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Sampling and Distribution

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In probability theory and statistics, a probability distribution is a mathematical function that predicts different possible outcomes for an experiment. It is a mathematical description of a random phenomenon in terms of its sample space and the probabilities of events (subsets of the sample space)

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  • June 16, 2021
  • 102
  • 2020/2021
  • Class notes
  • Dr. nasila
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keanuperumal
Chapter 1 – Probability, counting methods and binomial
coefficients
1.1 Definitions and theorems
Def. 1.1 A random experiment is an experiment whose outcomes
cannot be predicted with certainty.

Examples: Tossing a coin. Tossing a pair of dice. Drawing a card from
a deck of 52 cards. The sequence of winning numbers in the Lotto draw.

Def. 1.2 A sample space of a random experiment is the collection of
all possible outcomes.

Examples

1 Tossing a die. Outcomes = S = {1,2,3,4,5,6}.
2 Drawing a winning number from Lotto. Outcomes = S = {0,1,2,  ,49}
.
3 Drawing a card from a deck of 52 cards. Outcomes = { 2,3, . . . ,10,
jack, queen, king, ace for spades, club, diamonds and hearts } .

Def. 1.3 Each element of the sample space is called a sample point.

Def. 1.4 If the number of points are countable (finite) this samples
space is called a countable sample space.

Def 1.5 If the number of points are uncountable (infinite), then the
sample space is said to be a continuous sample space.

Examples

1 Countable sample space. Tossing a die where S = {1,2,3,4,5,6}.
Tossing a coin where S = {h, t} .

2 Continuous sample space. Measuring the temperature ( T ) in given
location, where S = {−5  T  32} .


1

,Def. 1.6 A subset A of the sample space S is said to be an event if it
belongs to a collection à of subsets of S satisfying the following three
rules:

(a) S Ã (b) If A Ã then Ac  Ã (c) ) If Aj  Ã then  Aj  Ã for
j  1.
The collection à is called an event space or a  -field.


Theorem 1.1

 Ã (  is an empty set which refers to an impossible event)

Proof

S Ã (a). Sc =   Ã (b).

Theorem 1.2

If A1c  Ã, A2c  Ã then A1  A2  Ã

Proof
A1c  Ã , A2c  Ã (ii) and A1c  A2c  Ã (iii). Using (ii), it follows that
( A1c  A2c ) c  Ã. Using de Morgan’s law ( A1c  A2c ) c = A1  A2  Ã.

Example

Describe the sample space S of rolling a pair of dice. Describe the event
A that the sum of numbers rolled is 7.

Solution

S = {(x, y) | x, y = 1, 2, 3, 4, 5, 6}

A = {(1, 6), (6, 1), (2, 5), (5, 2), (4, 3), (3, 4)}

Def. 1.7


2

,A function is a rule that associates each point in one set (A) of points
with one and only one point in another set of points (B). The sets A and
B are called the domain and counter domain respectively.

Let a  A and b  B . Then the function ( f ) can be written as b = f (a ) . The
set of all values of f (.) is called the range of f (.) .

Def. 1.8

Let S be the sample space of a random experiment. A probability
measure P : A* → [0,1] is a set function which assigns real numbers to
the various events of S satisfying

(P1) P ( A)  0 for all events A  A* ,
(P2) P( S ) = 1 ,
 
(P3) P( Ak ) =  Ak for A1 , A2 ,  mutually exclusive events of A* .
k =1 k =1



Theorem 1.3

P( ) = 0


Proof

Let A1 = S , Ai =  for i = 2,3,  . Then S =  Ai and Ai  A j =  for i  j.
i =1
   
1 = P( S ) = P( Ai ) = P( S ) + P( Ai ) = 1 +  P( Ai ) = 1 +  P( )
i =1 i =2 i =2 i =2




Therefore  P( ) = 0 . Since any probability is
i =2
 0, it follows that
P( ) = 0 .


Theorem 1.4




3

, Let A1 , A2 , , An be mutually exclusive events in à i.e. P ( Ai  A j ) =  for
n n
i j . Then P( Ai ) =  P( Ai ) .
i =1 i =1



Proof
n 
Let An+1 = An+2 =  =  . Then  Ai = Ai  Ã and
i =1 i =1
n   n
P( Ai ) = P( Ai ) =  P( Ai ) =  P( Ai ) since P( ) = 0 .
i =1 i =1 i =1 i =1



For n = 2 the above result gives P( A1  A2 ) = P( A1 ) + P( A2 ) when A1 and A2
are mutually exclusive (disjoint).

Theorem 1.5

For any event A of the sample space S , P( Ac ) = 1 − P( A) .


Proof

A  Ac = S and A  Ac = 
P( S ) = 1 = P( A  Ac ) = P( A) + P( Ac )  P( Ac ) = 1 − P( A)


Theorem 1.6

If A and B  Ã , then P( A  B c ) = P( A) − P( A  B) . The event A  B c is also
written as A − B .

Proof

A = ( A  B)  ( A  B c ) . Also ( A  B)  ( A  B c ) =  .
P( A) = P( A  B) + P( A  B c )  P( A  B c ) = P( A) − P( A  B) or
P( A − B ) = P( A) − P( A  B )


The event A  B can also be written as AB and therefore the result can
also be expressed as P( A − B) = P( A) − P( AB) .


4

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