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March 30th, 2016, 06:48 PM  #1 
Newbie Joined: Feb 2016 From: Colorado Posts: 28 Thanks: 0  Sample Space
Identify the sample space of the probability experiment: rolling a pair of 12sided dice (with sides numbered 112) and observing the total number of points of each roll. Is this question asking that I do (1,1) (1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,(1,9) and so forth? 
March 31st, 2016, 05:20 AM  #2 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77  
March 31st, 2016, 11:29 AM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 619 Thanks: 83 
There are 12*12 = 144 possible outcomes. Here are the frequencies for each point total: 2: 1 3: 2 4: 3 5: 4 6: 5 7: 6 8: 7 9: 8 10: 9 11: 10 12: 11 13: 12 14: 11 15: 10 16: 9 17: 8 18: 7 19: 6 20: 5 21: 4 22: 3 23: 2 24: 1 Note that it's symmetrical. Regardless of how many sides the dice have, the mean, median, and mode are the same and are the sum of 1 and the highest number. 
March 31st, 2016, 01:10 PM  #4  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
If you flip a coin, A sample space is {H,T}. If you are interested in seeing whether the phases of the moon (call them Q1, Q2, Q3, and Q4) affect the con flip, A PERFECTLY ACCEPTABLE sample space is {HQ1, HQ2, HQ3, HQ4, TQ1, TQ2, TQ3, TQ4}. All that is important to be a sample space, is that every possibility is included in exactly one member of that space. But the point of having a sample space is to define a probability for each member. Some choices might make that easier. A sample space for your experiment is the set of integers from 2 to 24. But it isn't immediately obvious what the probabilities are. Another is the set of 12*12 combinations for the "first die" and the "second die." These are all equally probable so they have probabilities 1/144. You can use this to establish the probabilities for the first one I mentioned, if you can figure out how many of the second correspond to each of the first.  

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