i dont know how to solve this problem.) The probability is then 0.60. Thats all that is given.

This is not an assignment question or anything so im not cheating. I have the answers from the back of the book, but cant make sense of it..

Any suggestions? Thanks alot guys!

2. Originally Posted by bboydocument
Okay, im doing some revision questions for probability and this question has me stuck.. ive spent hours on this one problem and i cant get my head around it. I have the answers to it, but they dont show the working. So here goes..

Question: A supplier has shipped a lot of 6 parts to a company. The lot contained 3 defective parts. Suppose the customer decided to randomly select two parts and test them for defects. How large a sample space is the customer potentially working with? List the sample space. Using the sample space list, determine the probability that the customer will select a sample with exactly I defect.

What I've tried: I've tried finding a sample list of 15 pairs of parts - each part chosen with each other (ie. with replacement i think?). So ive got a list like: (PART1,PART2),(PART1,PART3),and so on for part1....,then (PART2,PART3),(PART2,PART4).. and etc for each part.. So I end up with 15 possible pairs that can be chosen from the 6, you know?
But this sample list doesnt contain information about whether the part is faulty. So how do i do this?
I also tried listing each pair of faulty parts, ie. let 0 = not faulty, and 1 = faulty. Then for each possible pair i have is 00, 01, 10, 11. Then i tried doing combinations of the 6 parts, ie. 111000, 011100, 001110, 000111, 101100, etc etc but the sample space is greater than 15?? (as you woould expect!).

Answer: The sample space is 15 (but the answer wont list the samples so i dont know how to solve this problem.) The probability is then 0.60. Thats all that is given.

This is not an assignment question or anything so im not cheating. I have the answers from the back of the book, but cant make sense of it..

Any suggestions? Thanks alot guys!
As far as I can see, the sample space is as you say:

00, 01, 10, 11.

Pr(01) = (3/6)(3/5) = 9/30 = 3/10.

Pr(10) = (3/6)(3/5) = 9/30 = 3/10.

3/10 + 3/10 = 0.6.

3. Hi
Hypergeometric distribution.

Sample space=6Cn2=15

P(x=1)=(3Cn1)*(3Cn1)/(6Cn2)=0.6

Note: in quality control this is called an AcceptSamplingPlan. For my students I used to use Poisson approximation.

4. Originally Posted by bboydocument
What I've tried: I've tried finding a sample list of 15 pairs of parts - each part chosen with each other (ie. with replacement i think?). So ive got a list like: (PART1,PART2),(PART1,PART3),and so on for part1....,then (PART2,PART3),(PART2,PART4).. and etc for each part.. So I end up with 15 possible pairs that can be chosen from the 6, you know?
But this sample list doesnt contain information about whether the part is faulty. So how do i do this?
You already have the sample space here. All you need to do is to add "Without loss of generality, let parts 1, 2 and 3 be the 3 defective parts." Simply count the number of pairs with 1 defective part and you have your probability.

The faulty/not faulty representation is also a plausible sample space. Though calculating the probability is not as straightforward as in the first construction.