The prefix ‘bi’ means two or twice. A binomial distribution is considered as the probability of a trail with only two possible outcomes. It is a type of distribution that has two different outcomes which are ‘success’ and ‘failure’. For example, if we toss the coin then there is an equal chance of outcome it can be heads or tails. Therefore there is a 50% chance of the outcomes. It is applicable to discrete random variables only. The origin of Binomial distribution can be taken as towards Bernoulli’s trial. In this article, we will discuss the Binomial distribution formula with examples. Let us begin learning!
Binomial Distribution Formula
What is Binomial Distribution?
The binomial distribution formula helps to check the probability of getting an “x” number of successes in the “n” independent trials of a binomial experiment. As we know that binomial distribution is a type of probability distribution in statistics that has two possible outcomes. In probability theory, the binomial distribution has two parameters n and p.
The probability distribution becomes a binomial probability distribution if it satisfies the following requirements.
- Each trail can have only two outcomes. These outcomes may be either a success or a failure.
- The trails should be a fixed number.
- All the outcomes each trial must be independent of each other.
- Also, the success of probability must remain the same for each trail.
The binomial distribution summarized the number of trials, survey or experiments conducted. It is very useful when each outcome has an equal chance of attaining a particular value.
The formula for Binomial Distribution in Probability:
The formula for the binomial probability distribution is given below:
P(x) = [\(\frac{n!}{r!(n-r)!}\)] \(p^r (1-p)^{n-r}\)
Where,
P(x) | Binomial probability Distribution |
X | Random variable |
N | Total number of events |
R | The total number of successful events. |
P | Probability of success on a single trial. |
1-p | Probability of failure on a single trial. |
\(binomial{n}{r}\) | Combinatorial value |
Mean and Variance of a Binomial Distribution
Calculation of Binomial distribution value sometimes needs mean and variance values. These two terms will give more stability and reliability. Formulas are as given below:
\(\mu = np\)
\(\sigma^2 = npq\)
\(\mu\) | Mean |
\(\sigma^2\) | Variance |
p | Probability of success |
q | Probability of failure |
Also, it should be noted that,
- The variance of a Binomial Variable will always be less than its mean. i.e. npq < np.
- For Maximum Variance, p and q both should be 0.5 and hence
\(\sigma= \frac{1}{4}\)
Solved Examples
Q.1: Toss a coin for 12 times. Then what will be the probability of getting exactly 7 times head?
Solution: As given,
Number of trails I.e. n = 12
Number of success i.e. r = 7
Also, Probability of single trail i.e. p = \(\frac {1}{2}\) = 0.5
Formula for Binomial Distribution is,
P(x) = \(binom{n}{r}p^r (1-p)^{n-r}\)
\(binom{n}{r}\)
= \(\frac{n!}{r!(n-r)!}\)
= \(\frac{12!}{7!(5)!}\)
= 792
\(p^r= 0.5 ^ 7\)
= 0.0078125
To Find \((1−p)^{n-r},\)
calculate (1-p) and (n-r).
1 – p = 1 – 0.5 = 0.5
n – r = 12 – 7 = 5
So, \((1−p)^{n-r} = 0.5^5 = 0.03125\)
Thus,
P(x) = \(binom{n}{r}p^r (1-p)^{n-r}\)
= 792 x 0.0078125 x 0.03125
= 0.193359375
Therefore, probability of getting exactly 7 heads is 0.19
I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26