Question 485450: Three screws are drawn at random from a lot of 200 screws, 10 of which are defective. Find the probability that the screws drawn will be nondefective in drawing (a) with replacement, (b) without replacement.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
a) with replacement. The probability on each draw is constant, so use the Binomial distribution.
The probability of  successes in  trials where  is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where ) is the number of combinations of things taken at a time and is calculated by !})
You would be looking for:
\ =\ \left(3\cr 0\right\)\left(\frac{1}{20}\right)^0\left(\frac{19}{20}\right)^{3})
Noting that pick 0 (and pick for that matter) equals 1 for all positive integers . Also  for all real  , so the calculation reduces to:
\ =\ \left(\frac{19}{20}\right)^{3})
You can do your own arithmetic.
b)
The probabilities for each draw vary because you don't replace the drawn item.
The probability of a non-defective item on the first draw where there are 10 out of 200 defective is 
Assuming that a good item was drawn on the first draw, the new probability for the second draw becomes  . You decreased the total number of items by one, so the denominator is decremented, but since you did not select a defective item, the number of non-defective items remains at a constant 190. (If you had selected a defective item on the first draw, game over -- we are calculating the probability of drawing non-defectives.)
Again, assuming a non-defective on draw 2, decrement the denominator and keep the numerator constant.
Then multiply the three probabilities:
\left(\frac{190}{199}\right)\left(\frac{190}{198}\right))
Again, you are on your own with the arithmetic.
John
My calculator said it, I believe it, that settles it
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