Probability Finding Binomial Average Problem

[caljuice]

Homework Statement

In each batch of manufactured articles, the proportion of defective articles is p. From each
batch, a random sample of nine is taken and each of the nine articles is examined. If one article is found to be defective, the batch is rejected; otherwise, it is accepted. If rejected, a second sample of nine is taken and also rejected if an error is found. Otherwise it is accepted.

Evaluate the average number sampled per manufactured batch over a large number of batches when p has the value 0.1. [4 marks] (Note: Just to be sure you understand what this is asking, you are asked to calculate the average number of articles sampled per manufactured batch over the long run.)

I condensed the question here is the full incase I worded it poorly:

Spoiler

In each batch of manufactured articles, the proportion of defective articles is p. From each
batch, a random sample of nine is taken and each of the nine articles is examined. If two
or more of the nine articles are found to be defective, the batch is rejected; otherwise, it
is accepted. Show that the probability that a batch is accepted is (1−p)
8
(1+8p). [3 marks]
It is decided to modify the sampling scheme so that when one defective is found in the
sample, a second sample of nine is taken (each of the nine articles is examined) and the
batch rejected if this contains any defectives. With this exception, the original scheme is
continued. Find an expression in terms of p for the probability that a batch is accepted.
[3 marks].
For this modified scheme, evaluate the average number sampled per manufactured batch
over a large number of batches when p has the value 0.1. [4 marks] (Note: Just to be
sure you understand what this is asking, you are asked to calculate the average number of
articles sampled per manufactured batch over the long run.)

Attempt:
Any hints on the process?

My guess:

The probability for the # of error is easy using binomials.

It's discrete binomial question so I was thinking about making a table to find E(X) where X = # of error,

average # error = E(X) = X1*P1 + X2*P2 ... X18*P18 , 18 since 2 samples and professor hint was answer is between 9 - 18.

E(X) = np since binomial, where n = 18?
E(X) / p = n , which is the answer?

Thanks for any help.

 

[HallsofIvy]

Homework Statement

In each batch of manufactured articles, the proportion of defective articles is p. From each
batch, a random sample of nine is taken and each of the nine articles is examined. If one article is found to be defective

Do you mean "at least one"?

, the batch is rejected; otherwise, it is accepted. If rejected, a second sample of nine is taken and also rejected if an error is found. Otherwise it is accepted.

What happens if the batch is rejected a second time? Do you do only one or two samples?

Evaluate the average number sampled per manufactured batch over a large number of batches when p has the value 0.1. [4 marks] (Note: Just to be sure you understand what this is asking, you are asked to calculate the average number of articles sampled per manufactured batch over the long run.)

I condensed the question here is the full incase I worded it poorly:

Spoiler

In each batch of manufactured articles, the proportion of defective articles is p. From each
batch, a random sample of nine is taken and each of the nine articles is examined. If two
or more of the nine articles are found to be defective, the batch is rejected; otherwise, it
is accepted. Show that the probability that a batch is accepted is (1−p)
8
(1+8p). [3 marks]
It is decided to modify the sampling scheme so that when one defective is found in the
sample, a second sample of nine is taken (each of the nine articles is examined) and the
batch rejected if this contains any defectives. With this exception, the original scheme is
continued. Find an expression in terms of p for the probability that a batch is accepted.
[3 marks].
For this modified scheme, evaluate the average number sampled per manufactured batch
over a large number of batches when p has the value 0.1. [4 marks] (Note: Just to be
sure you understand what this is asking, you are asked to calculate the average number of
articles sampled per manufactured batch over the long run.)

Attempt:
Any hints on the process?

My guess:

The probability for the # of error is easy using binomials.

It's discrete binomial question so I was thinking about making a table to find E(X) where X = # of error,

average # error = E(X) = X1*P1 + X2*P2 ... X18*P18 , 18 since 2 samples and professor hint was answer is between 9 - 18.

E(X) = np since binomial, where n = 18?
E(X) / p = n , which is the answer?

Thanks for any help.

 

[caljuice]

Sorry I meant at least one. I was a little confused on the initial wording too, but I believe it is a max of two samples.

 

[Jotamide]

I see someone else is also doing the STAT 251 HW late. :tongue: