Lecture 4,5 slide about probability and statistics

This is a maths lecture about the topic probability and statistics

1. Introduction to Probability
Chapter 2: Conditional Probability
Dr. Nitin Gupta
Department of Mathematics
Indian Institute of Technology Kharagpur,
Kharagpur - 721 302, INDIA.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 1 / 17

2. Outline
1
Conditional Probability
2
Independence
3
Theorem of total probability
4
Bayes' theorem
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 2 / 17

3. References
1
Probability and statistics in engineering by Hines et al (2003) Wiley.
2
Mathematical Statistics by Richard J. Rossi (2018) Wiley.
3
Probability and Statistics with reliability, queuing and computer
science applications by K. S. Trivedi (1982) Prentice Hall of India
Pvt. Ltd.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 3 / 17

4. Conditional Probability
Consider a family having two children, then =fBB;GB;BG;GGg,
n() = 4. Consider the eventA: both the children are girls. Then
P(A) = 1=4.
If some information in the form of "E: at least one of the children is a
girl" is known. Then reduced sample space isE=fGB;BG;GGg. Then
the probability ofAgiven the conditionEisP(AjE) = 1=3.
Note thatP(AjE)P(A).
P(AjE) =
n(A\E)
n(E)
=
n(A\E)=n()
n(E)=n()
=
P(A\E)
P(E)
, providedP(E)>0.
Denition
Let probability model be (;f;P). Then the conditional probability of
A2fgivenBis dened as
P(AjB) =
P(A\B)
P(B)
;P(B)>0:
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 4 / 17

5. Multiplication rule
The probability that n eventsA1;A2; : : : ;An2foccur in a sequence is
P(\
n
i=1Ai) =P(A1)P(A2jA1)P(A3jA1\A2) P(AnjA1\A2 \An1);
providedP(A1)>0;P(A1\A2)>0; : : : ;P(A1\A2 \An1)>0.
Example
A bag contains 5 red, 5 white and 4 blue balls. If someone draws 3 balls
one by one without replacement, then the probability that three balls will
be drawn in the sequence red-white-blue is
P(R1\W2\B3) =P(B3jW2\R1)P(W2jR1)P(R1)
=P(R1)P(W2jR1)P(B3jW2\R1)
=
5
14

5
13

4
12
:
Here note thatR1;W2;B3are dependent events.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 5 / 17

6. Example
Example
A bag contains 5 red, 5 white and 4 blue balls. If someone draws 3 balls
one by one with replacement, then the probability that three balls will be
drawn in the sequence red-white-blue is
P(R1\W2\B3) =P(R1)P(W2)P(B3)
=
5
14

5
14

4
14
:
Here note thatR1;W2;B3are independent events.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 6 / 17

7. Example
Example
In a war game, submarineS1targetsS2, and bothS2andS3targetS1.
The probabilities ofS1;S2andS3hitting their targets are 1/2, 2/3 and
1/3 respectively. They shoot simultaneously. We want to determine the
conditional probability thatS2hits the target andS3does not given that
S1is hit. Let eventEidenote that submarineSihitting its targets. Here
the required probability is
P(E2\E3jS1is hit) =
P(E2)P(E3)
P(E2[E3)
=
2
3

2
3
2
3
+
1
3

2
3

1
3
=
4
7
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 7 / 17

8. Independence
Two events are independent if the occurrence of one does not eect the
occurrence or nonoccurrence of the other.
Denition
EventsAandBare independent ifP(AjB) =P(A). Hence
P(A\B) =P(A)P(B) and alsoP(BjA) =P(B).
IfAandBare independent, thenAand

Bare independent.
IfAandBare independent, then

AandBare independent.
IfAandBare independent, then

Aand

Bare independent.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 8 / 17

9. Example
Example
Suppose thatP(A) = 0:4;P(B) = 0:5, and A and B are independent
events. DetermineP(A
c
[B
c
):Note that
P(A
c
[B
c
) =P(A
c
) +P(B
c
)P(A
c
\B
c
)
= 1P(A) + 1P(B)P(A
c
)P(B
c
)
= 10:4 + 10:5(10:4)(10:5)
= 0:8;
it follows sinceAandBare independent, thenA=A
c
andB=B
c
are
also independent.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 9 / 17

10. Denition
TheneventsA1; : : : ;Anare mutually independent if and only if the
probability of intersection of any 2;3; : : : ;nof these sets is product of their
respective probabilities, i.e., forr= 2;3; : : : ;n,
P(Ai1
\Ai2
\ \Air) =P(Ai1
)P(Ai2
) P(Air):
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 10 / 17

11. Example
Consider the following electronic system (see diagram), which shows the
probabilities of the system components operating properly (i.e, the
reliability of the components). Assume that each component operates
independently. Find the system reliability, i.e., the probability that the
entire system operates?
0.9
0.8
0.7
Solution: Since components are mutually independent. Hence
System reliability = (1(10:9)(10:8))0:7 = 0:686
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 11 / 17

12. Example
Consider the experiment of rolling two fair dice repeatedly and
independently until a total of 5 or a total of 7 appears. We want to
determine the probability that a total of 5 is rolled before a total of 7 is
rolled.
Solution: Let eventAidenote that 5 is rolled before a 7 in theith trail
and experiment terminates.
P(Ai) =P
0
@
8
<
:
i1
\
j=1
(5[7)
c
9
=
;
\5
1
A=
indep

26
36

i1

4
36
NowP(5 before 7) =P

1
[
i=1
Ai
!
=
disjoint
1
X
i=1
P(Ai)
=
1
X
i=1

26
36

i1

4
36
=
4=36
1
26
36
=
2
5
:
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 12 / 17

13. Theorem of total probability
EventsE1; : : : ;Enare mutually exclusive and exhustive, and eventAis
caused by happening ofE1; : : : ;En, then
P(A) =
n
X
i=1
P(AjEi)P(Ei);
hereP(Ei)>0;i= 1;2; : : : ;n.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 13 / 17

14. Bayes' theorem
EventsE1; : : : ;Enare mutually exclusive and exhustive, and eventAis
caused by happening ofE1; : : : ;En, then fori= 1;2; : : : ;n
P(EijA) =
P(AjEi)P(Ei)
P
n
j=1
P(AjEj)P(Ej)
;
hereP(A)>0 andP(Ei)>0;i= 1;2; : : : ;n.
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 14 / 17

15. Example
In a town there are 200 car drivers, 500 two-wheeler drivers and 20 bus
drivers. There is a probability 0.01, 0.03 and 0.15 respectively for an
accident involving car, two-wheeler and bus. One of the drivers meets with
an accident, what is the probability that he/she was driving a car?
Solution: Let event A,B,C denote the events that the chosen driver drives
a car, a two-wheeler, a bus, respectively. Let eventEdenote an accident.
HereP(A) =
200
200+500+20
=
20
72
;P(B) =
50
72
;P(C) =
2
72
. Also
P(EjA) = 0:01;P(EjB) = 0:03 andP(EjC) = 0:15. Now the required
probability, using Bayes' theorem, is
P(AjE) =
P(A)P(EjA)
P(A)P(EjA) +P(B)P(EjB) +P(C)P(EjC)
=
20
72
0:01
20
72
0:01 +
50
72
0:03 +
2
72
0:15
= 0:1
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16. Example
Consider the experiment of rolling two fair dice repeatedly and
independently until a total of 5 or a total of 7 appears. We want to
determine the probability that a total of 5 is rolled before a total of 7 is
rolled.
Solution: Let eventAdenote that 5 is rolled before a 7. Then
A=
S
1
i=1
(A\Bi), whereBiis the event that the game terminates inith
roll. Now the required probability, using theorem of total probability, is
P(A) =
1
X
i=1
P(A\Bi)
=
1
X
i=1

26
36

i1

4
36
=
4=36
1
26
36
=
2
5
:
N. Gupta (IIT Kharagpur) Chapter 2: Conditional Probability 16 / 17

17. Summary
Since there may be some information available about the outcome of the
trail in a given experiment, hence we introduced the concept of the
conditional probability. Also if this information is irrelevant to the event
under considration from there comes the deniton of the independence of
the events. These denitions can be used to nd the reliabilities of the
series-parallel or parallel-series structures. Hence examples are provided for
the same. In the last theorem of total probability and Bayes' theorem were
presented.
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