Preliminary Course
Applied Mathematics
Probabilities. Binomial Law
Binomial Law
Introductive Example
Let us consider again the example of stem growth with a probability p = 0.8 of phytomer appearance (success) every month and observe the part of stem created after 3 months.
This part of stem can consist of 0, 1, 2, or 3 phytomers. What are the probabilities of obtaining each of these cases?
- 0 phytomer: it means that the stem has been in pause through the 3 months.
The probability is then (1-p)3.
- 1 phytomer: it means that the stem has been in pause through 2 month and produced
1 phytomer during the other month.
The probability of having such a sequence is p(1-p) 2.
But several sequences can generate this observation: {Phytomer, pause, pause},
{pause, phytomer, pause} or {pause, pause, phytomer} i.e. 3 sequences.
So the probability of observing only one phytomer on the stem is equal
to 3p(1-p)2.
- 2 phytomers: it means that the stem has been in pause through one month and produced phytomers during the 2 other months. There are again 3 possible sequences that can generate this observation: the pause can be located at the first, the second or the third month. The probability is then 3(1-p)p2.
- 3 phytomers: it means that the stem has been producing phytomers through the 3 months. The probability is p3.
Possible stems created after 3 months (Drawing V. Letort-Chevalier, ECOLE CENTRALE PARIS)
Left: where is only one way to have no phytomers: 3 consecutive pauses
Middle: where are three ways to have a stem of 1 or 2 phytomers
Right: where is only one way to have three phytomers: 3 consecutive elongations
Definition
Frequently, we are interested in the total number of successes produced in a succession of n Bernoullli trials (but not in their order).
The number of successes can be 0, 1, 2, ... , n.
Let Sn be the random variable of the number of successes in n trials.
The event -n trials result in k successes and n-k failures- can happen in as many ways as k letters S can be distributed among n places: this number is equal to the binomial coefficient
As we have seen before, the probability of each case is pk(1-p)n-k.
The probability that n Bernoulli trials with probabilities p for success and q = 1 - p for failure result in k successes and n-k failures is then:
And this defines the binomial distribution B(n,p).
Note: The Bernoulli distribution corresponds to the binomial distribution where n = 1.
Definition
Binomial distribution
Mathematics. In the theory of probability and statistics, the binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. Such a success/failure experiment is also called a Bernoulli experiment or Bernoulli trial; when n = 1, the binomial distribution is a Bernoulli distribution.Exercises
Exercise 1
The cumulative distribution function is defined by :
Select the expressions that are equal to it for any n, p.
Exercise 2
Let S = B(n,p) be a binomially distributed random variable.
Find the expressions of the mean and variance of S when the probability is p or 1-p.