COMPUTER ORIENTED

STATISTICAL METHODS
Permutations and
Combinations
 What's the Difference?
• In English we use the word "combination" frequently, without
thinking if the order of things is important.
• In real life, on many occasions, we need to deal with a number of
items at the same time.
• And we need to decide how many among them to be considered for
a given purpose.
• "My fruit salad is a combination of apples, grapes and
bananas"
• "The combination of 472 is rigid".
• So, in Mathematics we use more precise language:
• When the order doesn't matter, it is a Combination.
• When the order does matter it is a Permutation.
Fundamental principle of counting

• If I have 4 apples and 3 oranges in a basket,

• there are 3+4=7 ways I can pick one fruit from the basket,
• and there are 3⋅4 ways I can pick one apple and one
• orange from the basket.
 Example

1. In a race with eight competitors, how many different possibilities are

there for who finishes finish first, second and third?
8P3

2. Three major new roads are to be constructed and eight companies

have tendered for the three projects. If at most one construction
project is to be given to any one company, what is the total number
of ways in which the three contracts can be awarded?
8P3
• Example 2.19: In a college football training session, the defensive
coordinator needs to have 10 players standing in a row.Among these
10 players, there are 1 freshman, 2 sophomores, 4 juniors, and 3
seniors, respectively.
• How many different ways can they be arranged in a row if only their
class level will be distinguished?
 Reasons for studying probability

• Essential component or inductive logic (more useful in critical thinking)

• Essential component for scientific reasoning
• Philosophically important
• Essential in decision making
• In research investigation
• Along with statistics we can make future analysis (predicting accidents,
weather forecasting, natural calamite destructions etc)
• PROBABILITY
• SAMPLE SPACE AND EVENTS:
• 1. Random Experiment:
• A random experiment is an experiment whose outcome or result is not unique
and therefore cannot be predicted with certainty.
• Ex: In tossing a coin, one is not sure whether a head or tail will occur.
. Trail:
Each performance in a random experiment
is called a trial.
Ex: Tossing a coin first time is first trail,
second time is second trial.
3. Outcome:
The result of a trial in a random experiment
is called an outcome. In coin tossing
experiment getting head and tail are
outcomes.
4. Sample Space:
The set of all possible outcomes of an
experiment is called a sample space.
Ex: In throwing a die, {1,2,3,4,5,6}will form
the sample space.
Discrete Sample Space:- A sample space is
said to be a discrete sample space if it has
finitely many or a countable infinity of
elements.
Ex:- a) A sample space consists of finite no
of elements.
-2000 students.
b) The sample space consists of countable
infinity of elements -- the whole set of
natural nos
Continuous Sample Space:-
If the elements of a sample space
constitute a continuum – for example, all
the points on a line, all the points on a line
segment or all the points in plane – the
sample space is said to be continuous .
Event:-
Every non empty subset of a sample space
of a random experiment is called an event.
Ex: In throwing a die{1,2,3,4,5,6}
-- Sample Space
Event: A = {2,4.6} -- Set of even numbers
B = {1,3.5} -- Set of odd numbers
C= {1,2,3,4,5,6} -- Set of all elements
Mutually exclusive events:

• Events are said to be mutually

exclusive if the happening of anyone of
them prevents the happening of all the
others i.e. if no two or more of them
can happen simultaneously in the same
trail.
• Ex: In tossing a coin the events Head
turning up and tail turning up are
mutually exclusive.
Some Formulae and Laws
1. A A = A 2. A A=A

3. A B=B A

4. ( A B) C=A (B C)

5. A B=B A 6. A  =A

7. A  = 8. A U =U
9. A U=A 10. A Ac = U

11. A Ac =  12. ( Ac )c = A

13. Uc =  ,  c = U
Demorgan Laws:

1. (A B) C
= AC BC

2. (A B)C = AC BC

Distributive Laws:

A (B C) = (A B) (A C)

A (B C) = (A B) (A C)

If A  B, A B=A

A B=B
PROBABILITY:
The classical probability concept:
Ex:-What is the probability of drawing an
ace from a well shuffled deck of 52 Playing
cards?
Solution:
There are S=4 aces among the n=52 cards,
so we get s/n= 4/52 = 1/13.
Ex: A bag contain 5 red balls, 8 blue balls
and 11 white balls. Three balls are drawn
together from the box. Find the probability
that.
1) One is red, one is blue and one is white.
2) Two whites and one red.
3) Three white.
Solution:
There are 24C3 = 2024 equally likely
ways of choosing 3 of 24 balls, so n=2024.
1. The number of possible cases
5C1. 8C1.11C1 = 440.
Required probability = s/n = 440/2024 =
55/253
2. No. of possible cases = 11C2.5C1 = 275.
Required probability = 275/2024 = 25/184
3. No. of possible cases = 11C3 = 165
Required probability = 165/2024 = 15/184
EXAMPLE
AXIONS OF PROBABILITY:
Given a finite sample space S and an
event A in S we define P(A), the probability
of A is a function defined on a sample
space satisfies the following three
conditions.
Axiom : 1 0 ≤ P(A) ≤ 1 for each event A in S
Axiom : 2 P(S) = 1
Axiom: 3. If A and B are mutually
exclusive events in S, then
P (AUB) = P (A) + P (B).
Ex: Find out whether the following

probabilities are permissible.

 
1) S =A A A 
 1, 2, 3 
   
P  A  = -1/2, P  A  = ¼, P  A  = ¼, .
 1,   2,   3
 
Can’t be permissible since PA  is
 1, 

negative.
1) S = (A1, A2, A3)
P(A1) = 1/3, P (A2) = 1/3, P(A3) = 1/6.
Not permissible since P(A1) +P(A2)
+P(A3) = P(S)  1.

2) S= (A1, A2, A3, A4)

P(A1) = ½, P(A2) = 1/6, P(A3) = 1/6,

P(A4) = 1/6.
4
Permissible  P (Ai) = P(S) = 1
i =1
Some Elementary theorems
Generalization of the third axiom of
Probability
Theorem
If A1, A2 ,-- - - - - - - - -An are mutually

exclusive events in a sample space S,

then P(A1 A2 - - - ----- An) = P (A1) + P (A2)

+ - - --- - --- P (An)

Problem

The probability that a consumer testing

service will rate a new antipollution device

for cars very poor , poor, fair , good , very

good or excellent are .07 , 0.12 ,0.17 ,0.32 ,

0.21 and 0.11. What the probabilities that

it will rate the device

a) very poor, poor, fair or good
(A1) (A2) (A3) (A4)

b) Good, very good or excellent?

(A4), (A5) (A6)
Solution

Since the probabilities are all

mutually exclusive, we have

a) P(A1 A2 A3 A4) = P(A1) + P (A2) +

P (A3) + P ( A4) =

0.07 + 0.12 + 0.17+ 0.32 = 0.62

b) P( A4 A5 A6 ) = P (A4) + P ( A5 ) + P ( A6)

= 0.32 + 0.21 + 0.11 = 0.64

Note

A sample space of n outcomes has 2n

subsets.
Rules for calculating probability of an event

Theorem

If “A” is an event in the finite

sample space S, then P (A ) equals the sum

of the probabilities of the individual

outcomes comprising A
Proof

Let E1, E2, E3 - - - - - En be the “ n ”

outcomes comprising event “A ” , so that

we have A = E1 E2 - -- -- -- En

Since E ’
s are individual outcomes they

are mutually exclusive , hence

P ( A ) = P (E1 E2 E3 - - - - - En) =

P ( E1) + P( E2 ) + P ( E3 ) - - - - - - + P ( En )
Using Venn diagrams to visualize
Probability Calculations

Let us consider the Venn diagram

which concerns the job offers received by

recent engineering graduates. The letter I

and G stand for a job offer from industry

and a job offer from government

respectively.
From Venn diagram

P(I) = 0.18+0.12 = 0.30

P(G)= 0.12+0.24 = 0.36

P(I G)= .18+ .12+.24= .54

Here I,G and I G are mutually exclusive.

So we can add various probabilities. But if

they are not mutually exclusive, this is not

possible.
General addition rule for probability

Theorem:- If A and B are any events in S,

then P(A B) = P(A) + P (B) – P(A B).

Proof.

From the figure A = (A BC) (A B).

B = (AC B) (A B).
Note that (A BC), ( A B), (AC B) are

mutually exclusive.

Applying Axiom (3)

we have,
P(A) = P[ (A BC ) (A B)]

P(A) = P[(A BC ) + P(A B) ----------------- (1)

P(B)= P[ (AC B) (A B)]

P(B)= P (AC B) + P(A B) ----------------- (2)

Adding (1) and (2)

P(A) + P(B) = P[(A BC ) + P(A B)

+ P (AC B) + P(A B) --------- (3)

Note from the figure that

A B = (A BC ) (A B) (AC B)

P (A B) = P (A BC ) + P (A B) +P (AC B)
------ (4)

substituting (4) in (3)

P (A) + P (B) = P (A B) + P (A B)

ie, P (A B) = P (A) + P (B) – P (A B)

 → If A and B are mutually exclusive

P(A B) =0

ie P (A B) = P (A) + P (B) -------------- (Third

axiom of probability)
Problem

A card is drawn at random from a

well shuffled deck of 52 cards. Find the

probability of getting a spade or a king ?

Solution

P(S K)=P(S)+P(K)–

P(S K)

= 13/52 + 4/52 - 1/52 = 4/13

Probability rule of the complement

Theorem
If A is any event in S ,

then P ( Ac ) = 1 – P ( A )
Proof

We know that

S=A Ac .

Here A and Ac mutually exclusive

P(S)=P(A Ac )

P ( S ) = P ( A ) + P ( AC )
  1 =P(A)+P(A C
)
( By Axiom 3 )

 P ( AC ) = 1 – P ( A )
Also P ( φ ) = 1 – P ( S ) , since the

empty set φ is the complement of S .

Problem

Determine the probability for the

following event

A non defective bolt will be found if out of

600 bolts already examined 12 were

defective
Solution

Probability of a defective bolt

P ( D ) = 12/ 600 = 1/ 50

 Probability of finding a non

defective bolt P ( DC ) = 1- P ( D ) =

= 1- ( 1/50 ) = 49/50
Problem

If P ( A ) = 1/2 , P ( B ) = 1/3 ,

P(A B ) = 1/5

Find 1) P ( A B)

2) P ( AC B) 3) P ( A BC )

4) P ( AC BC ) 5) P ( AC BC )
Solution
1) P ( A B ) = P ( A ) + P ( B ) –P ( A B)

= (1/2) + ( 1/3 ) – ( 1/5 ) = 19/30

2) B = (AC B) (A B)

P( B ) = P (AC B) + P ( A B)

 P (AC B) = P ( B ) – P ( A B)

= ( 1/3 ) – ( 1/5 ) = 2/15

3) A=(A BC ) ( A B)

P(A)= P(A BC ) + P(A B)

 P(A BC ) = P ( A ) - P(A B)

= ( 1/2) - ( 1/5 ) = ( 3/10 )

4 ) P ( AC BC ) = P ( A B)C

= 1 – ( P (A B))=

= 1 – ( 19/ 30 ) = ( 11/30 )