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May 9th, 2013, 07:02 PM   #1
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probability

Is it right to break up this into two cases?
-a computer prints out three digits at random, 0-9 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
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May 10th, 2013, 01:06 AM   #2
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Your method is correct.

Alternatively, P(1st digit differs from 3rd digit)P(2nd digit differs from 3rd digit) = (9/10) = 81/100.
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