Mathematics.Statistics - ChatArea.com

Mathematics Forum

Author	Message / Information
robinclp Quote \| Reply \|	Simple Question... hard to prove posted on: 11/17/2006 6:00:38 PM 8 contractors, 3 chosen at random. You are one of the 8. Shouldn't the answer that comes straight to mind be 3/8? But my teacher insists that I wrote (1C1*7C2)/8C3. This is ridiculous. My teacher insists that I come up with a working to prove my answer but my way of thinking should be right but I dont know how to prove it. Thanks
mathgeek Quote \| Reply \|	Simple Question... hard to prove replied on: 3/15/2007 11:36:49 AM One way you could prove your answer is like this: First ask what is the probability of not getting picked? You have a 7/8 probability of not getting picked first. You have a 6/7 probability of not getting picked second. And you have a 5/6 probability of not getting picked third. Using the multiplication prinicple: 7/8 * 6/7 * 5/6 = (7 * 6 * 5)/(8 * 7 * 6) = 5/8. Since the probability of not getting picked is 5/8 then the probability of getting picked must be 1 - 5/8 = 3/8.
Pythagoras Quote \| Reply \|	Simple Question... hard to prove replied on: 3/17/2007 10:12:29 AM (nice smooth answer) (i should have been an expert at Probability by now!...)
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