Defective Sample

[Rose_01298]

A delivery of 25 parts is evaluated for quality purposes according to the following scheme: A sample of size 5 is drawn (without replacement of drawn parts). If at least one part is faulty, then the delivery is rejected. If all 5 parts are o.k., then they are returned to the lot, and a sample of size 10 is randomly taken from the original 25 parts. The delivery is rejected if at least 1 part out of the 10 is faulty.
Determine the probabilities that a delivery is accepted on condition that (1) the delivery contains 2 defective parts,
(2) the delivery contains 4 defective parts.

 

[Deleted member 4993]

A delivery of 25 parts is evaluated for quality purposes according to the following scheme: A sample of size 5 is drawn (without replacement of drawn parts). If at least one part is faulty, then the delivery is rejected. If all 5 parts are o.k., then they are returned to the lot, and a sample of size 10 is randomly taken from the original 25 parts. The delivery is rejected if at least 1 part out of the 10 is faulty.
Determine the probabilities that a delivery is accepted on condition that (1) the delivery contains 2 defective parts,
(2) the delivery contains 4 defective parts.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

 

[Rose_01298]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

This is where I have gotten so far.
Event A: delivery is accepted. Event A is the intersection of samples of (a_1) 5 and (a_2) 10 being fault free. So in other words event A = a_1 U a_2.
Event B: delivery has 2 defective parts
Event C: delivery has 4 defective parts
So part 1 wants to know P(A|B) and part 2 wants to know P(A|C). But now I am stuck here because I am not sure how to calculate P(A|B) and P(A|C) with the given information or how to plug the correct numbers in

 

[Dr.Peterson]

A delivery of 25 parts is evaluated for quality purposes according to the following scheme: A sample of size 5 is drawn (without replacement of drawn parts). If at least one part is faulty, then the delivery is rejected. If all 5 parts are o.k., then they are returned to the lot, and a sample of size 10 is randomly taken from the original 25 parts. The delivery is rejected if at least 1 part out of the 10 is faulty.
Determine the probabilities that a delivery is accepted on condition that (1) the delivery contains 2 defective parts,
(2) the delivery contains 4 defective parts.

This is where I have gotten so far.
Event A: delivery is accepted. Event A is the intersection of samples of (a_1) 5 and (a_2) 10 being fault free. So in other words event A = a_1 U a_2.
Event B: delivery has 2 defective parts
Event C: delivery has 4 defective parts
So part 1 wants to know P(A|B) and part 2 wants to know P(A|C). But now I am stuck here because I am not sure how to calculate P(A|B) and P(A|C) with the given information or how to plug the correct numbers in

I wouldn't think too deeply in terms of conditional probabilities. Instead, for part 1, just think like this: Suppose a delivery contains two defective parts out of 25. We draw a sample of 5; what is the probability that at least one is faulty? What is the probability that all 5 are good, AND a sample of 10 contains at least one that is faulty? Then put it together.

If I named events, it would be A = at least one of the sample of 5 is faulty, and B = at least one of the sample of 10 is faulty. You have misstated the event you want; it is not quite the intersection (as the second sample is not always taken), and an intersection is not written as "U".

 

[Steven G]

This is where I have gotten so far.
Event A: delivery is accepted. Event A is the intersection of samples of (a_1) 5 and (a_2) 10 being fault free. So in other words event A = a_1 U a_2.

You wrote AND but in the set notation you wrote U which is used for or.
You really should learn the difference between AND and OR