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Title: PROBABILITY
Description: THE CONCEPT OF PROBABILITY FROM ADVANCED MATHEMATICS BOOK
Description: THE CONCEPT OF PROBABILITY FROM ADVANCED MATHEMATICS BOOK
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Introduction to
Probability
and Statistics
Syllabus
Introduction to Probability and Statistics
Course ID:MA2203
Lecture-2
Course Teacher: Dr
...
...
...
1/14
Introduction to
Probability
and Statistics
Syllabus
Continue
...
Mathematically, if for n trials an event A
happens m times, then
m
P(A) = lim
...
E
...
The number of heads found were: 502,
511, 529, 504, 476, 507, 520, 504, 529
...
5079 ≈
...
...
5
...
We also call the probability as the counterpart of relative
frequency
...
Then Relative frequency of A is denoted
by
f (A)
frel (A) =
...
For
two mutually
exclusive events, say A and B we have,
∪
frel (A B) = frel (A) + frel (B)
...
This we call as Axiomatic definition of
Probability, which we will discuss in the next
...
...
...
3/14
Introduction to
Probability
and Statistics
Syllabus
Axiomatic Definition
Let S be a given sample space, and S be a σ-field on it
...
1 For every event A ∈ S, 0 ≤ P(A) ≤ 1
...
3 For mutually exclusive events A and B,
∪
P(A B) = P(A) + P(B)
...
that is,
∪ ∪
P(A1 A2
...
...
...
...
Observe that,
∪ the ∅ does not contain any elements, hence
we have S ∅ = S
...
Using the
third axiom, we have P(S ∅) = P(S) = P(S) + P(∅), which
implies that P(∅) = 0
...
In practice, probability ‘0’ is assigned to the events which
are so rare that they happen only once in a life time
...
2 Probability of the complementary event Ac of A is obtained as
P(Ac ) = 1 − P(A)
...
Hence using axiom (3),
have P(A A ) = P(S) = 1, this implies P(A) + P(Ac ) = 1,
and the result
...
...
...
5/14
Introduction to
Probability
and Statistics
Syllabus
Some More Results
∩
3 For any
(i) P(Ac B) = P(B) −
∩ two events A and B, we have ∩
P(A B)
...
To prove (ii), the events
A B c are
∪ B∩and
c
mutually
∩ exclusive, hence P(A) =∩P(B (A B )) = P(B) +
P(A ∩ B c ), this implies that P(A B c ) = P(A) − P(B)
...
4 Addition Theorem∪
of Probability: For events A ∩
and B in a sample space S, P(A B) = P(A)
+
P(B)
−
P(A
B)
...
∩Hence applying
∩ axiom (3), we
have
P(A
B)
=
P(A
−
(A
B))
+
P(A
B) +
∩
∪
P(B − (A B)) = P(A) + P(B) − P(A B)
...
Ex
...
Hint:
Use set theory approach
...
...
...
6/14
Introduction to
Probability
and Statistics
Boole’s Inequality
For n events A1 , A2 ,
...
i=1
Ai )
n
∑
≤
i=1
P(Ai )
...
Verify the result
∩ for n = 2 that is, P(A1 ∩ A2 ) =
P(A1 ) + P(A2 ) − P(A1 A2 ) ≤ 1, this implies P(A1 A2 ) ≥
P(A1 ) + P(A2 ) − 1
...
Assume that
the result holds true for n = k , that is
P(
k
∩
i=1
Ai ) ≥
k
∑
P(Ai ) − (k − 1)
...
...
...
...
Syllabus
k∩
+1
L
...
S : P(
Ai ) =
P((
i=1
k
∩
Ai )
∩
Ak+1 )
i=1
≥ P(
k
∩
Ai ) + P(Ak +1 ) − 1
i=1
≥
k
∑
P(Ai ) − (k − 1) + P(Ak +1 ) − 1
i=1
=
k +1
∑
P(Ai ) − k : R
...
S
i=1
...
...
...
Proof (ii): Now applying the above inequality, with the events
Ac1 , Ac2 ,
...
i=1
Hence, we have
n
∑
i=1
∩
P(Ai ) ≥ 1 − P( Aci )
∪
= 1 − P([ Ai ]c )
∪
= P( Ai )
...
...
...
9/14
Introduction to
Probability
and Statistics
Syllabus
Bonferroni’s Inequality
Given n events, A1 , A2 ,
...
1≤i≤j≤n
Proof: This can be proved by the method of mathematical
induction
...
P(
3
∪
)
=
i=1
3
∑
P(Ai ) −
i=1
≥
3
∑
∑
P(Ai
∩
Aj ) + P(
1≤i
P(Ai ) −
i=1
∑
3
∩
Ai )
i=1
P(Ai
∩
Aj )
...
To prove the result for
n = k + 1
...
...
...
10/14
Introduction to
Probability
and Statistics
Proof Continue
...
i=1
...
...
...
Syllabus
From Boole’s inequality, we have
P(
k
∪
(Ai
∩
Ak+1 )) ≤
k
∑
i=1
P(Ai
∩
Ak +1 )
i=1
Using this, we get
k∪
+1
P(
Ai ) ≥
i=1
k+1
∑
P(Ai ) −
i=1
=
k+1
∑
i=1
∑
P(Ai
∩
Aj ) −
k
∑
1≤i
P(Ai ) −
∑
P(Ai
∩
Ak +1 )
i=1
P(Ai
∩
Aj )
...
...
...
12/14
Title: PROBABILITY
Description: THE CONCEPT OF PROBABILITY FROM ADVANCED MATHEMATICS BOOK
Description: THE CONCEPT OF PROBABILITY FROM ADVANCED MATHEMATICS BOOK