Going crazy over probability problem

#1
I'm working through the book Probability Demystified, which has been an okay refresher on basic probability, but is known to contain a lot of errors. Problem is, they don't publish errata. I'm wondering if someone would post a solution to this problem, because I think the one in the book is incorrect. The question is:

A phone extension consists of 3 digits. If all the digits have a probability of being selected, what is the probability that the extension consists of the digits 1, 2, and 3 in any order? Repetitions are allowed.

The book answer is 0.006, which looks like they did permutations of the digits 1-3 divided by 10^3. This doesn't seem to fit the problem to me. I think it would be (3/10)^3, which is 3^3 (the # of events where each slot has 1-3 with replacement) divided by 10^3 (the total # of events in the sample space). I'm probably overlooking something simple, but I don't trust this book anymore either. Thanks for any help.
 

Dason

Ambassador to the humans
#2
The way I read it I agree with your solution. It should be .027.

However, if the book intended that "the extension consists of the digits 1, 2, and 3 in any order" means that you need to have each of the digits 1, 2, and 3 in the number itself then the "repetitions are allowed" is just superfluous and slightly misleading but not incorrect and the book's solution is correct.

However, I'm going to say you're right and the book is wrong.
 
#3
Ok, thanks. I have probably just been staring at numbers too long the past few days, or I wouldn't have dwelled on this. Sometimes word problems throw me for a loop, though, so I appreciate an outside perspective. :)