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 Latest Posts Is there anybody out there? by True 01/07/20 09:26 AM
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 #10254 - 11/08/05 12:35 AM Re: IQ and Intelligence Count Iblis II Senior Member Registered: 01/21/05 Posts: 375 Here is how I did it (I tried to post this before but I got an error message due to the symbol for ''smaller'' or ''larger'' being recognised as an illegal HTML command.):Write the probability as an integral over the arrival times denoted as t1, t2 and t3 for the three ladies, respectively. Imposing an ordering t1 smaller than t2 smaller than t3 would reduce the probability by a factor of 6 by symmetry so you can solve theproblem using this ordering and then multiply by 6. Then you split the problem for t1 below 5/6 and t1 between 5/6 and 1. In the first case t2 goes from t1 to t1 + 1/6 and t3 goes from t2 to t1 + 1/6 and in the second case t2 goes from t1 to 1 and t3 goes from t2 to 1.To simplify the calculations you partially undo the imposed ordering that led to the factor 6. If you allow t2 to be come larger than t3 (i.e. you add contributions in the two cases for t2 and t3 interchanged), then you must replace the factor 6 by a factor 3 and the two cases become:1) t2 goes from t1 to t1 + 1/6 and t3 goes from t1 to t1 + 1/62) t2 goes from t1 to 1 and t3 goes from t1 to 1.The first case yields a probability of 5/6*1/6*1/6*3The second case yields 1/3 (1/6)^3 * 3Which gives the result of 2/27 Top
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