Binomial experiment is a kind of probability distribution which expresses the probability of a set of dichotomous alternatives i.e. success or failure.
This distribution has been used to describe a wide variety of processes
in business and social sciences as well as other areas. This type of
process which gives rise to this distribution is usually referred to as a
Bernoulli process. This is developed under a very specific set of
assumption involving the concept of a series of experimental trials.**A binomial experiment is a statistical experiment which has the following properties: **

- The experiment would consist of n repeated trials.
- Each trial can be have a result with just two possible outcomes. These outcomes are better known as success and failure.
- The probability of success, denoted by p, is the same on every trial for a given experiment.
- The outcome of one trial does not affect the outcome of the other trials.

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In a Binomial experiment, the probability of $X$ success in n trial is given by the following formula,

$P(x)$ = $c^{n}_{x}\ p^x\ q^{n\ -\ x}$, $x$ = $0,\ 1,\ 2,\ ...........,\ n$

$0\ <\ p\ <\ 1\ :\ q$ = $1\ -\ p$

Here the variable $X$ is discrete and it is called the Binomial variate. Below are examples of the binomial experiment formula:

(i) all the bombs destroy the bridge

(ii) 2 bombs hit the bridge

(iii) the bridge is destroyed.

The probability that a bomb hits is 0.4

=> p = 0.4

Since 8 bombs are dropped n = 8

X : number of bombs hitting the bridge

Then X is B(n = 8, p = 0.4). That is X is a binomial variate with parameters n = 8 and p = 0.4

The probability mass function is

P(X) =

(i) The probability that all the bombs hit the bridge is

P(X = 8) = p(8) =

(ii) The probability that 2-bombs hit the bridge is p(X = 2)

P(2) =

(iii) The probability that bridge is destroyed = P(bridge is destroyed)

= p(2 or more bombs hit the bridge) = P(X $\geq$ 2)

= 1 - p(X < 2)

= 1- { P(0) + P(1)} = 0.8936

(i) Heads only

(ii) 3 heads 3 tails

(iii) 5 or more heads

X = Number of heads obtained in 6 tosses.

Then X is B(n = 6,p = $\frac{1}{2}$)

The probability mass function is p(x) =

(i) Prob (heads only) = P(X = 6) = p(6)

=

(ii) Prob (3 heads 3 tails) = P(X = 3) = p(3)

=

(iii) Prob (5 or more heads) = p(5) = P(X $\geq$ 5)

= p(5) + p(6) =

(i) 2 games

(ii) At least one game

X = number of games team A plays.

Then X is B(N = 4,P = $\frac{2}{3}$)

The probability mass function is p(x) =

=

(ii) P(A wins at least one game) = 1 - P( A wins none)

= 1 - P( X = 0)

= 1 - p(0) = 1 -

= 0.9877

X = Number of heads obtained in a throw of 7 coins.

Then X is B(n = 7, p = $\frac{1}{2}$)

The probability mass function is p(x) =

P(at least 3 heads) = 1 - P(at most 2 heads)

= 1- {p(0) + p(1) + p(2)}

= 1- {

The number of tosses resulting in at least 3 heads = n* P(at least 3 heads)

= 128 $\times$ $\frac{99}{128}$ = 99

Thus in 99 tosses we expect at least 3 heads to occur.

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