Find the probability of a defective item in a small sample

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Homework Help Overview

The problem involves calculating the probability of drawing one non-defective item from a box containing both defective and non-defective items. The context is centered around probability theory and combinatorial analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of calculating probabilities using combinations and the implications of drawing items without replacement. There are attempts to share calculations and reasoning about the probability of drawing one non-defective and one defective item.

Discussion Status

The discussion is ongoing, with participants providing calculations and questioning the accuracy of their numbers. Some guidance has been offered regarding the need to consider the dependence of events in the context of drawing without replacement.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a mention of practical examples to illustrate the problem, which may influence their reasoning.

lorik
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Homework Statement


In a box there are 25 non defective details and 28 defective details find propability from two random extraction that
c) One of them is non-defective


Homework Equations





The Attempt at a Solution


I think I know the answer but I need explanation.
 
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If you think you know the answer, how about sharing your solution with us?
 


awkward said:
If you think you know the answer, how about sharing your solution with us?
permutation


C25(1) * C28(1)/ C53(2) = 700/(2756/2) =1400/2756

But still need explenation though.
 


You have the right idea but you need to check your numbers.

There are C(53,2) ways to draw 2 items from the box. We assume all of these are equally likely. Of these, how many consist of one defective and one non-defective?
 


awkward said:
You have the right idea but you need to check your numbers.

There are C(53,2) ways to draw 2 items from the box. We assume all of these are equally likely. Of these, how many consist of one defective and one non-defective?

It should be about 50% chance of one consisting non defective and the other being defective I THINK

in practice assume we have 50 balls 25 good and 25 bad ,we draw 2 balls and it should be G1/2 * B1/2 ?
 


lorik said:
It should be about 50% chance of one consisting non defective and the other being defective I THINK

in practice assume we have 50 balls 25 good and 25 bad ,we draw 2 balls and it should be G1/2 * B1/2 ?
Close but not quite right. The complication is that the draws are not independent events. Since you are drawing balls without replacement, if you draw a good ball on the first draw it changes the probability that you will draw a bad draw on the second draw. Numerically, there won't be much difference, in this case, though-- you could say the draws are "almost independent".

One way to approach this type of problems is the way I outlined previously. All C(50,2) combinations of 50 balls taken two at a time are equally likely. So count the number of ways to draw one good and one bad, then divide by C(50,2).
 

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