How use probability distribution to generate a binary array with k number 1s acording a range of lower and upper bounds I need to generate a binary array of size N with K numbers 1 and (n-k) numbers 0 on it,does not matter in which position.     The amount of K numbers 1s belongs to an interval  Min <= K <= Max ,  and  Max <= N .
Since these values is represented by a binary variable, the probability of array position 
has 0 is P(x=0) = 0.5 and it has 1 is P(x=1) = 1-P(x=0) = 0.5.
Let's suppose an array size of N=100 must have K numbers 1s on it according  20 <= K <= 30 ,i.e.  Min=20 and Max=30.
My first idea to use a probability function to generate the values for this array was developed according to the following considerations, which I'm not sure if it is correct :
(1) for the MIN constraint K>=20, since the probability of 1s is P(x=1)=0.5 and it is required that at 
    least 20 positions been filled with 1s then , at least 20/100 =20% , (Min/N) positions must have 1s, 
    However, for a random binary variable P(X=1) = 0.5 , so how to combine this with the constraint of 
    20% must be 1 ?
(2) for the MAX constraint K<=30, it means that the (100 - 30) remaining positions must be 0, i.e. 
    70/100 = 70%,  (1 - Max/N) positions must have zero.  However, for a random binary variable 
    P(X=0)=0.5, how to combine this with the constraint of 70% must be 0 ? 
(3) the interval  20 <= K <= 30 , can be either 0 or 1,  i.e. (30 - 10)/100= 20% , (Max - Min)/N 
    positions can be filled 0 or 1.
After found the probability of 1s "prob_one", I'd like to write my code using a "rand" function (which returns a random number between 0 and 1) to fill each position of the binary array like this :
"if rand <= prob_one then myarray[position] := 1 else myarray[position] = 0" . 
My Questions :
Q1) Do considerations (1) , (2) and (3) make sense ?
Q2) Is there any probability distribution function that can solve this problem and give a direct answer 
    like P(X=1) = prob_one   and P(X=0)= prob_zero  ?
I appreciate solutions, explanations, examples and material references on how to solve this. 
 A: The number of arrays which contains exactly $k$ times the digit $1$ is given by:
$$n(k) = {N \choose k} = \frac{N!}{k!(N-k)!},$$
where $a!$ is the factorial of the number $a$.
If you want to find the number of arrays with $20 \leq k \leq 30,$ then you must evaluate:
$$\sum_{k=20}^{30} n(k) = ...$$
In order to arrive to the same result using a probabilistic approach, 
you should first notice that the number of $N$-arrays composed by "0" and "1" digits only is $2^N$.
Moreover, the probability that one of these array contains exactly $k$ digits $1$ is given by the 
binomial distribution:
$$P(k) = {N \choose k}p^k (1-p)^{N-k},$$
where $p = \frac{1}{2}$ is the probability of having a $1$.
Therefore, the number of arrays which contains exactly $k$ times the digit $1$ is given by:
$$n(k) = 2^N \cdot P(k) = 2^N  {N \choose k}p^k (1-p)^{N-k} = 2^N  {N \choose k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{N-k} = {N \choose k} 2^{N - k - N + k} = {N \choose k},$$
which is exactly the same result presented at the beginning of this answer.
If you want to find the probability to extract one array with $20 \leq k \leq 30$, then you should evaluate:
$$\sum_{k=20}^{30} P(k) = \frac{1}{2^N}\sum_{k=20}^{30}n(k) = ...$$
