MHB Help with this double sampling scheme problem

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The discussion revolves around calculating the probability of having at least 8 tightly packed items in a combined sample of 10 from a food production process. The first sample consists of 5 items, and a second sample is taken if the first sample has 2 or fewer tightly packed items. Participants clarify the conditions under which the second sample is taken and the necessary binomial probabilities to consider for the calculations. The correct approach involves summing the probabilities of different combinations of tightly packed items from both samples. The conversation highlights the importance of accurately interpreting the sampling conditions to arrive at the correct probability.
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Please help with a hint of doing his question, I have drawn the diagram but still confused.


Q)In a food production process, packaged items are sampled as they come off a
production line. A random sample of 5 items from each large production batch is
checked to see whether each item is tightly packed. A batch will be accepted
immediately if all of these 5 items are tightly packed, and rejected immediately if at
least 3 items among the 5 are not tightly packed. Otherwise a second sample is taken
before making a decision.
Suppose that 80% of the items produced by the machine are tightly packed

a)When a second sample is to be taken, it also consists of 5 items. What is the
probability that in a combined sample of 10 items there are at least 8 tightly
packed?

Thanks.
 
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statrw said:
Please help with a hint of doing his question, I have drawn the diagram but still confused.


Q)In a food production process, packaged items are sampled as they come off a
production line. A random sample of 5 items from each large production batch is
checked to see whether each item is tightly packed. A batch will be accepted
immediately if all of these 5 items are tightly packed, and rejected immediately if at
least 3 items among the 5 are not tightly packed. Otherwise a second sample is taken
before making a decision.
Suppose that 80% of the items produced by the machine are tightly packed

a)When a second sample is to be taken, it also consists of 5 items. What is the
probability that in a combined sample of 10 items there are at least 8 tightly
packed?

Thanks.

When a second sample is taken there are 3 or 4 tightly packed items in the first sample, so for at least 8 to be tightly packed we need 4 or 5 to be tightly packed in the second sample if only one was not tightly packed in the first sample and 5 if two were not tightly packed, so the required probability is:

\(p = (b(4;5,0.8) + b(5;5,0.8)) \times b(4;5,0.8) + b(5;5,0.8) \times b(3;5,0.8)\)

where \(b(n;N,x)\) denotes the binomial probability on \(n\) successes in \(N\) trials with a success probability of \(x\) in each trial.

CB
 
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CaptainBlack said:
When a second sample is taken there are 4 tightly packed items in the first sample, so for at least 8 to be tightly packed we need 4 or 5 to be tightly packed in the second sample, so the required probability is:

p = b(4;5,0.8) + b(5;5,0.8)

where b(n;N,x) denotes the binomial probability on n successes in N trials with a success probability of x in each trial.

CB

Thanks but not sure how you got the 4 or 5 tightly packed to add up to 8. Remember for second sample to be taken there must be 0 or 1 or 2 tightly packed.

I may have missed the point,but could you explain.
Thanks.
 
statrw said:
Thanks but not sure how you got the 4 or 5 tightly packed to add up to 8. Remember for second sample to be taken there must be 0 or 1 or 2 tightly packed.

I may have missed the point,but could you explain.
Thanks.

Sorry, misread the condition for drawing a second sample, previous post has now been corrected.

CB
 
Last edited:
CaptainBlack said:
When a second sample is taken there are 3 or 4 tightly packed items in the first sample, so for at least 8 to be tightly packed we need 4 or 5 to be tightly packed in the second sample if only one was not tightly packed in the first sample and 5 if two were not tightly packed, so the required probability is:

\(p = (b(4;5,0.8) + b(5;5,0.8)) \times b(4;5,0.8) + b(5;5,0.8) \times b(3;5,0.8)\)

where \(b(n;N,x)\) denotes the binomial probability on \(n\) successes in \(N\) trials with a success probability of \(x\) in each trial.

CB

Not sure about your suggestion above, here is what I think, see below.
Any opinions?To get a total of 8 tightly packed from the first and second selection we need to add up binomial probabilities.

1) x =0 from n1 and x=8 from n2
2) x=1 from n1 and x=7 from n2
3) x=2 from n1 and x=6 from n2

For each case use binomial B(5, 0.8), where B(n,p).
 
statrw said:
Not sure about your suggestion above, here is what I think, see below.
Any opinions?To get a total of 8 tightly packed from the first and second selection we need to add up binomial probabilities.

1) x =0 from n1 and x=8 from n2
2) x=1 from n1 and x=7 from n2
3) x=2 from n1 and x=6 from n2

For each case use binomial B(5, 0.8), where B(n,p).

The second sample is of size 5.

CB
 
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