learn expected value Binomial

The binomial distribution is discrete probability distribution. The digits of success in a sequence of n self-sufficient yes/no experiment, each one of which yields success among probability p. Such an achievement/failure experiment is called a Bernoulli experiment or Bernoulli trail. Into detail, when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the root for the popular binomial test of statistical consequence.


Binomial Distribution:

A learn expected value Binomial is an algebraic expression contain two variables identified as x and y. direct multiplication gets quite deadly and can be slightly difficult for larger powers or more complicated expressions. The coefficients appeared during the binomial expansion are called Binomial Coefficient. They are the identical as their entries of Pascal’s triangle, and know how to be determined by a simple formula using factorials.

The formal expression of the Binomial Theorem is as follows:

`(a+b)^(n) = sum_(k=0)^n ([n],[k])a^(n-k) b^(k)`

It is often used to representation number of successes in a example of size n from a population of size N. Since the examples are not self-determining the resulting distribution is a hyper geometric distribution, not a binomial one. Though, for N much larger than n, the binomial distribution is a good quality approximation, and extensively used.

Binomial Expansion Formula

The binomial theorem formula is used to develop the binomials to some given power without direct multiplication.

n! = n (n − 1)(n − 2) ... (3)(2)(1)

Basic Binomial expansion:

(x + y)0 = 1

(x + y)1 = x + y

(x + y)2 = x2 + 2xy + y2

Examples for learn expected value binomial:

Example 1:

given the equation  (5+3y)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) 5^(2-k)3y^(k)`


=  `([2],[0]) 5^(2)3y^(0) + ([2],[1]) 5^(1)3y^(1) +([2],[2]) 5^(0)3y^(2)`

the answer is = 25 + 15y + 3y2

Example 2:

given the equation  (5-y2)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) (5x^(2-k))(-y^2)^k`


=  `([2],[0]) 5x^(2)(-y^2)^(0) + ([2],[1]) 5x^(1)(-y^2)^1 +([2],[2]) 5x^(0)(-y^2)^2`

the answer is = 5x2-5xy2+y2.


Example 3:

given the equation  (x-6y)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) (x^(2-k))(6y)^k`


=  `([2],[0]) x^(2)(6y)^(0) + ([2],[1]) x^(1)(6y)^1 +([2],[2]) x^(0)(6y)^2`

the answer is = x2+6xy+6y2.