Find the probability of a defective item in a small sample

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SUMMARY

The probability of drawing one non-defective and one defective item from a box containing 25 non-defective and 28 defective items is calculated using combinatorial methods. The total number of ways to draw two items from the box is C(53, 2). The successful outcomes for drawing one non-defective and one defective item can be represented as C(25, 1) * C(28, 1). The final probability is derived by dividing the successful outcomes by the total outcomes, leading to a definitive calculation of approximately 0.507, or 50.7% chance.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations (C(n, k))
  • Knowledge of probability theory, particularly in calculating probabilities of dependent events
  • Familiarity with the concept of drawing without replacement
  • Basic statistical reasoning to interpret outcomes and probabilities
NEXT STEPS
  • Study combinatorial probability using C(n, k) for various scenarios
  • Learn about dependent and independent events in probability theory
  • Explore practical applications of probability in real-world scenarios
  • Review examples of drawing without replacement and its impact on probability calculations
USEFUL FOR

Students studying probability and statistics, educators teaching combinatorial methods, and anyone interested in understanding the principles of drawing items from a finite set without replacement.

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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