If X denotes the number of successes in n trials, then X is a random variable which can take the values 0, 1,........, n. Since in n trials, we may get no success, one success, ........., n success. Total probability is unity. Since the probability are the successive terms in the binomial expansion (q + p)$^n$, it is called the binomial distribution.

Binomial distribution is also known as Bernoulli distribution after the Swiss mathematician James Bernoulli who discovered it in 1700 & was first published in 1713, 8 years after his death.

## Binomial Probability Formula

General expression for the probability of r successes is
p(r) = P(X  = r) = $\binom{n}{r}p^{r}q^{n - r}$, r = 0, 1, 2, .....
where,
n = Total number of trials
r = Total number of successful trials
p = Probability of success in a single trial
q = Probability of failure in a single trial = 1 - p

## Binomial Probability Function

Important discrete distribution is the binomial. Only two possible outcomes are possible when making a single observation. The number '1' is used to represent a success and a failure is represented by '0'. A common convention 'p' represent the probability of the success and q = 1 - p be the probability of failure. In binomial probability function, the random variable X represents the total number of successes among n observations. It represented as a P(X = x).

p(r) = P(X  = r) = $\binom{n}{r}p^{r}q^{n - r}$, r = 0, 1, 2, ...

## Binomial Probability Distribution

An experiment has a binomial probability distribution, if the following conditions are satisfied.
1. n, the number of trials is fixed.
2. Each trial results in 2 mutually exclusive and exhaustive outcomes termed as success and failure.
3. Trials are independent.
4. p, the probability of success is constant for each trial. Then, q = 1 - p is the probability of failure in any trial.

The trials satisfying the above condition are also known as bernoulli trials.

## Binomial Probability Table

The binomial table has the most common probabilities listed for n. The table gives the probability of obtaining atmost x successes in n independent trials, each of which has a probability p of success.

P(X$\leq$ x)=  $\sum_{r=0}^{x}C(n,r)p^{r}(1-p)^{n-r}$

## Binomial Probability Experiment

If we toss a coin n times, then the outcome of any trial is one of the mutually exclusive events head and tail. Further all the trials are independent.
Since the result of any throw of a coin does not affect, n is not affected by the result of other throws. Moreover, the probability of success in any trial is half which is constant for each trial. Hence, the coin tossing problem will give rise to binomial distribution. In flipping coins, like determining the probability of getting at least 2 heads in 5 throws. Dice throwing problem will also confirm to binomial distribution.

## Binomial Probability Examples

Given below are some of the examples on binomial probability.

### Solved Examples

Question 1: The number of tosses of a coin that are needed so that the probability of getting atleast one head be on 0.875 is
1. 2
2. 3
3. 4
4. 5

Solution:

Let the required number of tosses of the coin be n.
Then, P[Atleast one head in n tosses of a coin] = 1 - P (Number of heads in n tosses of a coin)

= 1 - ($\frac{1}{2}$)$^{n}$

We want n so that, this probability is 0.875

1 - ($\frac{1}{2}$)$^{n}$ = 0.875

($\frac{1}{2}$)$^{n}$ = 1 - 0.875

= 0.125
= (0.5)$^{3}$

= ($\frac{1}{2}$)$^{3}$

n = 3.
Therefore, option 2 is the correct solution.

Question 2: Assume that 0.5 of the population is vegetarian so that, the chance of an individual being a vegetarian is 0.5. Assuming that 100 investigators each take the sample of 10 individuals to see whether they are vegetarian, how many investigators would expect to report that three or less were vegetarian?
Solution:
n = 10
p = Probability that an individual is vegetarian is 0.5, q = 1 - p = 0.5
Then, by binomial probability law, the probability that there are 'r' vegetarian in a sample of 10 is

p(r) = $\binom{10}{r}p^{r}q^{10-r}$

= $\frac{1}{1024}\binom{10}{r}$

Thus, the probability that in a sample of 10, three or less people are vegetarian is
P(0) +  P(1) + P(2) + P(3)

= $\frac{1}{1024}$ ($\binom{10}{0}\binom{10}{1}\binom{10}{2}\binom{10}{3}$)

= $\frac{1}{1024}$(1 + 10 + 45+ 120)

= $\frac{176}{1024}$

= $\frac{11}{64}$

Hence, out of 100 investigators, the number of investigators who will report three or less vegetarians in a sample of 10 is

= 100 x $\frac{11}{64}$

= $\frac{275}{16}$

= 17.2 $\simeq$  17 (Since the number of investigators cannot be a fraction.)