## What is Binomial in Math?

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

We have learned to find the squares and cubes of binomials like (a-b) and (a+b). Using these we can calculate the numerical values of numbers like 96^{2}=(100-4)^{2}, 998^{2}=(1000-2)^{2}. But for higher powers like 98^{6}, 999^{7}, etc. the calculation becomes difficult by repetitive multiplication.

To make this calculation easy, we use the binomial theorem. The theorem gives an easier way to expand (a+b)^{n}, where “n” is a positive integer. The expansion of the term gives (n+1) terms.

## Binomial Theorem for Positive Integral Indices

$\begin{array}{l}{\left(a+b\right)}^{0}=1\\ {\left(a+b\right)}^{1}=a+b\\ {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ {\left(a+b\right)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\end{array}$

From these, we observe that,

The total number of terms in the expansion is one more than the index. For example, in the expansion of (a+b)^{3}, the number of terms is 4 whereas the index is 3.

The power of the first quantity "a" goes on decreasing by 1 whereas the power of the second quantity "b" increases by 1 in the successive terms.

In each term of the expansion, the sum of the indices of "a" and "b" is the same and is equal to the index of (a+b).

So, the binomial theorem for any positive integer “n” is given by:

${\left(a+b\right)}^{n}{=}_{n}{C}_{0}{a}^{n}{+}_{n}{C}_{1}{a}^{n-1}b{+}_{n}{C}_{2}{a}^{n-2}{b}^{2}+\mathrm{...}{+}_{n}{C}_{n-1}a{b}^{n-1}{+}_{n}{C}_{n}{b}^{n}$

Or ${\left(a+b\right)}^{n}={\displaystyle \sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){a}^{n-k}{b}^{k}}$

## What is Binomial Distribution?

A binomial distribution is such a distribution that will always have only two possible outcomes. In simple terms, the binomial distribution gives the probability of a success or failure outcome in an experiment or trial that is being repeated several times.

We use binomial distribution in probability, however, not every probability question requires the application of binomial distribution. For example, the probability of a dice showing up 1 on throwing can be directly written as 1/6.

So, when will we use binomial distribution? We will use binomial distribution after the Bernoulli trials.

## What is Bernoulli Trial?

Many experiments are dichotomous in nature, which means it has exactly two outcomes. For example, a tossed coin shows a "head" or "tail"; the decision of something is "yes" or "no". In such cases, it is customary to call one of the outcomes a "success" and the other a "failure". This consideration depends totally on our convenience.

Each time we perform any experiment, we call it a trial. If we tossed a coin 5 times, then the number of trials is 5, each having exactly two outcomes, namely success or failure. The outcomes of any trial are independent of the outcomes of any other trial. In each such trials, the probability of success or failure remains constant. Such independent trials having only two outcomes "success" or "failure" are called Bernoulli trials.

**Conditions for Bernoulli Trials**

1. There should be a finite number of trials.

2. The trials should be independent.

3. Each trial has only two outcomes: success and failure.

4. The probability of success remains the same for each trial.

Consider an example where 5 balls are drawn from a bag containing 10 white and 20 black balls. Predict whether the trials are Bernoulli trials if the ball drawn is replaced and if it is not replaced.

When a ball is drawn with replacement, the probability of success (say, white ball) is p= 10/30 = 1/3 which is the same for all five trials. Hence, the trials involving the drawing of balls with replacement are the Bernoulli trials.

When a ball is drawn without replacement, the probability of success varies with the number of trials. For the first trial, the probability of success will be p=10/30 , for the second trial, the probability of success may be p= 9/29 or p= 10/29 , which is not equal to the first trial. Hence, the trials involving drawing balls without replacements are not Bernoulli trials.

Now, after knowing that the trial is a Bernoulli trial, how can we calculate the probability? That’s where binomial distribution comes into play.

Consider an experiment of tossing a coin, in which each trial results in success (heads) or failure (tails).

Let S and F denote the success and failure respectively of each trial.

Now we are interested in finding the ways in which we get six successes and one failure in seven trials.

So, six different cases are as follows

SSSSSSF, SSSSSFS, SSSSFSS, SSSFSSS, SSFSSSS, SFSSSSS, FSSSSSS

Similarly, if we want to show 3 successes and 4 failures, it will be a long process to list all of these ways. To avoid such lengthy processes and listing of all the possible cases, for the probabilities of the number of successes in n-Bernoulli trials, a formula has been derived.

Let us take four Bernoulli trials with probabilities p for success and q = (1 – p) for failure in each trial. So, the sample space of the experiments is the set:

S = {SSSS, SSSF, SSFS, SFSS, FSSS, SSFF, SFFS, FFSS, SFSF, FSFS, FSSF, SFFF, FSFF, FFSF, FFFS, FFFF}

Assume, the number of success is a random variable X and can take values 0,1, 2, 3, and 4.

The probability distribution of the number of successes as:

When there is no success:

$P\left(X=0\right)=P\left(\left\{FFFF\right\}\right)$

$=P\left(F\right)P\left(F\right)P\left(F\right)P\left(F\right)$

$=q\times q\times q\times q$

$={q}^{4}$

When there is one success:

$P\left(X=1\right)$

$=P\left(\{SFFF,FSFF,FFSF,FFFS\}\right)$

$=P\left(\left\{SFFF\right\}\right)+P\left(\left\{FSFF\right\}\right)+P\left(\left\{FFSF\right\}\right)+P\left(\left\{FFFS\right\}\right)$

$=P\left(S\right)P\left(F\right)P\left(F\right)P\left(F\right)+P\left(F\right)P\left(S\right)P\left(F\right)P\left(F\right)+P\left(F\right)P\left(F\right)P\left(S\right)P\left(F\right)$

$+P\left(F\right)P\left(F\right)P\left(F\right)P\left(S\right)$

$=\left(p\times q\times q\times q\right)+\left(q\times p\times q\times q\right)+\left(q\times q\times p\times q\right)+\left(q\times q\times q\times p\right)$

$=4{p}^{1}{q}^{3}$

When there are two successes:

$P\left(X=2\right)$

$=P\left(\{SSFF,SFFS,FFSS,SFSF,FSFS,FSSF\}\right)$

$=P\left(\left\{SSFF\right\}\right)+P\left(\left\{SFFS\right\}\right)+P\left(\left\{FFSS\right\}\right)+P\left(\left\{SFSF\right\}\right)+P\left(\left\{FSFS\right\}\right)+P\left(\left\{FSSF\right\}\right)$

$=P\left(S\right)P\left(S\right)P\left(F\right)P\left(F\right)+P\left(S\right)P\left(F\right)P\left(F\right)P\left(S\right)+P\left(F\right)P\left(F\right)P\left(S\right)P\left(S\right)+P\left(S\right)P\left(F\right)P\left(S\right)P\left(F\right)$

$+P\left(F\right)P\left(S\right)P\left(F\right)P\left(S\right)++P\left(F\right)P\left(S\right)P\left(S\right)P\left(F\right)$

$=\left(p\times p\times q\times q\right)+\left(p\times q\times q\times p\right)+\left(q\times q\times p\times p\right)+\left(p\times q\times p\times q\right)+\left(q\times p\times q\times p\right)+\left(q\times p\times p\times q\right)$

$=6{p}^{2}{q}^{2}$

When there are three successes:

$P\left(X=4\right)$

$=P\left(\{SSSF,SSFS,SFSS,FSSS\}\right)$

$=P\left(\left\{SSSF\right\}\right)+P\left(\left\{SSFS\right\}\right)+P\left(\left\{SFSS\right\}\right)+P\left(\left\{FSSS\right\}\right)$

$=P\left(S\right)P\left(S\right)P\left(S\right)P\left(F\right)+P\left(S\right)P\left(S\right)P\left(F\right)P\left(S\right)+P\left(S\right)P\left(F\right)P\left(S\right)P\left(S\right)$

$+P\left(F\right)P\left(S\right)P\left(S\right)P\left(S\right)$

$=\left(p\times p\times p\times q\right)+\left(p\times p\times q\times p\right)+\left(p\times q\times p\times p\right)+\left(q\times p\times p\times p\right)$

$=4{p}^{3}{q}^{1}$

When there are four successes:

$P\left(X=4\right)=P\left(\left\{SSSS\right\}\right)$

$=P\left(S\right)P\left(S\right)P\left(S\right)P\left(S\right)$

$=p\times p\times p\times p$

$={p}^{4}$

The probability distribution of X is

No Success | One Success | Two Successes | Three Successes | Four Successes |

q^{4} | 4p^{1}q^{3} | 6p^{2}q^{2} | 4p^{3}q^{1} | p^{4 } |

Also the binomial expansion of ${\left(q+p\right)}^{4}$ is ${q}^{4}+4p{q}^{3}+6{p}^{2}{q}^{2}+4{p}^{3}q+{p}^{4}$

Now, if we see the above table and compare it with the expansion, we can conclude that the probabilities of 0, 1, 2, 3, and 4 successes are the 1st, 2nd, 3rd, 4^{th}, and 5th term of the expansion of (q+p)^{4} respectively. Also, q + p = 1, which follows the sum of these probabilities.

Therefore, in an experiment of n-Bernoulli trials, having x successes (S) and (n-x) failures (F)

can be obtained in ${}_{n}{C}_{x}$ ways or $\frac{n!}{x!\left(n-x\right)!}$ ways.

The probability of x successes and (n-x) failures is:

P (x successes) $\times $ P((n-x) failures)

= P(S) $\times $P(S). . .x times $\times $ P(F) $\times $P(F). . . (n-x) times

=${p}^{x}{q}^{n-x}$

The probability of x successes in n-Bernoulli trials is $\frac{n!}{x!\left(n-x\right)!}\times {p}^{x}{q}^{n-x}$ or ${}_{n}{C}_{x}{p}^{x}{q}^{n-x}$

Thus, P (x successes) = _{n}C_{x}p^{x}q^{n-x}, where x = 0, 1, 2, 3, …, n and q = (1 – p).

Suppose there are 5 red balls and 5 blue balls in a bag. Among them, 6 are drawn successively with replacement from the bag. What is the probability that exactly 2 red balls were drawn and what is the probability that at least 4 red balls were drawn?

Lets’ take drawing a red ball from the bag as success. So, p is $\frac{3}{6}=\frac{1}{2}$

Clearly, X is the binomial distribution, where n = 6, p = $\frac{1}{2}$, and q = 1 - $\frac{1}{2}$= $\frac{1}{2}$

So, the general solution is $P\left(X=x\right){=}_{6}{C}_{x}{\left(\frac{1}{2}\right)}^{6-x}{\left(\frac{1}{2}\right)}^{x}{=}_{6}{C}_{x}{\left(\frac{1}{2}\right)}^{6}$

In the first part of the question, X = 2.

So,

$P\left(X=2\right){=}_{6}{C}_{2}{\left(\frac{1}{2}\right)}^{6-2}{\left(\frac{1}{2}\right)}^{2}{=}_{6}{C}_{2}{\left(\frac{1}{2}\right)}^{6}$

$=\frac{6!}{6!\left(6-2\right)!}{\left(\frac{1}{6}\right)}^{6}$

$=\frac{15}{64}$

Therefore, the probability that exactly 2 times red balls were drawn is $\frac{15}{64}$.

In the second part of the question, X ≥ 4.

$P\left(X\ge 2\right)=P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)$

${=}_{6}{C}_{4}{\left(\frac{1}{2}\right)}^{6}{+}_{6}{C}_{6}{\left(\frac{1}{2}\right)}^{6}{+}_{6}{C}_{6}{\left(\frac{1}{2}\right)}^{6}$

$=\frac{11}{32}$

Therefore, the probability that at least 4 times the red balls were drawn is $\frac{11}{32}$.

## Formula

The probability of x successes in n-Bernoulli trials is _{n}C_{x}p^{x}q^{n-x}, where n denotes the total number of trials, p denotes the probability of success in a trial and q denotes the probability of failure in a trial.

## Context and Application

We have already seen in various examples how binomial distribution helps in our daily lives.

- Bernoulli distribution is sometimes used to find the production in a factory; the good product indicates success and the defective product indicates failure.
- If you purchase a lottery ticket, you're either going to win the money or lose.
- Giving an examination—either you pass the examination or fail.
- Sports teams playing against other teams either win or lose.

## Common Mistakes

Some of the common mistakes that learners do while dealing with the binomial distribution include:

- Identifying the intent of the problem statement, whether the problem involves the Bernoulli trial or not.
- Finding proper fractions of success and failures for every Bernoulli trial.
- Making errors in calculations. Do not forget to cross-check calculations.

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