Conditional Probability
• The probability of an event B occurring when
it is known that some event A has occurred is
called a conditional probability.
• Definition:
• The conditional probability of B, given A and
P A  B 


P
B
/
A
denoted by
is defined by PB / A  P A
provided PA  0
Consider an event B of getting a perfect square when a die is
tossed.
The die is constructed so that even numbers are twice as likely to
occur as odd numbers.
That is P(odd)=p , P(even)=2p
so p=1/9
Now suppose that the toss of the die resulted in a number greater
that 3, which reduced our sample space
S={4,5,6}
P(B occurs relative to reduced sample space A) is
P(4)+P(5)+P(6)=1
2w+w+2w=1
w=1/5
now P(odd)=1/5 and P(even)=2/5
P(B/A i.e number is a perfect square given that it is greater that
3)=2/5
PB / A
• Suppose sample space S is the population of adults in a small
town who have completed the requirements for a college
degree.
Employed
Unemployed
Total
Male
460
40
500
Female
140
260
400
Total
600
300
900
The data is categorized according to gender and employment
status. One of these individuals is selected at random for a tour
throughout the country to publicize the advantages of
establishing new industry in the town.
• M: a man is chosen
• E: The one chosen is employed.
Find P(M/E), probability that a man is chosen given that he is
employed.
nM  E  nE  M  / nS 
PM / E  

n E 
nE  / nS 
460 23
P E  M  

as
900 45
600 2
P E  

900 3
23
PM  E 
23
45
 PM / E  


2
P E 
30
3
The probability that a
regularly scheduled flight departs on time is P(D)=0.83;
The probability that it arrives on time is
P(A)=0.82;
The probability that it departs and arrives on time is
PD  A  0.78
Find the probability that a plane
a) Arrives on time given that it departed on time.
P A / D  
P  D  A
P D 
0.78

 0.94
0.83
b) Departed on time given that it has arrived on time.
PD  A 0.78
P  D / A 

 0.95
P  A
0.82
Consider an industrial process in the textile industry in which
strips of a particular type of cloth are being produced. These
strips can be defective in two ways, length and nature of
texture. It is known from the historical information on the
process that 10% of strips fail the length test, 5% fail the
texture text, and only 0.8% fail both tests. If a strip is
selected randomly from the process and a quick
measurement identifies it as failing the length test, what is
the probability that it is texture defective?
Sol: Let L: length defective, T: Texture defective.
Given that the strip is length defective, the probability that the
strip is texture defective is given by
PT / L  
PT  L  0.008

 0.08
P L 
0.1
Independent Events: Two events A and B are independent if and only
if
P B / A  P B , or P A / B  P A .
assuming the existences of conditional probability.. Otherwise, A and B
are dependent.


 


 
Multiplicative Rules
Theorem: If in an experiment the events A and B can both occur, then
P A  B  P A P B / A . provided P(A)>0.


Since the events
  

A  B and B  A are equivalent, we can also write
PB  A  PBP A / B.
Suppose that we have a fuse box containing 20 fuses, of which 5
are defective. If 2 fuses are selected at random and removed
from the box in succession without replacing the first, what is
the probability that both fuses are defective?
Sol:
Let A be the event that first fuse is defective.
B the event that second fuse is defective.
Then A  B is that A occurs, and B occurs after A has occurred.
Probability of selecting first defective bulb
5 1 (probability of selecting first defective bulb).
P  A 

20 4
4 (probability of selecting second defective
P B / A 
19 bulb from remaining four)
⇒
 1 


P A  B   
4 1

 4  19  19
(probability of selecting
both defective fuses).
Theorem:
two events A and B are independent iff
P A  B  P( A) P( B)
Therefore to obtain the probability of that two events both occur, we simply
find the product of their individuals.
Example:
A small town has one fire engine and one ambulance available for engineers.
The probability that the fire engine is available when needed is 0.08 and
probability that ambulance is available when called is 0.92. In the event of
an injury resulting from a burning find the probability that the both
ambulance and the fire engine will be available.
Sol:
P A  B  P( A) P( B)
=(0.08)(0.92)=0.916
An electrical system consists of four components . The system works if
components A and B work and either of the components C or D
work. The reliability of each component is given. Find the
probability that
a) Entire system works,
b) The component C does not work, given that entire system works.
Assume that four components work independently.
Sol:
a) Entire system work if either of C or D works.
P[A ∩ B ∩ (CUD)]=P(A)P(B)P(CUD)
=P(A)P(B)[1-P(C′ ∩ D′)]
Since P(C′)=P(D′)=1-0.8=0.2
so probability becomes
=(0.9)(0.9)[1-(0.2)(0.2)]=0.7776
b) Component c does not work given entire system works
=P(A ∩ B ∩ C′ ∩ D)/P(entire system works)
=P(A)P(B)P(C′)P(D)/0.7776
=(0.9)(0.9)(0.2)(0.8)/0.7776
Two cards are drawn from a well shuffled ordinary deck of 52
playing cards. Find the probability that they are both aces if the
first card is (i) replaced, (ii) not replaced.
Sol:
Let A be the event Ace on the first draw and B be the event Ace
on the second draw.
In Case of replacement, event A and B are independent, thus
P(Both are Aces)=P(A∩B)=P(A)P(B)=4/52)(4/52)
If the first card is not replaced, then events are dependents so
P(Both are Aces)=P(first card is an Ace)P(Second card is an Ace
given that the first card is an Ace)
so
P(A∩B)=P(A)P(B/A)=(4/52)(3/51)
Theorem: If, in an experiment, the events
A1 , A2 ,..., Ak
can occur, then
P A1  A2  ...  Ak   P A1 P A2 / A1 P A3 / A1  A2 ...
P Ak / A1  A2  ...  Ak 1 
If events
A1 , A2 ,..., Ak are independent, then
P A1  A2  ...  Ak   P A1 P A2 P A3 ...P Ak .
Two fair dice one red and one green are thrown. Let A denote the
event that the red die shows an even number and B, the event the
green die shows a 5 or 6. Show that the events A and B are
independent.
Sol:
The sample space S contains 36 equally likely outcomes.
A= event that red shows an even number
B= event that green shows 5 or 6
A∩B= event that as both events
P(A)=18/36
P(B)=12/36
P(A∩B)=6/36
Since P(A∩B)=1/6=(1/2)(1/3)=P(A)P(B)
Therefore both events are independent
A random sample of 200 adults are classified below by sex and their
level of education attained.
Education
Male
Female
Elementary
38
45
Secondary
28
50
College
22
17
If a person is picked at random from this group, find the probability
that
(a) the person is a male, given that the person has a secondary
education;
(b) the person does not have a college degree, given that the person
is a female.
If R is the event that a convict committed armed robbery and D is
the event that the convict pushed dope, state in words what
probabilities are expressed by
(a) P(R|D);
(b) P(D’|R);
(c) P(R|’D).
A class in advanced physics is composed of 10 juniors, 30 seniors,
and 10 graduate students. The final grades show that 3 of the
juniors, 10 of the seniors, and 5 of the graduate students received
an A for the course. If a student is chosen at random from this
class and is found to have earned an A, what is the probability that
he or she is a senior?
In an experiment to study the relationship of hypertension and
smoking habits, the following data are collected for 180
individuals:
Nonsmokers
Moderate
Smokers
Heavy
Smokers
Total
H
21
36
30
87
NH
48
26
19
93
Total
69
62
49
180
Where H and NH stand for hypertension and non hypertension,
respectively. If one of these individuals is selected at random,
find the probability that the person is
a) experiencing hypertension, given that the person is a heavy
smoker.
b) a non smoker, given that the person is experiencing no
hypertension.
For married couples living in a certain suburb, the probability
that a husband will vote on a bond referendum is 0.21, the
probability that his wife will vote in referendum is 0.28, and
the probability that both husband and wife will vote is 0.15.
Find the probability that
a) At least one member of a married couple will vote?
b) A wife will vote, given that her husband will vote?
c) A husband will vote given that his wife does not vote?
Before the distribution of certain statistical software every fourth
compact disk (CD) is tested for accuracy. The testing process
consists of running four independent programs and checking
the results. The failure rate for the four testing programs are,
respectively, 0.01, 0.03, 0.02 and 0.01.
a) What is the probability that a CD was tested and failed
anyway?
b) Given that a CD was tested, what is the probability that it
failed program 2 or 3?
c) In a sample of 100, how many CDs would you expect to be
rejected?
The probability that a person visiting his dentist will have an Xray is 0.6; the probability that a person who has an X-ray will
also have a cavity filled is 0.3; and the probability that a person
who has had an X-ray and a cavity filled will also have a tooth
extracted is 0.1.
What is the probability that a person visiting his dentist will have
an X-ray, a cavity filled, and a tooth extracted?