What are the rules of binomial distribution

Binomial distribution

The binomial distribution is the most important distribution in high school. The prerequisite for the use of the binomial distribution is that a) the experiment consists of identical and mutually independent experiments and b) the experiments may have either "success" or "failure" as a result.

  • The binomial distribution is skewed to the left if p> 0.5, skewed to the right if p <0.5 and symmetric for p = 0.5 (see the comparison between binomial and normal distribution in the figure above right).
  • If n is sufficiently large, the normal distribution can be used as an approximation to the binomial distribution, since the skewness gets smaller with increasing n (for further comparisons with the normal distribution and rules of thumb on when the normal distribution can be used instead of the binomial distribution, see the article Normal distribution).

Example with explanation

According to the 2012 Federal Education Report, 33.9% of all German students in a given year graduate from university. How high is the probability that exactly 2 out of a group of 5 randomly selected students have obtained the university entrance qualification?

First we need to determine how many different ways there are two people can choose from a group of five. One possibility is that the first two selected students have graduated from high school (A) and the last three have not (N). Then we come to the following probability:

(0,339)(0,339)(0,661)(0,661)(0,661) = (0,339)²  ·  (0,661)³ ≈ 0.03319 = 3,319%

But there are nine more - so a total of 10 - different ways in which we can arrange two people within a group of five. We have to take all these arrangements into account:

  1. AANNN = (0.339) (0.339) (0.661) (0.661) (0.661) = (0.339) 2 x (0.661) 3
  2. ANANN = (0.339) (0.661) (0.339) (0.661) (0.661) = (0.339) ² * (0.661) ³
  3. ANNAN = (0.339) (0.661) (0.661) (0.339) (0.661) = (0.339) ² · (0.661) ³
  4. ANNNA = (0.339) (0.661) (0.661) (0.661) (0.339) = (0.339) 2 x (0.661) 3
  5. NAANN = (0.661) (0.339) (0.339) (0.661) (0.661) = (0.339) ² * (0.661) ³
  6. NANAN = (0.661) (0.339) (0.661) (0.339) (0.661) = (0.339) 2 x (0.661) 3
  7. NANNA = (0.661) (0.339) (0.661) (0.661) (0.339) = (0.339) 2 x (0.661) 3
  8. NNAAN = (0.661) (0.661) (0.339) (0.339) (0.661) = (0.339) 2 x (0.661) 3
  9. NNANA = (0.661) (0.661) (0.339) (0.661) (0.339) = (0.339) 2 x (0.661) 3
  10. NNNAA = (0.661) (0.661) (0.661) (0.339) (0.339) = (0.339) 2 x (0.661) 3

What is noticeable is that they all have the same probability. So we can simply multiply (0.339) ² · (0.661) ³ by 10. The probability of randomly choosing 2 high school graduates from a group of 5 students is therefore:

10 · (0,339)² · (0,661)³ ≈ 0.3319 = 33,19%

This task fulfills all requirements to be solved with the binomial distribution. For a problem to be solvable with the binomial distribution, some conditions must be met:

  1. There must be a fixed number of attempts (n)
  2. The probability p must remain constant
  3. The trials must be independent
  4. Each attempt may only have two different results: "Success" or "Failure"

In our example, it is a success when the student has graduated from high school.

definition

If a binomial distributed experiment consists of n attempts, each attempt having a probability of p, then the probability of k successes is:

The binomial coefficient calculates for us the number of possibilities how k objects can be arranged in a group with no repetition.

Interactive binomial distribution

Calculator for the binomial distribution

The calculator can be used to calculate exact values ​​for the binomial distribution. Is calculated

  • P (X = k) ["exactly"],
  • P (X ≤ k) ["at most"] and
  • P (X ≥ k) ["at least"].

$$ \ large P (X = k) \, = \, f (k; \, n, \, p) \, = \, {n \ choose k} \ cdot p ^ k \ cdot (1-p) ^ {nk} $$


Calculation result

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


Calculation result

$$ \ large P (X \ ge k) \, = \, \ sum_ {i = \ lfloor k \ rfloor} ^ {n} {n \ choose i} \ cdot p ^ i \ cdot (1-p) ^ {ni} $$


Calculation result