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# Binomial distribution

A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q respectively, is called a Bernoulli trial.

Distribution

The following are the two types of Theoretical distributions :

1. Discrete distribution  2. Continous distribution

Discrete distribution

The binomial and Poisson distributions are the most useful theoretical distributions for discrete variables.

Binomial distribution

Binomial distribution was discovered by James Bernoulli(1654_1705) in the year 1700 and was first published posthumously in 1713 , eight years after his death.

A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q respectively, is called a Bernoulli trial.

Some examples of Bernoulli trials are :

(i) Tossing of a coin (Head or tail)

(ii) Throwing of a die (getting even or odd number)

Consider a set of n independent Bernoullian trails (n being finite) in which the probability ‘p’ of success in any trial is constant , then q = 1–p , is the probability of failure. The probability of x successes and consequently (n–x) failures in n independent trials, in a specified order (say) SSFSFFFS….FSF is given in the compound probability theorem by the expression

P(SSFSFFFS…..FSF) = P(S)P(S)P(F)P(S)x…….xP(F)P(S)P(F)

p.p.qp………………q.p.q

p.p. p.p.       ……q.q.q.q.q.q

{x factors}     {(n–x) factors}

pxq(n−x)

x successes in n trials can occur in nCx ways and the probability for each of these ways is same namely px qn−x .

The probability distribution of the number of successes, so obtained is called the binomial probability distribution and the binomial expansion is (q + p)n

### Definition 7.1

A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non- negative value and its probability mass function in given by Note

Any random variable which follows binomial distribution is known as binomial variate i.e X ~ B(n,p) is a binomial variate.

The Binomial distribution can be used under the following conditions :

1. The number of trials ‘n’ finite

2. The trials are independent of each other.

3. The probability of success ‘p’ is constant for each trial.

4. In every trial there are only two possible outcomes – success or failure.

## Derivation of the Mean and Variance of Binomial distribution : Variance = E(X2) – E(X)2

= n2p2 np2 + np n2p2

= np(1–p) = npq

Hence, mean of the BD is np and the Variance is npq.

Properties of Binomial distribution

1. Binomial distribution is symmetrical if p = q = 0.5. It is skew symmetric if pq. It is positively skewed if p < 0.5 and it is negatively skewed if p > 0.5

2. For Binomial distribution, variance is less than mean

Variance npq = (np)q < np

Example 7.1

A and B play a game in which their chance of winning are in the ratio 3:2 Find A’s chance of winning atleast three games out of five games played.

Soltion:

Let ‘p’ be the probability that ‘A’ wins the game. Then we are given n = 5, p = 3/5, q = 1 – 3/5 = 2/5 (since q = 1–p)

Hence by binomial probability law, the probability that out of the 5 games played, A wins ‘x’ games is given by The required probability that ‘A’ wins atleast three games is given by

P ( X 3) = P(X = 3) + P(X = 4) + P(X = 5) Example 7.2

A fair coin is tossed 6 times. Find the probability that exactly 2 heads occurs.

Solution :

Let X be a random variable follows binomial distribution with probability value p = 1/2 and q = 1/2

Prob: that exactcy 2 heads occur are as follows Example 7.3

Verify the following statement: The mean of a Binomial distribution is 12 and its standard deviation is 4.

Solution:

Mean: np = 12 Since p + q cannot be greater than unity, the Statement is wrong

Example 7.4

The probability that a student get the degree is 0.4 Determine the probability that out of 5 students (i) onewill be graduate (ii) atleast  one will be graduate

Solution: Example 7.5

In tossing of a five fair coin, find the chance of getting exactly 3 heads.

Solution :

Let X be a random variable follows binomial distribution with p = q = 1 /2 Example 7.6

The mean of Binomials distribution is 20 and standard deviation is 4. Find the parameters of the distribution.

Solution

The parameters of Binomial distribution are n and p

For Binomial distribution Mean = np =20

Standard deviation = √npq = 4

npq = 16

npq/np = 16/20 = 4/5

q = 4/5

p = 1–q = 1– (4/5) = 1/5

Since np = 20

n = 20/p

n= 100

Example 7.7

If x is a binomially distributed random variable with E(x) =2 and van (x) = 4/3 . Find P(x = 5)

Solution:

The p.m.f. Binomial distribution is

p(x) = nCx px qn−x

Given that E(x) =2

For the Binomial distribution mean is given by np = 2 ... (1)

Given that var (x) = 4/3

For Binomial distribution variance is given by npq= 4/3 ... (2)

(2)/(1) = 4/6 = 2/3

q = 2/3 and p = 1–2/3 = 1/3

Substitute is (1) we get

n = 6 Example 7.8

If on the average rain falls on 9 days in every thirty days, find the probability that rain will fall on atleast two days of a given week.

Solution :

Probability of raining on a particular day is given by p = 9/30 = 3/10 and

q = 1–p = 7/10.

The BD is P(X = x) = nCx px qn−x

There are 7 days in a week P(X = x) = The probability of raining for atleast 2 days is given by Therefore the required probability = 1– [P(x = 0) +P(x = 1)]

= 1– {0.082+ 0.247]

= 0.6706

Example 7.9

What is the probability of guessing correctly atleast six of the ten answers in a TRUE/ FALSE objective test?

Solution :

Probability p of guessing an answer correctly is p = 1/2

q = 1/2

Probability of guessing correctly x answers in 10 questions The required probability P(X ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10) Example 7.10

If the chance of running a bus service according to schedule is 0.8, calculate the probability on a day schedule with 10 services : (i) exactly one is late (ii) atleast one is late

Solution :

Probability of bus running late is denoted as p = 1-0.8 = 0.2

Probability of bus running according to the schedule is q = 0.8

Also given that n = 10

The binomial distribution is p(x) = 10Cx (0.2)x (0.8)10-xt.

(i) probability that exactly one is late P(y=1) = 10C1pq9

= 10C1 (0.2) (0.8)9

(ii) probability that at least one is late

= 1 – probability that none is late

= 1 – p(x=0)

= 1– (0.8)10

Example 7.11

The sum and product of the mean and variance of a binomial distribution are 24 and 128. Find the distribution.

Solution :

For Binomial Distribution the mean is np and varaiance is npq

Given values are np + npq = 24

np(1 + q) = 24 – (1)

Other term np × npq = 128

n2p2q=128 - (2)

From (1) we get np = 24/(1+q) which implies n2p2 = (24/(1+q))2

Substitute this value in equation (2) we get Example 7.12

Suppose A and B are two equally strong table tennis players. Which of the following two events is more probable:

(a) A beats B exactly in 3 games out of 4 or

(b) A beats B exactly in 5 games out of 8 ?

Solution :

Here p = q = 1/2

(a) probability of A beating B in exactly 3 games out of 4 = 1/4 = 25%

(b) probability of A beating B in exactly 5 games out of 8 = 7/32 = 21.875%

Clearly, the first event is more probable.

Example 7.13

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes.

Solution :

In a throw of a pair of dice the doublets are (1,1) (2,2) (3,3) (4,4) (5,5) (6,6)

Probability of getting a doublet p = 6/36 = 1/6

q = 1 – p = 5/6 and also n = 4 is given

The probabitliy of successes Therefore the probability of 2 successes are Tags : Definition, Properties, Derivation, Formulas, Solved Example Problems , 12th Business Maths and Statistics : Chapter 7 : Probability Distributions
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12th Business Maths and Statistics : Chapter 7 : Probability Distributions : Binomial distribution | Definition, Properties, Derivation, Formulas, Solved Example Problems