Binomial Distribution Formula

Binomial Distribution Formula

The article deals with Binomial distribution, binomial distribution formula, applications, and solved examples.

In probability and statistics, the binomial distribution theorem plays a vital role. A binomial distribution formula is a discrete probability function with several successive sequences with their value and outcomes. The single success or failure trial is the Bernoulli experiment or Bernoulli trials, i.e., n=1, a binomial distribution used in statistics and probability.

In a binomial distribution, the formula helps to ensure that the probability gets x success with n has its independent trials. It has two possible outcomes with two parameters, n and p.

The following criteria are required for a binomial distribution in probability.

All the trials should have only two outcomes, or the obtained result can be reduced to two products, either success or failure.

The trails must have a fixed number

The outcomes should be independent 

The success of the probability will remain the same.

Application

• Used in the finding of the number of materials used.

• Helpful in taking a survey from the public for particular work

• To find the number of staff and students in the public and private sectors.

• It helps to count the votes during the election.

Binomial Distribution Formula

The binomial distribution formula for random variable 

X = P(x:n,p) = ⁿC_x px (1-p)n-x Or P(x:n,p) = ⁿC_x px (q)n-x

P = probability of success

q = probability of failure 

n = number of trials

Binomial distribution in Bernouli’s distribution is

ⁿC_x = n!/x!(n-x)! or

P(x:n,p) = n!/[x!(n-x)!].px.(q)n-x

Example 1

If a coin is tossed five times, find the probability of obtaining at least two heads.

Solution:

P(at most 2 heads) = P(X ≤ 2) 

= P (X = 0) + P (X = 1)

P(X = 0) 

= (½)5 

= 1/32

P(X=1) 

= ⁵C₁ (½)5

= 5/32

Therefore, P(X ≤ 2) = 1/32 + 5/32 = 3/16

Example 2

If a coin is tossed five times, find the probability of getting at least four heads?

Solution

P(at least 2 heads) = P(X ≤ 4) 

P(x ≥ 4) = P(x = 4) + P(x=5)

P(x = 4) 

= ⁵C₄ p4 q5-4 

= 5!/4! 1! × (½)4× (½)1 

= 5/32

P(x = 5) 

= ⁵C₅ p5 q5-5

= (½)5 

= 1/32