Chapter 2: Probability

 An Experiment: is some procedure (or process) that we do

and it results in an outcome.
A random experiment: is an experiment we do not
know its exact outcome in advance but we know the
set of all possible outcomes.

2.1 The Sample Space:

Definition 2.1:
 The set of all possible outcomes of a statistical experiment
is called the sample space and is denoted by S.
 Each outcome (element or member) of the sample space S
is called a sample point.
2.2 Events:

Definition 2.2:
An event A is a subset of the sample space S. That is AS.
 We say that an event A occurs if the outcome (the result)
of the experiment is an element of A.
 S is an event ( is called the impossible event)
 SS is an event (S is called the sure event)

Example:
Experiment: Selecting a ball from a box containing 6 balls
numbered 1,2,3,4,5 and 6. (or tossing a die)
 This experiment has 6 possible outcomes
The sample space is S={1,2,3,4,5,6}.
Consider the following events:
E1= getting an even number ={2,4,6}S
E2 = getting a number less than 4={1,2,3}S
E3 = getting 1 or 3={1,3}S
E4 = getting an odd number={1,3,5}S
E5 = getting a negative number={ }= S
E6 = getting a number less than 10 = {1,2,3,4,5,6}=SS
Notation:
•n(S)= no. of outcomes (elements) in S.
•n(E)= no. of outcomes (elements) in the event E.
Example:
Experiment: Selecting 3 items from manufacturing process;
each item is inspected and classified as defective (D) or non-
defective (N).
· This experiment has 8 possible outcomes
S={DDD,DDN,DND,DNN,NDD,NDN,NND,NNN}

· Consider the following events:

A={at least 2 defectives}= {DDD,DDN,DND,NDD}S
B={at most one defective}={DNN,NDN,NND,NNN}S
C={3 defectives}={DDD}S
Some Operations on Events:
Let A and B be two events defined on the sample space S.
Definition 2.3: Complement of The Event A:
 Ac or A' S
 Ac = {x S: xA }
 Ac consists of all points of S that are not
in A.
 Ac occurs if A does not.

Definition 2.4: Intersection: S

· AB =AB={x S: xA and xB}

· AB Consists of all points in both A and
B.
· AB Occurs if both A and B occur
together.
Definition 2.5: Mutually Exclusive (Disjoint) Events:

Two events A and B are mutually exclusive (or disjoint) if and

only if AB=; that is, A and B have no common elements (they
do not occur together).

A B   A B = 
A and B are not A and B are
mutually exclusive mutually exclusive
(disjoint)
 Definition 2.6: Union:
· AB = {x S: xA or xB } S

· AB Consists of all outcomes in

A or in B or in both A and B.
· AB Occurs if A occurs,
or B occurs, or both A and B occur.
That is AB Occurs if at least one of
A and B occurs.
2.3 Counting Sample Points:
· There are many counting techniques which can be used to
count the number points in the sample space (or in some
events) without listing each element.
· In many cases, we can compute the probability of an event
by using the counting techniques.
Combinations:
In many problems, we are interested in the number of ways of
selecting r objects from n objects without regard to order.
These selections are called combinations.
· Notation:
n factorial is denoted by n! and is defined by:
n!  n   n  1    n  2      2   1  for n  1, 2, 
0!  1

Example: 5!  5  4  3  2  1  120
Theorem 2.8:
The number of combinations of n distinct objects taken r at a
time is denoted by  n  and is given by:
 
r 
n  n!
   ; r  0 , 1, 2, , n
r  r ! n  r !
 Notes:
· n  is read as “ n “ choose “ r ”.
 
r 
· n  ,n  , n  , n   n 
   1    1    n     
n  0  1  r  n  r 
· n  = The number of different ways of selecting r objects
  from n distinct objects.
r 
· n  = The number of different ways of dividing n distinct
  objects into two subsets; one subset contains r
 
r
objects and the other contains the rest (nr) bjects.
Example:
If we have 10 equal–priority operations and only 4 operating
rooms are available, in how many ways can we choose the 4
patients to be operated on first?

Solution:
n = 10 r = 4
The number of different ways for selecting 4 patients from 10
patients is

10  10 ! 10 ! 10  9  8  7  6  5  4  3  2  1
    
 4  4! 10  4 ! 4!  6!  4  3  2  1    6  5  4  3  2  1 
 210 ( different ways )
2.4. Probability of an Event:

· To every point (outcome) in the sample space of an

experiment S, we assign a weight (or probability), ranging
from 0 to 1, such that the sum of all weights (probabilities)
equals 1.
· The weight (or probability) of an outcome measures its
likelihood (chance) of occurrence.
· To find the probability of an event A, we sum all
probabilities of the sample points in A. This sum is called the
probability of the event A and is denoted by P(A).

Definition 2.8:
The probability of an event A is the sum of the weights
(probabilities) of all sample points in A. Therefore,
1) 0  P  A   1
2) P  S   1
3) P    0
Example 2.22:
A balanced coin is tossed twice. What is the probability that at
least one head occurs?

Solution:
S = {HH, HT, TH, TT}
A = {at least one head occurs}= {HH, HT, TH}
Since the coin is balanced, the outcomes are equally likely; i.e.,
all outcomes have the same weight or probability.

Outcome Weight
(Probability)

HH P(HH) = w
4w =1  w =1/4 = 0.25
HT P(HT) = w
P(HH)=P(HT)=P(TH)=P(TT)=0.25
TH P(TH) = w
TT P(TT) = w
sum 4w=1
The probability that at least one head occurs is:
P(A) = P({at least one head occurs})=P({HH, HT, TH})
= P(HH) + P(HT) + P(TH)
= 0.25+0.25+0.25
= 0.75

Theorem 2.9:
If an experiment has n(S)=N equally likely different outcomes,
then the probability of the event A is:

n( A) n( A) no . of outcomes in A
P ( A)   
n(S ) N no . of outcomes in S
Example 2.25:
A mixture of candies consists of 6 mints, 4 toffees, and 3
chocolates. If a person makes a random selection of one of
these candies, find the probability of getting:
(a) a mint
(b) a toffee or chocolate.

Solution:
Define the following events:
M = {getting a mint}
T = {getting a toffee}
C = {getting a chocolate}
Experiment: selecting a candy at random from 13 candies
n(S) = no. of outcomes of the experiment of selecting a candy.
= no. of different ways of selecting a candy from 13 candies.

13 
    13
1
 The outcomes of the experiment are equally likely because the
selection is made at random.
(a) M = {getting a mint}
n(M) = no. of different ways of selecting a mint candy
from 6 mint candies
6 
    6
1  n M  6
P(M )= P({getting a mint})= 
n S  13

(b) TC = {getting a toffee or chocolate}

n(TC) = no. of different ways of selecting a toffee
or a chocolate candy
= no. of different ways of selecting a toffee
candy + no. of different ways of selecting a
chocolate candy
 4  3 
       4  3  7
 1  1 
 = no. of different ways of selecting a candy
from 7 candies
7 
    7
1  n T C  7
P(TC )= P({getting a toffee or chocolate})= 
n S  13
Example 2.26:
In a poker hand consisting of 5 cards, find the probability of
holding 2 aces and 3 jacks.
Solution:
Experiment: selecting 5 cards from 52 cards.
n(S) = no. of outcomes of the experiment of selecting 5 cards
from 52 cards.
 52  52 !

    2598960
 5  5!  47 !
The outcomes of the experiment are equally likely because the
selection is made at random.
Define the event A = {holding 2 aces and 3 jacks}
n(A) = no. of ways of selecting 2 aces and 3 jacks
= (no. of ways of selecting 2 aces)  (no. of
ways of selecting 3 jacks)
= (no. of ways of selecting 2 aces from 4 aces)  (no.
of ways of selecting 3 jacks from 4 jacks)
4  4 
     
2  3 
4! 4!
   6  4  24
 2!
2! 3! 1!
P(A )= P({holding 2 aces and 3 jacks })
n A  24
   0 . 000009
n S  2598960