Find the probability of a defective item in a small sample

  • Thread starter Thread starter lorik
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
To find the probability of drawing one non-defective and one defective item from a box containing 25 non-defective and 28 defective items, the total combinations of drawing two items from 53 is calculated using C(53,2). The correct approach involves determining the combinations that consist of one non-defective and one defective item, which requires checking the calculations for accuracy. The draws are not independent due to the lack of replacement, meaning the outcome of the first draw affects the second. Ultimately, the probability can be approximated by counting the favorable outcomes and dividing by the total combinations.
lorik
Messages
38
Reaction score
0

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.
 
Physics news on Phys.org


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).
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top