If I know the probability of success for an event, the number of time that I want to observe the event at a certain confidence level how do I calculate the number of trials needed?


Let's say your Success probability is $p=0.3,$ your target number of Successes is $k = 10$ and your target probability of fulfillment is 0.95.

Then you seek $n$ sufficiently large that $P(X \ge 10) = 0.95,$ given that $X \sim \mathsf{Binom}(n, p=0.3).$

For most values of $p,$ you could get a close answer by obtaining a normal approximation of $P(X \ge 10)$ and solving for $n.$

However, it is simple to search for $n$ using R. First, we guess that $n = 100$ might we large enough. So we use R to find that $P(X \ge 10) = 1 - P(X \le 9) = 0.9999996 > 0.95,$ using pbinom, which denotes a binomial CDF in R:

1 - pbinom(9, 100, .3) 
[1] 0.9999996

Also it is obvious that $n \ge 10.$ So we look at all values of $n$ between 10 and 100 inclusive, and pick the smallest $n$ that works. [In R, n is a numerical vector of length 91, thus bpr is a numerical vector of length 91, and bpr >= .95 is a logical vector of length 91, containing TRUEs and FALSEs, You can read the bracket notation [ ] as "such that." ]

n = 10:100;  k = 10;  p= .3
bpr = 1 - pbinom(k-1, n, p)
min(n[bpr >= .95])
[1] 49

Thus $n = 49$ should be barely large enough and $n = 48$ not quite large enough. We verify that this is true:

1 - pbinom(9, 49, .3)
[1] 0.9520446
1 - pbinom(9, 48, .3)
[1] 0.9430372
  • $\begingroup$ Thank you for the detailed answer. Does this problem/test have a name? $\endgroup$ – nex Jul 18 at 16:19

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