Question 485450
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a) with replacement.  The probability on each draw is constant, so use the Binomial distribution.


The probability of *[tex \Large k] successes in *[tex \Large n] trials where *[tex \Large p] is the probability of success on any given trial is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE \left(n\cr k\right\)] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


You would be looking for:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3\left(0,\frac{1}{20}\right)\ =\ \left(3\cr 0\right\)\left(\frac{1}{20}\right)^0\left(\frac{19}{20}\right)^{3}]


Noting that *[tex \Large n] pick 0 (and *[tex \Large n] pick *[tex \Large n] for that matter) equals 1 for all positive integers *[tex \Large n].  Also *[tex \Large a^0\ =\ 1] for all real *[tex \Large a], so the calculation reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_3\left(0,\frac{1}{20}\right)\ =\ \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 *[tex \Large \frac{190}{200}]


Assuming that a good item was drawn on the first draw, the new probability for the second draw becomes *[tex \Large \frac{190}{199}].  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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{190}{200}\right)\left(\frac{190}{199}\right)\left(\frac{190}{198}\right)]


Again, you are on your own with the arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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