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Bernoulli Trials and Binomial Distribution

 

Table of Content


Bernoulli Trial

A random experiment which has only two mutually exclusive outcomes, namely ‘success’ and ‘not success’ or ‘failure’ is called Dichotomous Experiment.

If a dichotomous experiment is repeated n times and if in each trial the probability of success p (0 < p < 1) is constant, then all such trials are called Bernoulli trails.

This term is named after Jacob Bernoulli, a 17th century Swiss Mathematician.

Since, Bernoulli trial has only 2 possible outcomes, it can be framed as ‘yes’ or ’no’ question.

Example

  • Is the top card of the shuffled deck an ace?

  • Was the new born child a girl?
     

Binomial Distribution

An experiment of n repeated independent Bernoulli trails is called a Binomial Experiment.

Consider a Binomial experiment. Let the outcomes in each trial be a success (S) or Failure (F) with probability p and (1-p) = q respectively.

Let a random variable X = No. of successes

0 ≤ X ≤ n.

We want to find the probability of an event that the n trial result in exactly x successes and (n-x) failures is 

Let A = {S S S…. S x times}

B= {F F F……F (n-x) times }

Since trials are independent,

P( A∩B) = P(a) . P(b)

= P(S).P(S)…..x times. P(F). P(F)…. (n-x) times

= p.p.p…..x times . q.q.q  ….. (n-x) times

= px (q)n-x

Now, the number of possible orders of x successes and (n-x) failures in independent trials is nCand these are mutually disjoint orders. Hence, probability that n trials result in exactly x successes and (n-x) failures in any order is given by

P [X = x] = nCx  px   qn-x  , 0 ≤ x ≤ n

This is known as Binomial Distribution

Binomial Distribution

A discrete random variable X is said to follow a Binomial distribution with parameters n and p if its probability mass function (pmf) if given by

P[X = x] = p(x) = nCpx   qn-x , x = 0,1 , 2….n

0< p< 1, q = 1 – p,

n = 1, 2, 3….

Otherwise

P[X = x] for every x is xth term in the expansion of (q+p)

Binomial distribution can be represented in tabular form as follows:

X = x 0 1 2 ….. x ….. n
P(x) nCqn   nCpqn-1   nC2 p2qn-2   ….. nCx pxqn-x   ….. nCn pn  

From the above table we can see that,

  • P(x) ≥ 0 for all  x

  • ∑ p(x) = ∑ nCpx   qn-x

= (q + p)n

= 1

Probability Mass function of Binomial Distribution

Fig: Probability Mass function of Binomial Distribution
 

Conditions for Binomial Distribution

The binomial distribution is used whenever each of the following is satisfied:

  • Trials of experiment are Bernoulli trials

  • The performance of a Bernoulli trial results in an outcome that can be classified either as a success or a failure

  • The number of trials (n) is finite

  • All n trials are independent

  • The probability of success is constant from trial to trial, i.e. the trials are repeated under identical conditions. Further, if P(success) = p and P(failure) = q, then p + q =1, 0 < p < 1

  • The random variable X denotes the number of successes x in n independent Bernoulli trials.

Notations:

  • X ~ B(n,p) denotes that X follows Binomial Distribution with parameters n and p.

  • If X ~ B(n,p), then Y= (n – X) ~ B(n,q)

Types of Probability Distribution

Fig: Types of Probability Distribution
 

Mean and Variance of Binomial Distribution

The mean of random variable X denoted as E(X) is given by E(X) = np

The variance of random variable X denoted by Var (X) is given by Var(X) = npq

The standard derivation of random variable X is defined as the positive random variable X denoted by SD(X) is given by SD(X)= √Var(X) = √npq

Note: Mean > Variance that is, np > npq
 

Poisson Distribution

The Poisson binomial distribution, which describes the number of successes in a series of independent Yes/No experiments with different success probabilities.

A Poisson distribution is the probability distribution that results from a Poisson experiment.
 

Attributes of a Poisson Experiment

Poisson experiment is a statistical experiment that has the following properties:

  • The experiment results in outcomes that can be classified as successes or failures.

  • The average number of successes (μ) that occurs in a specified region is known.

  • The probability that a success will occur is proportional to the size of the region.

  • The probability that a success will occur in an extremely small region is virtually zero.

Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.

Poisson Random Variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson Distribution.

Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula:

 Formula of Poisson 

Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is µ. Then, the Poisson probability is:

P(x; μ) = (e) (μx) / x!

Where x is the actual number of successes that result from the experiment and is approximately equal to 2.71828

The Poisson distribution has the following properties:

  • The mean of the distribution is equal to μ.

  • The variance is also equal to μ.

Example

A fair coin is tossed 8 times. Find the probability that it shows heads

  • Exactly 5 times

  • A larger number of times than tails

  • At least once

Solution

Let X= Number of heads

P = Probability of getting head

P = 1/2

Q = 1- p = 1-1/2 = ½

Given: n = 8

X ~ B(8,1/2)

The p.m.f of X is given as

P(X = x) = p(x)  = nCpx   qn-x  

that is, p(x) = 8C(1/2)x   (1/2)8-x  , x=0,1,2….8

  • P (exactly 5 heads) = P[X = 5]

= P (5) = 8C(1/2)(1/2)8-5

 8C(1/2)(1/2)3  [since nCx  =  nCn-x ]

= 7/32

Hence, the probability of getting exactly 5 heads is 0.21875

  • P( getting heads larger number of times than tails )

= P[ X ≥ 5] = p(5) + p(6) + p(7) + p(8)

= 8C5 (1/2)5  (1/2)8-5 + 8C6 (1/2)6 (1/2)8-6 + 8C7 (1/2)(½)8-7 8C8 (1/2)8 (1/2)8-8

= 8C3 (1/2)5  (1/2)3 + 8C2 (1/2)6 (1/2)2 + 8C1 (1/2)(½) + 8C8 (1/2)8

= (½)8 [56 + 28 + 8 + 1] = 93/256

= P[X ≥ 5] = 0.36328

  • P( getting heads at least once )

= P [ X ≥ 1] = 1-P[X = 0]

= 1- p(0)

= 1- 8C0 (1/2)0 (½) 8-0

= 1 – (1/2)8 = 1 - 1/256 = 255 / 256

P[ X ≥ 1] = 0.996

Hence, the probability of getting heads at least once is 0.996

Example

Suppose that 80% of all families own a television set. If 10 families are interviewed at random, find the probability that 

  • Seven families own a television set

  • At most three families own a television set

Solution

Let X = Number of families who own a television

P = Probability of family who own a television set

P = 80% = 80/100 = 4/5

Q = 1- p = 1-4/5 = 1/5

Given: n =10

X ~ B(10,4/5)

The p.m.f of X is given as

P(X = x) = p(x)  = nCpx   qn-x  

that is, p(x) = 10C4/5x   1/510-x  

  • P (seven families own a television set) = P[X = 7] = p(7)

= 10C7  (4/5)7   (1/5)10-7

= 10C3  (4/5)7   (1/5)3 [since nCx  =  nCn-x ]

= 0.2013

Hence, the probability that seven families own a television set is 0.2013

  • P (at most three families own a television set )

= P [X ≤ 3] = P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]

= p(0) + p(1) + p(2) + p(3)

= 10C0 (4/5)0 (1/5)10-0 + 10C1 (4/5)1 (1/5)10-1 + 10C2 (4/5)2 (1/5)10-2 + 10C3 (4/5)3 (1/5)10-3  

= [1 + 40 + 45 * 16 + 120 * 64](1/5)10

= 8441/9765625

P [ X ≤ 3 ] = 0.00086

Hence, the probability that at most three families own a television set is 0.00086
 

Applications of Binomial Distribution

  • To find the probability of getting x heads in the experiment of tossing a fair coin n times. Here, X ~ B (n, p = ½)

  • To find the probability of getting success in the experiment of throwing a fair die or a pair of fair dice n times

  • To find the probability of correct answers in a multiple choice test consisting of n questions

  • To find the probability of success in any dichotomous experiment.
     

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Bernoulli Trials and Binomial Distribution

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