Dear GAP-Forum, and Stas,

as an alternative to Sven's solution (which will almost surely come up
with a matrix of a prescribed weight; I don't know if that's what you
want?), you could `flip an unfair coin' for every entry:

r:=[1..10];
p:=2;
coin:=function() if Random(r)>p then return 0*Z(2); else return Z(2); fi; end;
n1:=5;
n2:=5;
A:=List([1..n1],i->List([1..n2],j->coin()));

This gives a matrix in which on average 20% of the entries are ones.

This is probably less efficient than Sven's solution, but the underlying
probability model is easier. So it depends on your application which
solution to use.

Best regards,

Jan