Question 1123451
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Let's say you have *[tex \Large n] things, *[tex \Large k] of which would be considered successful selections.  Then the probability of making a successful selection on the first try would simply be *[tex \Large \frac{k}{n}].  But if you don't replace the thing selected on the first try, you no longer have *[tex \Large n] things *[tex \Large k] of which would be considered successful, you now have *[tex \Large n\ -\ 1] things *[tex \Large k\ -\ 1] of which would be successful, and the probability for the second try would be *[tex \Large \frac{k-1}{n-1}].


Since by adjusting the number of items and the number of successes you have made the two tries completely independent events, the probability of both is the product of the two individual probabilities.


So, for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(2)\ =\ \(\frac{125}{350}\)\(\frac{124}{349}\)\ =\ \(\frac{5}{14}\)\(\frac{124}{349}\)]


You can do your own arithmetic


Extra Credit: *[tex \LARGE P(3)\ =\ ?]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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