Download PROBABILITY Definition 2.1 The set of all possible outcomes of a

Topic: Probability
PROBABILITY
Definition 2.1 The set of all possible outcomes of a
statistical experiment is called a sample space and is
represented by the symbol S.
Element - each outcome in a sample space, or also known
as member of a sample space or sample point
Observation - refer to any recording of information,
whether it is numerical or categorical
EVENTS
Definition 2.2 An event is a subset of a sample space.
Or An event is a set of outcomes in a sample space.
An event can be specified
1) by giving a list of the outcomes that make it up, or
2) by giving a descriptive phrase or condition that serves to
characterize those outcomes
An event may be a subset that includes the entire sample
space S, or a subset of S called the null set and denoted by
the symbol ∅, which contains no elements at all.
Definition 2.2a Events A and B are equivalent, written A =
B, if and only if the outcomes in A are precisely the same as
the outcomes in B.
ENGSTAT Notes of AM Fillone
Topic: Probability
Definition 2.3 The complement of an event A with respect
to S is the set of all elements of S that are not in A. We
denote the complement of A by the symbol A’.
Definition 2.4 The intersection of two events A and B,
denoted by the symbol A ∩ B, is the event containing all
elements that are common to A and B.
Definition 2.5 Two events A and B are mutually exclusive
or disjoint if A ∩ B = O, that is, if A and B have no
elements in common.
Definition 2.6 The union of the two events A and B,
denoted by the symbol A ∪ B, is the event containing all
the elements that belong to A or B or both.
Definition 2.7 Event A is contained in event B, written A
⊂ B, if and only if every element of A is also in B.
ENGSTAT Notes of AM Fillone
Topic: Probability
Venn Diagram - a graphical illustration of the relationship
between events and corresponding sample space. In a Venn
diagram we let the sample space be a rectangle and
represent events by circles drawn inside the rectangle. For
example
A
B
2
7
6
4
1 3
5
C
In the Figure, we see that
A ∩ B = regions 1 and 2,
B ∩ C = regions 1 and 3,
A ∪ C = regions 1, 2, 3, 4, 5, and 7,
B’ ∩ A = regions 4 and 7,
A ∩ B ∩ C = region 1,
(A ∪ B) ∩ C’ = regions 2, 6, and 7, etc.
ENGSTAT Notes of AM Fillone
Topic: Probability
Counting Sample Points
Theorem 2.1 If an operation can be performed in n1 ways,
and if for each of these a second operation can be performed
in n2 ways, then the two operations can be performed in
n1*n2 ways.
Theorem 2.2 If an operation can be performed in n1 ways,
and if for each of these a second operation can be performed
in n2 ways, and for each of the first two a third operation can
be performed in n3 ways, and so forth, then the sequence of
k operations van be performed in n1*n2 *, …,*nk ways.
Theorem 2.3 The number of permutations of n distinct
objects is n!
Theorem 2.4 The number of permutations of n distinct
objects taken r at a time is
nPr = n(n-1)*…*(n-r+1)
or, in factorial form,
n!
nPr = ------------(n-r)!
ENGSTAT Notes of AM Fillone
Topic: Probability
Theorem 2.5 The number of permutations of n distinct
objects arranged in a circle is (n-1)!
Theorem 2.6 The number of distinct permutations of n
things of which n1 are of one kind, n2 are of a second kind,
…, nk are of a kth kind, and n1 + n2 …+ nk = n, is
n!
------------------------n1!n2!…nk!
Theorem 2.7 The number of ways of partitioning a set of n
objects into r cells with n1 elements in the first cell, n2
elements in the second, and so forth, is

n

n!
 n1, n2, …, nr  = -------------------------n1! n2! …nr!
where n1 + n2 + … + nr = n.
Theorem 2.8 The number of combinations of n distinct
objects taken r at a time is
n
n(n-1)(n-2) *… * (n-r+1)
 r  = ------------------------------------r!
or, in factorial form,
n
n!
 r  = ----------------------r! (n-r)!
ENGSTAT Notes of AM Fillone
Topic: Probability
PROBABILITY OF AN EVENT
Definition 2.8 The probability of an event A is the sum of
the weights of all sample points in A. Therefore,
0 ≤ P(A) ≤ 1,
P(∅) = 0, and P(S) = 1
Theorem 2.9. If an experiment result in any one of N
different equally likely outcomes, and if exactly n of these
outcomes correspond to event A, then the probability of
event A is
n
P(A) = -------.
N
ADDITIVE RULES
Theorem 2.10 If A and B are any two events, then
P(A ∪B ) = P(A) + P(B) - P(A ∩B)
Corollary 1 If A and B are mutually exclusive, then
P( A ∪ B) = P(A) + P(B)
Corollary 2 If A1, A2, A3, …, An are mutually exclusive,
then
P(A1∪A2∪…∪An) = P(A1) + P(A2) + … + P(An).
ENGSTAT Notes of AM Fillone
Topic: Probability
Corollary 3 If A1, A2, …, An is a partition of a sample
space S, then
P(A1∪ A2∪…∪An) = P(A1)+P(A2)+…+P(An)
= P(S)
=1
Theorem 2.11 For three events A, B, C
P(A∪B∪C) = P(A) + P(B) + P( C) - P(A∩B) - P(A∩C) P(B∩C) + P(A∩B∩C)
Theorem 2.12 If A and A’ are complementary event, then
P(A) + P(A’) = 1.
CONDITIONAL PROBABILITY
Definition 2.9 The conditional probability of B, given A,
denoted by P(B|A), is defined by
P(A∩B)
P(B|A) = -------------- if P(A) > 0.
P(A)
Independent Events
Definition 2.10 Two events A and B are independent if and
only if
P(B|A) = P(B) and P(A|B) = P(A).
Otherwise, A and B are dependent.
ENGSTAT Notes of AM Fillone
Topic: Probability
MULTIPLICATIVE RULES
Theorem 2.13 If in an experiment the events A and B can
both occur, then
P(A∩B) = P(A)P(B|A) or
P(B∩A) = P(B)P(A|B).
Theorem 2.14 Two events A and B are independent if and
only if
P(A∩B) = P(A)P(B).
Theorem 2.15 If, in an experiment, the events A1, A2, A3,
…, Ak can occur, then
P(A1∩ A2∩ A3∩…∩Ak) =
P(A1)P(A2|A1)P(A3|A1∩A2)…P(Ak|A1∩A2∩…∩Ak-1)
If the events A1, A2, A3, …, Ak are independent, then
P(A1∩ A2∩ A3∩…∩Ak) = P(A1)P(A2)P(A3)…P(Ak).
ENGSTAT Notes of AM Fillone
Topic: Probability
BAYES’ RULE
Theorem 2.16 If the events B1, B2, …, Bk constitute a
partition of the sample space S such that P(Bi) ≠ 0 for i = 1,
2, …, k, then for any event A of S,
k
k
i=1
i=1
P(A) = Σ P(Bi ∩ A) = Σ P(Bi)(A|Bi)
Theorem 2.17 (Bayes’ Rule) If the events B1, B2, …, Bk
constitute a partition of the sample space S, where P(Bi) ≠ 0
for I = 1, 2, …, k, then for any event A in S such that
P(Br) P(A|Br)
P(Br ∩ A)
P(Br|A) = -------------------- = ---------------------k
Σ P(Bi ∩ A)
i=1
ENGSTAT Notes of AM Fillone
k
Σ P(Bi) P(A|Bi)
i=1