What is the probability that a randomly selected three-digit number is divisible by 11?
I know that the divisibility rule for 11 is to take the alternating sum of the digits in the number and see if it's divisible by 11, but how could this help?
+2862 First we try to find the number of 3 digit numbers.
100-999 that is 900 three digit numbers.
You do not need the divisibility rule.
In these problems justs divide 900 by 11, and you will get 81.81818181818181...
round
82
then
So your answer is 82/900 -> 41/450
There are: 999 - 100 + 1 =900 3-digit numbers.
There are: [990 - 110] / 11 + 1 =81 numbers that are multiples of 11 between 100 and 999
So, the probability is: 81 / 900 = 9%
+118609 Yes 900 3 digit numbers
1000/11 = 90.9
round down to 90
9 of those are less than 100
so 81 are between 100 and 999
so that is
P(that a 3 digit number is divisable by 11 ) = 81/900 = 9/100 = 9%
Just as guest said. 
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You rounded up instead of down so
instead of ending with the las multiple of 11 LOWER than 1000,
you ended with the first multiple of 11 AFTER 1000.
Oh, I do not like your 900/11 bit either. That could also have put the answer out by a little bit.
In this case it makes no answer difference but sometimes it would. Another thing for you to think about. 
