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}

p^{x}q^{(n−x)}

x successes in n trials
can occur in ^{n}C_{x} ways and the probability for each of
these ways is same namely p^{x} q^{n−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}

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.

∴
Variance = *E*(*X*^{2}) – *E*(*X*)^{2}

*= n*^{2}*p*^{2}* *–* np*^{2}*
*+* np *–* n*^{2}*p*^{2}

*= 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 *p* ≠
*q*. 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) = ^{n}C_{x}
p^{x} q^{n−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) = ^{n}C_{x}
p^{x} q^{n−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*) = 10C_{x} (0.2)^{x} (0.8)^{10-x}t.

(i) probability that
exactly one is late *P*(*y*=1) = 10C_{1}*pq*^{9}

= 10C_{1} (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

n^{2}p^{2}q=128 *- (2)*

From (1) we get np =
24/(1+q) which implies n^{2}p^{2} = (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

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