Question 1163157
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Presuming you meant "four" black socks rather than "for" black socks as you wrote, it should be clear that, for the first draw, there are four outcomes that you would consider a success out of sixteen total outcomes -- giving you the probability that you draw a black sock on the first draw.  Presuming that you were successful on the first draw, how many socks would remain in the drawer and how many of those would be black.  Those numbers will give you the probability of drawing a black sock on the second draw.  Since these two events are independent because you accounted for the possibility of the first sock being black when you calculated the probability of the second sock being black, the total probability is the product of the probabilities of the two individual events.
								
								
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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