It is incorrect to write binomial, it is correct to write binomial.
This article implies that you are already familiar with the basics of statistics, in particular, what is the law of distribution and what parameters it is characterized.
Binomial distribution is a description of «N» independent experiments, the result of which is always either strictly positive or strictly negative. It is important that the experiments are conducted under the same conditions. As an example, you can consider tossing a coin or guessing the test results, in the case of a coin a positive result will be an "eagle" and when guessing the test, the correct answer.
The conducted experiments should be independent, which means that the result of the subsequent one does not depend on the result the previous one.
№ | Result | No. | Result | No. | Result | |||
---|---|---|---|---|---|---|---|---|
1 | heads | 2 | tails | 3 | heads | |||
4 | heads | 5 | tails | 6 | tails | |||
7 | tails | 8 | tails | 9 | heads | |||
10 | heads | 11 | tails | 12 | tails | |||
13 | tails | 14 | tails | 15 | heads | |||
16 | tails | 17 | heads | 18 | heads | |||
19 | heads | 20 | heads | 21 | heads | |||
22 | heads | 23 | heads | 24 | heads | |||
25 | heads | 26 | tails | 27 | tails | |||
28 | tails | 29 | heads | 30 | heads | |||
31 | tails | 32 | heads | 33 | tails | |||
Table 1. Bernoulli tests. The results of the coin toss. The eagle fell out 18 times |
As a result of a series of experiments, we will get the probability of a positive initial event, in the above experiment, the eagle fell out 18 times, so the probability of an eagle falling out is equal to:
P(x) = Nsuccessful/Ntotal = 18/33 = 0.55
Binomial distribution formula
The binomial distribution depends on two parameters: the number of trials and the number of successful events.
The formula of the probability function of the binomial distribution:
P(X) = (nk)·pk·qn-k, (nk) - binomial coefficient
(nk) = (n!)/[(n-k)!k!]
And, since the distribution is discrete, the formula of the distribution function is written as the sum of probabilities:
k ∈ [0, ⌊y⌋], F(X) = Σ(nk)·pkqn-k
The values of the mathematical expectation and the variance of the Bernoulli distribution are equal, respectively:
E[X] = np
D[X] = npq
Example
Using the data in Table 1, we will evaluate the coin and the next time we throw, we will guess the side we need :) .
Suppose we flip a coin five times, whose side fell out more - he wins, what is the probability that will the eagle win?
Source data:
n = 5
k = 3
p = 0.55
q = 0.45
Probabilities:
P(n = 5, k = 3) = 120 / [2 · 6] · 0.553 · 0.452 = 0.34
P(n = 5, k = 4) = 120 / [1 · 24] · 0.553 · 0.452 = 0.21
P(n = 5, k = 5) = 120 / [1 · 120] · 0.553 · 0.452 = 0.05
F(X) = Σk=3,4,5 P(n = 5, k = k) = 0.6
Thus, the probability that the eagle will win is 60%