Binomial Distribution

The binomial distribution is the probability distribution that describes the number of successes x out of n binomial trials. The trials must satisfy three conditions.

1. There is a fixed number of independent trials.
2. Each trial has exactly two possible outcomes — success and failure.
3. The probability of success is constant for each trial.

Let X be a random variable representing the total number of successes among n trials. The possible values of X are x=0, 1, 2, ..., n. The binomial probability formula can be used to determine the probability of x successes among n trials P(X=x).

In this formula, _nC_x is the binomial coefficient and p and q are the probabilities of success and failure, respectively. Additionally, the expected value of X can be determined by the product of the number of trials n and the probability of success p.

This means that the expected number of successes in n trials is given by np.

Example

Consider the experiment of drawing 1 card from a standard deck of cards with replacement.

If drawing a diamond is considered a success, let X be the number of diamonds drawn. Since there are 13 diamond cards in a standard deck, the probability of success p in each trial will be 1352= 14. The probability of failure q will be 1- 14= 34. Suppose the experiment is repeated 5 times. Then, the binomial distribution with n= 5 can be determined. P(X= x)= _5C_x( 1/4)^x( 3/4)^(5- x) In this situation, the possible values of x are 0, 1, 2, 3, 4, and 5. The distribution of the random variable X can be obtained by evaluating this formula for each of these values.

The following graph shows the probability distribution of X.

Binomial Distribution and Normal Distribution

The binomial distribution is discrete with a finite number of outcomes, allowing for the calculation of the probability of a specific outcome. In complex calculations, a normal distribution can approximate a binomial distribution if some conditions are met. A comparison of these distributions can be seen in the applet below, where the probability of success is 0.5.

If the number of trials increases and the probability of success is closer to 0.5, the binomial distribution will closely resemble the normal distribution.

Extra

Deriving the Expected Value of a Binomial Distribution

The expected value E(X) of a discrete random variable X is given by the sum of the products of every possible value x_i of the variable and its corresponding probability P(X=x_i). E(X) = ∑ _(i=1)^n x_i * P(X=x_i) Because the number of successes out of n trials can be x=0, 1, ..., n in a binomial experiment, the expected value of the binomial random variable can be simplified as follows. E(X) = ∑ _(x=0)^n x * P(X=x) The binomial probability formula gives that P(X=x) is equal to _nC_xp^xq^(n-x). This expression can be substituted into the simplified form of the expected value of a binomial distribution. E(X) = ∑ _(x=0)^n x * P(X=x) ⇓ E(X) = ∑ _(x=0)^n x * _nC_xp^xq^(n-x) Note that when x is equal to 0, the expression x * _nC_xp^(x_i)q^(n-x)=0. This means that the first addend of the sum can be omitted and the first term of the sum is calculated for x=1. E(X) = ∑ _(x=1)^n x * _nC_xp^xq^(n-x) This formula can be manipulated to find the expected value of the binomial distribution.

E(X) = ∑ _(x=1)^n x * _nC_x* p^xq^(n-x)

C(n,k)=n!/k! (n-k)!

E(X) = ∑ _(x=1)^n x * n!/x!(n-x)!* p^xq^(n-x)

▼

Rewrite

Write as a product

E(X) = ∑ _(x=1)^n x * n*(n-1)!/x*(x-1)!(n-x)!* p^xq^(n-x)

CancelCommonFac

Cancel out common factors

E(X) = ∑ _(x=1)^n x * n*(n-1)!/x*(x-1)!(n-x)!* p^xq^(n-x)

SimpQuot

Simplify quotient

E(X) = ∑ _(x=1)^n n*(n-1)!/(x-1)!(n-x)!* p^xq^(n-x)

Rewrite

Rewrite x as (x-1)+1

E(X) = ∑ _(x=1)^n n*(n-1)!/(x-1)!(n-x)!* p^((x-1)+1)q^(n-x)

SumInExponentOne

a^(1+m)=a*a^m

E(X) = ∑ _(x=1)^n n*(n-1)!/(x-1)!(n-x)!* p* p^(x-1)q^(n-x)

MovePartNumLeft

a* b/c=a*b/c

E(X) = ∑ _(x=1)^n n*(n-1)!/(x-1)!(n-x)!* p* p^(x-1)q^(n-x)

CommutativePropMult

Commutative Property of Multiplication

E(X) = ∑ _(x=1)^n np*(n-1)!/(x-1)!(n-x)! * p^(x-1)q^(n-x)

FactorOut

Factor out np

E(X) = np∑ _(x=1)^n ((n-1)!/(x-1)!(n-x)!)p^(x-1)q^(n-x)

Rewrite

Rewrite n-x as n-1-(x-1)

E(X) = np∑ _(x=1)^n ((n-1)!/(x-1)!((n-1)-(x-1))!)p^(x-1)q^((n-1)-(x-1))

Now, substitute y=x-1 and m=n-1. If x=1, then y=0 If x=n, then y=n-1=m The sum in the last expression can now be rewritten in terms of y and m by using this information.

E(X) = np∑ _(x=1)^n ((n-1)!/(x-1)!((n-1)-(x-1))!)p^(x-1)q^((n-1)-(x-1))

SubstituteValues

Substitute values

E(X) = np∑ _(y=0)^m (m!/y!(n-y)!)p^yq^(n-y)

C(n,k)=n!/k! (n-k)!

E(X) = np∑ _(y=0)^m _mC_y p^yq^(n-y)

Note that the sum corresponds to the Binomial Theorem.

Next, recall that p+q=1 and that the powers of 1 are all equal to 1. Therefore, the sum will equal 1. ∑ _(y=0)^m _mC_y p^yq^(n-y)&=(p+q)^m &⇓ ∑ _(y=0)^m _mC_y p^yq^(n-y)&=1 Substituting this into the expression of the expected value results in the formula for the expected value of a binomial distribution.

E(X) = np∑ _(y=0)^m _mC_y p^yq^(n-y)

Substitute

∑ _(y=0)^m _mC_y p^yq^(n-y)= 1

E(X) = np* 1

MultByOne

a * 1=a

E(X) = np