Binomial Distribution

Binomial distribution is the distribution of a total number of successes in a given number of Bernoulli trials. The common notation is b(k; n, p), where k is the number of successes, n is the number of trials, p is the probability of success. We know that b(k; n, p) = C(n, k) pk(1 - p)n - k.


If you are reading this, your browser is not set to run Java applets. Try IE11 or Safari and declare the site as trusted in the Java setup.

Binomial Distribution

What if applet does not run?

For a fixed number of trials n, the binomial distribution always behaves in the same way: as a function of k, it is monotone increasing up to a certain point m after which (perhaps with an exception of the next point) it is monotone decreasing.


 b(k; n, p) / b(k-1; n, p)= C(n, k) pk(1 - p)n - k / C(n, k-1) pk-1(1 - p)n - k + 1
(1) = (n - k + 1) p / k(1 - p)
  = 1 + ((n + 1) p - k) / k(1 - p).

It is clear now that the right hand side in (1) is greater than 1 whenever k > (n + 1) p and it is less than 1 when k < (n + 1) p. It may happen, of course, that m = (n + 1) p is an integer in which case

 b(m; n, p) = b(m - 1; n, p).

In any event, there is only one integer m that satisfies

(2)(n + 1) p - 1 < m ≤ (n + 1) p.

Summing up, as a function of k, the expression b(k; n, p) is monotone increasing for k < m and monotone decreasing for k > m, with the exception of one case where (n + 1) p is an integer. In this case, there are two maximum values for m = (n + 1) p and m - 1.

The number m that satisfies (2) is known as the most probable (most likely) number of successes in n Bernoulli trials. As a matter of fact, b(m; n, p) is quite small for a large n, even for a reasonable value of p. It is also always different from the average number of successes. The latter could be found the following way. Letting q = 1 - p, we get

 ∑ k b(k; n, p)= ∑ k C(n, k) pkqn - k
  = pqn-1∑ k C(n, k) (p/q)k - 1
  = pqn-1 n (1 + p/q)n - 1
  = pqn-1 n (q + p)n - 1 / qn - 1
  = p n,

where we used the identity

 n (1 + x)n - 1= ∑ k C(n, k) xk - 1.

The result for the expected value np = ∑ k b(k; n, p) might have been anticipated given the interpretation of the probability as a relative frequency.

Note that the expected value np is always different from the most likely value (n + 1) p, provided of course p ≠ 0.

|Contact| |Front page| |Contents| |Probability| |Activities|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: