I know that the probability mass function of the binomial distribution is

$f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = {n\choose k}p^k(1-p)^{n-k}$

This functions returns the probability of getting exactly k successes within n trials, with p being the probability of success for each trial.

The cumulative function is

$F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i}$

where $\displaystyle \scriptstyle \lfloor k\rfloor \,$ is the "floor" under k, i.e. the greatest integer less than or equal to k. But this cumulative function gives, as I understood, the probability of having the number of successes between 0 and k.

How to get a cumulative function which gives me, at least one success, that is, the probability of having the number of successes between 1 and n.

For example, someone throws an even dice. After throwing 1000 times, what's the probability of getting at least once, the number 6?

  • $\begingroup$ Why do you use the floor function? In my opinion it is not necessary, since k is an integer $\endgroup$ – callculus Jul 18 '18 at 20:10


$$P(X\geq k) = 1-P(X \lt k)$$


$$P(X\geq 1) = 1-P(X\lt 1) = 1- {1000\choose 0}\left(1\over 6\right)^0\left(1-{1\over 6}\right)^{999} = 1-({5\over 6})^{999}\approx 1$$ This "trick" is pretty obvious as it stands and it's pretty common to use it when dealing with calculating the cumulative probability of at least $m$ successes of big data sets when $m\ll n$


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