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# Binomial Experiment

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.

 Related Calculators Binomial Experiment Calculator Binomial Calculator Binomial Confidence Interval Calculator Binomial Multiplication Calculator

## Binomial Experiment Formula

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

Binomial Experiment 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.

## Binomial Experiment Examples

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Below are examples of the binomial experiment formula:

### Solved Examples

Question 1: The probability that a bomb dropped on a bridge hits is 0.4. 8 bombs are dropped on it. 2 bomb-hits are enough to destroy the bridge. Find the probability that
(i) all the bombs destroy the bridge
(ii) 2 bombs hit the bridge
(iii) the bridge is destroyed.
Solution:

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) = 8cx (0.4)x (0.6)8 - x, x = 0, 1, 2, 3, ......., 8

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

P(X = 8) = p(8) = 8c8 (0.4)8(0.6)8 - 8 = 0.0006554

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

P(2) = 8c2(0.4)2(0.6)8 - 2 = 0.2090

(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

Question 2: A coin is tossed 6 times. Find the probability that the tosses result in
(i) Heads only
(ii) 3 heads 3 tails
(iii) 5 or more heads
Solution:

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) = 6Cx ($\frac{1}{2}$)x ($\frac{1}{2}$) 6-x x = 0,1,2......6

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

= 6C6 ($\frac{1}{2}$) 6 = 0.0156

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

= 6C3 ($\frac{1}{2}$) 6 = 0.3125

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

= p(5) + p(6) = 6c5 ($\frac{1}{2}$)6 + 6C6 ($\frac{1}{2}$) 6 = 0.1094

Question 3: Team A has probability $\frac{2}{3}$ of winning a game. If it plays 4 games, find the probability that it wins
(i) 2 games
(ii) At least one game
Solution:

X = number of games team A plays.

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

The probability mass function is p(x) = 4Cx ($\frac{2}{3}$)x ($\frac{1}{3}$)4 - x

(i) P( team A wins 2 games) = P(X = 2) = p(2)

= 4C2 ($\frac{2}{3}$)2 ($\frac{1}{3}$)4 - 2

= 0.2963

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

= 1 - P( X = 0)

= 1 - p(0) = 1 - 4C0 ($\frac{2}{3}$)0 ($\frac{1}{3}$)4 - 0

= 0.9877

Question 4: 7 Unbiased coins are thrown 128 times. In how many throws would you find at least 3 heads?

Solution:

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) = 7Cx ($\frac{1}{2}$)x ($\frac{1}{2}$)7 - x x = 0, 1, 2,.......,7

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

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

= 1- {7C0 ($\frac{1}{2}$)7 + 7C1 ($\frac{1}{2}$)7 + 7C2($\frac{1}{2}$)7 } = $\frac{99}{128}$

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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