Question 1163782
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The probability that the first box that is chosen has the right gift is the number of boxes that have the gift you want divided by the total number of boxes on the shelf.  Given that the first box actually has the right thing, then when you choose the second box, there is one less box that has the right gift and one less box total, changing the probability on the second draw.  Since you have accounted for the change in values from one draw to the other, the events are independent and the multiplication rule applies. 


																
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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