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May 9th, 2013, 07:02 PM  #1 
Member Joined: Mar 2012 Posts: 60 Thanks: 0  probability
Is it right to break up this into two cases? a computer prints out three digits at random, 09 inclusive. Find the probability that the third digit is different from the first two. Case 1 first two digits are the same, third is different 10 x 9 =90 Case 2 first two are different, and third is different 10 x 9 x 8 = 720 P third is different = case 1 + case 2 = 810 / (10 x 10 x 10) or 81/100 
May 10th, 2013, 01:06 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
Your method is correct. Alternatively, P(1st digit differs from 3rd digit)P(2nd digit differs from 3rd digit) = (9/10)² = 81/100. 

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