Centrino Posted October 14, 2005 Share Posted October 14, 2005 3 positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer. a) 9/28 b) 1/28 c) 1/3 d) 5/14 e) 9/29 D Quote Link to comment Share on other sites More sharing options...
chix475ntu Posted October 15, 2005 Share Posted October 15, 2005 Centrino .. where did you get this question.. seems to be HARDDDDDDDDDDDDDDDD BIN ... struggled for sometime but think this is the approach... i got the answer with this approach.. the 9 numbers are 3 are in the form 3n ( multiples of 3) ---- SET 1 3 are in the form 3n+1 ---- SET 2 3 are in the form 3n+2 ---- SET 3 if you take the 3 numbers from the same set then the sum is divisible by 3 .. so we got 3 combinations here for 3 sets .. the only other way is taking 1 from each set ... so 3C1 * 3C1 * 3C1 = 27 [ this is like 3n + 3n+1+3n+2 = 9n+3 which is divisible by 3] so total combinations = 3+27 = 30 probability = 30/84 = 5/14 D 1 Quote Link to comment Share on other sites More sharing options...
Centrino Posted October 15, 2005 Author Share Posted October 15, 2005 Many thanks Chix.........I liked your logic. Quote Link to comment Share on other sites More sharing options...
CTG1983 Posted October 16, 2005 Share Posted October 16, 2005 It's probably the hardest probability problem. Is there any easier approach? Quote Link to comment Share on other sites More sharing options...
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