TITLE: How can I determine the probability of a number in a random sequence being of a cerrtain type?
QUESTION [5 upvotes]: I would like to produce some kind of probability function/density for finding out the likelihood of a chosen number, $k$, to occur in a random sequence of length, $N$.

For Example:
Let's say we have a sequence of $N=1000$, random numbers in the range $\in [0,1]$. Furthermore, let's truncate the random numbers to have only 3 significant digits (0.001).
How can I calculate how many times a particular "even" modulo number is expected to occur, such that we have the probability of:  

(a) $k \in [0.1, 0.2, ..., 0.9]$ 
(b) $k \in [0.01, 0.02, ..., 0.09]$ 
(c) $k \in [0.02, 0.05, 0.08]$


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An integer is chosen at random from the first 1000 positive integers. Probability that is a multiple of both 6 and 8?
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PS. It's quite likely I have not stated the title of this question correctly, please feel free to adjust it.

REPLY [2 votes]: Since you are looking only at numbers with three decimal places, your sample space has just $1000$ possible outcomes. For any particular set of $n$ outcomes the probability that each choice is in that set is just $p = n/1000$. (Your examples have $n= 9$, $9$ and $3$.) Since the choices are independent, the number of successes is a binomial distribution.
The expected number of successes in $N$ tries is then just $Np$.