The answer has to be subtracted from 100% to get the probability that one or more will show up.
In fact, as MNdude ultimately explained his problem, he wanted the calculation of HOW MANY of 1000 'callees' he would ultimately reach, given a 0.16 (or so...he was vague) probability of reaching someone on a given call, and given that he would call them no more than six times total, and given (reasonably) that he would not call them again, having reached them once.
Well, of course, this is a different problem from the "fives" problem being thrown around here (and, in an AWFUL lot of cases, thrown around hilariously badly), but the solution to his actual problem is simple, if one is familiar with the actual discipline of probability.
Just FYI, I learned probability from Don Feder and Pete Weiner at Yale in 1970-1972, and have employed the science in my business enterprises ever since. I believe that I am quite competent offering a solution to simple problems such as these...and am entirely willing to back that view with some reasonable wager with ANYONE here. Proceeds to Free Republic, of course.
It really does become tiresome when FReepers either don't understand the question being asked or go off on some tangent or other when they post a 'solution'. Hence my 'sigh...' in the previous post.
You will have seen, I'm sure, the assorted calculations of 6^6 on this thread. That's fine, and they're accurate, but utterly irrelevant, and 6^6 plays no part whatever in ANY of the possible iterations of MNdude's original question or his later revision of what he 'really wanted to know'.
Best to you, and FReegards!