So Pr(r≤k;n,p)
=Pr(r;n,p)+Pr(r+1;n,p)+....+Pr(n;n,p)
=C(n,r)p^r(1-p)^(n-r) + C(n,r+1)p^(r+1)(1-p)^(n-1-r) + ... + C(n,n)p^n(1-p)^0 as you have.
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Explanation for the formula Pr(k;n,p)=(n k)•p^k•(1-p)^(n-k), i.e. for determining the probability of EXACTLY k success in n trials where the probability of success is p.
You are correct that if you n trials, you can denote them (T1,T2,...,Tn).
Each trial is either a success 0 or a failure 1 (this is the usual notation).
So we could write the result of n trials as a combination of 0 and 1s.
For example: (0,0,0,1,1,...,1) would mean that the first 3 trials were successes and the rest of the n-trials were failures.
The idea is that if you are counting the number of successes k in n trials then the following combinations would all be "good" for k=3: (0,0,0,1,...,1), (0,0,1,0,1,....,1), etc. That is, any combination with 3 "zeros" and (n-3) "ones" must be counted.
A. What is the probability of any one of these combinations?
Of course we look at the easiest one for calculating which is the first one.
Pr[(0,0,0,1,...,1)]=p³ • (1-p)^(n-3).
But you can see that this works for any combination with 3 "zeroes" and (n-3) "ones".
B. How many such combinations are there with 3 "zeros" and (n-3) "ones"?
This is combinatorics and the answer is the binomial coefficient:
"n choose 3" =(n 3)=n!/(3!•(n-3)!)
So now we multiply A•B to get the probability of 3 successes in n trials where the probability of success is p to be:
Pr(3;n,p)=p³ • (1-p)^(n-3)•n!/(3!•(n-3)!) = (n 3)•p³ • (1-p)^(n-3)
Now just substituting k for 3, we get the standard formula for EXACTLY k successes in n trials.
Pr(k;n,p)= (n k) •p^k • (1-p)^(n-k)