The probability of 
successes in 
trials where the probability of success on any given trial is 
is given by:
\ =\ {{n}\choose{k}}\(p\)^k\(1\,-\,p\)^{n\,-\,k})
Where 
the number of ways to choose things from a group of things where the order of selection does not matter. AKA, the binomial coefficient.
If the probability of being less than, less than or equal, greater than, or greater than or equal to some value 
is calculated by one of the following sums. (note that the only thing that changes is that the summation is re-indexed)
\ =\ \sum_{k=0}^m\,{{n}\choose{k}}\(p\)^k\(1\,-\,p\)^{n\,-\,k})
\ =\ \sum_{k=0}^{m-1}\,{{n}\choose{k}}\(p\)^k\(1\,-\,p\)^{n\,-\,k})
\ =\ \sum_{k=m}^n\,{{n}\choose{k}}\(p\)^k\(1\,-\,p\)^{n\,-\,k})
\ =\ \sum_{k=m+1}^n\,{{n}\choose{k}}\(p\)^k\(1\,-\,p\)^{n\,-\,k})
Since
\ =\ \sum_{k=0}^n\,{{n}\choose{k}}\(p\)^k\(1\,-\,p\)^{n\,-\,k}\ =\ 1)
That is, the probability of at least 0 successes is a certainty, i.e. = 1. The following relationships may provide some computational relief because the required sum may have fewer terms:
\ =\ 1\ -\ P(X{>}m,n,p))
\ =\ 1\ -\ P(X{\geq}m,n,p))
Also, note the following identities which will certainly ease the computational burden:
For your situation, 
and 
. That's the best I can do since you didn't actually state what "these probabilities" are. But in truth, this was the answer I would have given you anyway because I don't generally do arithmetic for people. If you have several of these to calculate and you have access to Excel or an equivalent spreadsheet program, note that, with the appropriate values inserted for 
and
:
=BINOMDIST(k,n,p,false) yields )
=BINOMDIST(k,n,p,true) yields )
=1 - BINOMDIST(k,n,p,true) yields \ =\ P(X > m,n,p))
John
My calculator said it, I believe it, that settles it
