Chapter 6 Rules of Probability
 6.1 Sample spaces and events
 6.2 Postulates of probabilities
 6.4 Additive rules
 6.5 Conditional probability
 6.6 Multiplication rules
 6.7 Bayes theorem
6.1 Sample Spaces and Events

In a probability experiment the possible
outcomes form the samples space, S
1. Pick a card and note the suit
S={spade, club, heart, diamond}
2. Pick a card and note the color
S={red, black}
3. Pick a card and note if it is an ace
S={ace, not ace}
More examples
4. Roll a die and note the dots
S={1, 2, 3, 4, 5, 6}
5. Roll 2 dice and note the sum
S={2, 3, 4, … …, 11, 12}
Event
 A subset of a sample space is called an
event
For example, roll one die.
Even={2, 4, 6}
Odd= {1, 3, 5}
Greater than 6=empty set=
Operations
 Union
A∪B=all outcomes in A or B
 Intersection
A∩B=all outcomes in both A and B
 Complement
A’=all outcomes not in A
(like S-A)
Operations of Events
 If two events have no outcomes in common they
are mutually exclusive (or disjoint)
i.e., A∩ B=
S={1, 2, 3, 4, 5, 6}
Odd ={1, 3, 5}=A
Even={2, 4, 6}=B
more than 2={3, 4, 5, 6}=C
‘Odd’ and ‘Even’ are mutually exclusive;
‘Even’ and ‘C’ are not mutually exclusive.
Example 6.1
 Roll a red die and a white die
S= {(i,j): i, j=1,2,…, 6}
A={(i,j): i+j=7}={sum=7}
B={(3,j): j=1,..,6}={red=3}
Get
A∪B
A∩B
A’
Another example
 S={52 cards in deck}
Club ∪Spade =
Black
Club ∩Spade=

Red Ace=
{Ace of hearts, ace of diamond}
Red’=
Black
Venn Diagram
A Venn Diagram shows events as potentially intersecting
circles.
e.g., R=Republican
F=Female
S
R
F
R U F
S
R
R ∩ F
F
S
R
F
6.2 Postulates of probabilities
 Roll a die
S={1, 2, 3, 4, 5, 6}
 P(1)=P(2)=…=P(6)=1/6
For any event A:
0  P(A)  1
P(S)=1
P(A’)=1-P(A)
Example 6.2
 A={1, 2}, B={ 3, 4}
A and B are disjoint.
(or mutually exclusive)
P(A∪B)=2/6+2/6=2/3
3, 4
B
1, 2
A
5, 6
P(A∪B)=P(A)+P(B)
if A and B are mutually exclusive
Example 6.3
 S={52 cards}
P(ace ∪ king)=P(ace or king)
=4/52+4/52=2/13
 P(rain)=0.7
then P(not rain)=1-0.7=0.3
Exercise
 Which of the following pairs of events are
mutually exclusive? Explain.
– A driver getting a ticket for speeding and a
ticket for going through a red light.
– Being foreign-born and being President of the
United States.
– A person wearing black shoes and green socks.
– Having rain and sunshine on the 4th of July,
2005.
Exercise
 If A and B are the events that Consumer
Union will rate a car stereo good or poor,
P(A)=0.24 and P(B)=0.35, determine the
following probabilities:
– P(A’)
– P(AUB)
– P(A ∩B)
Exercise
 Let E, T, and N be the events that a car
brought to a garage needs an engine
overhaul, transmission repairs, or new tires.
Draw a Venn diagram for the following
–
–
–
–
EUT
EUTUN
E ∩T ∩N
E ∩T ∩N’
6.3. Probability and odds-skip
6.4 Addition Rules
 For mutually exclusive events
P(A1∪A2 ∪ ••• ∪Ak)=P(A1)+P(A2)+•••+P(Ak)
P(ace or king or queen)
=P(ace)+P(king)+P(queen)
=4/52+4/52+4/52=3/13
When individual outcomes are mutually
exclusive,
The probability of an event A is the sum of the
probabilities for all outcomes in A
P( A)   P( x)
xA
For a dice,
S={1, 2, 3, 4, 5, 6}
P(even)=P(2)+P(4)+P(6)=1/6+1/6+1/6=1/2
General Addition Rule
 For any two events A and B
P(A∪B)=P(A)+P(B)–P(A∩B)
A A∩ B
B
Example 6.4
 Draw a card from a deck. What is the
probability of getting an ace or a spade?
U
P(ace or spade)
=P(ace)+P(spade) – P(ace and spade)
=4/52+13/52 –1/52
=16/52
6.5 Conditional Probability
 Roll a die. The probability of getting any number
in the set of {1,2,3,4,5,6} is 1/6 !
Try to predict the outcome when I roll the die. We
predict that any number will have the same
probability of showing up.
 How about if you know some prior information?
If you know the number is even
 Re-evaluate the probabilities:
0 probabilities for all 3 odd numbers
equal probabilities for 3 even numbers
(each even number has a probability of 1/3)
This is called the conditional probability.
Notation: P(A|B)
 The conditional probability of event A,
given event B (occurs)
P(1|even)=P(3|even)=P(5|even)=0
P(2|even)=P(4|even)=P(6|even)=1/3
Conditional Probability
 S={1,2,3,4,5,6}
 E=even
 L=less than or equal to 3
 P(E|L) is the probability of “E given L”.
 P(even given that point total≤3) =P(E|L)=1/3.
# of points in E and L
# of points in L
# of points in E and L / n
=
# of points in L /n
P ( E L)
=
P ( L)
P ( E | L) 
E
L
4, 6, 2 1, 3
5
Conditional Probability
P( A  B)
P( A | B) 
P( B)
 Verify: P(1|even)=0
P(2|even)=1/3
Example 6.5
 Hair
Eye
Probability
brown
blue
0.3
brown
brown
0.4
blond
blue
0.2
blond
brown
0.1
-------------------------------------------P(blue eyes|blond hair)
=P(blue eyes and blond hair)/P(blond hair)
=0.2/(0.2+0.1)=2/3=0.67
P(blond hair|blue eyes)
= P(blond hair and blue eyes)/P(blue eyes)
=0.2/(0.3+0.2)=2/5=0.40
A simpler way: tabulate the probabilities
Hair brown
blond
Subtotal
Eye
blue
0.3
0.2
0.5
brown
0.4
0.1
0.5
Subtotal 0.7
0.3
1.0
P(blue eyes|blond hair)
=0.2/0.3
Exercise
 Candidates for a committee of 3 people
include 6 women and 4 men. What is the
probability that the # of women chosen is 2?
Exercise
 Sometimes a laboratory assistant has lunch at the cafeteria
where she works, sometimes she brings her own lunch,
sometimes she has lunch at a nearby restaurant, sometimes
she goes home for lunch, and sometimes she skips lunch to
lose weight. If the corresponding probabilities are 0.23,
0.31, 0.15, 0.24 and 0.07, find the probabilities that she
will
–
–
–
–
Have lunch at the cafeteria or the nearby restaurant;
Bring her own lunch, go home for lunch, or skip lunch altogether;
Have lunch at the cafeteria or go home for lunch;
Not skip lunch to lose weight.
Exercise
 The probability that it will rain in Tucson,
Arizona, on a day in mid-August, that there
will be a thunderstorm on that day, and that
there will be rain as well as a thunderstorm
are 0.27, 0.24 and 0.15. What is the
probability that there will be rain and /or a
thunderstorm in Tucson on such a day?
Exercise
 Roll a die. What is the probability of getting
a number less than 5 if we know that it is an
even number?
Exercise
 The probability that Henry will like a new
movie is 0.70 and the probability that Jane,
his girlfriend, will like it is 0.60. If the
probability is 0.28 that he will like it and
she will dislike it, what is the probability
that he will like it given that she is not
going to like it?
6.6 Multiplication Rules
 Conditional probability formula
P( A  B)
P( A | B) 
P( B)
 This is equivalent to
P(A∩B)=P(A|B)P(B)
Similarly
P(A∩B)=P(B|A)P(A)
Example 6.6
Pick up two cards from a deck without
replacement
What is the probability of getting two aces?
Method 1: like in Example 5.13
# of ways to pick 2 aces
P(both aces)=
# of ways to pick 2 cards
 4
 
2
43
6

=


 0.0045
 52  52  51 1326
 
2 
Method 2
 A=first card is an ace
 B=second card is an ace
P(A)=4/52
P(B|A)=3/51
P( A B)  P(1st ace and 2nd ace)
=P(1st ace)P(2nd ace|1st ace)
=P(A)P(B|A)
4 3
=   0.0045
52 51
 Toss 2 dice
 A= 6 first die
 B= 6 second die
P( A B)  P(6 first and 6 second)
=P(6 first)P(6 second| 6 first)
1 1 1
=  =
6 6 36
For more than 2 events multiply
conditional probabilities in a similar way
 P(A∩B∩C)=P(A)P(B|A)P(C|A∩B )
Example 6.7
 Pick up 3 cards without replacement. Find
the probability of getting 3 aces.
A=1st card is an ace; B=2nd card is an ace;
C=3rd card is an ace.
P(A)=4/52
P(B|A)=3/51
P(C|A∩B)=2/50
P(A∩B∩C)=(4/52)(3/51)(2/50)
Example 6.8
 Probability of getting 4 aces in 4 cards
4 3 2 1
  
52 51 50 49
Independent events vs Dependent events
 Roll a red die and white die.
A=red is a 6;
B= white is a 6
Then P(A)=P(B)=1/6;
P(A∩B)=1/36;
P(B|A)=P(A∩B )/P(A)=1/6
=P(B)
P(white is 6|red is 6) = P(white is 6)
The probability that white is a 6 is independent of whether
red is a 6.
“red is a 6” and “white is 6” are independent events.
Independence
If the conditional probability of B given A is equal
to the unconditional probability of B
P(B|A)=P(B)
then, A and B are said to be independent
This is equivalent to
P(A|B)=P(A)
Or
P(A∩B )=P(B)P(A|B)=P(B)*P(A)
O.W., A and B are independent
For independent events A and B
P(A∩B )=P(A)*P(B)
P(A|B)=P(A)
P(B|A)=P(B)
If any of these 3 conditions holds, then A &
B are independent.
If any of these 3 conditions does not hold,
then A & B are dependent.
Example 6.9
 Flip two fair coins. Probability of getting
two heads?
Method 1: Sample space: {HH, HT, TH, TT}
P(HH)=1/4
Method 2: A=head on 1st flip={HH, HT}
B=head on 2nd flip={HH, TH}
Show P(B|A)=P(B)
 P(HH)=P(A∩B)=P(A)P(B)=(1/2)(1/2)=1/4
Example 6.10
 Pick a card, set
H=heart A=ace K=king
P(H)=13/52=1/4
P(H|K)=1/4  H and K are independent
P(A)=4/52
P(A|K)=0
 A and K are dependent
Independent and mutually exclusive
 They are totally different concepts! Do not
mix these up.
 Difference?
How to get the probability?
 The solutions of most probability problems
use the following facts.
 For equally likely outcomes
#of outcomes in A
P(A)=————————————
total # of possible outcomes
 P(not A)=1–P(A)
More
 For mutually exclusive events
P(A1 or A2 or … or Ak)= P(A1)+P(A2)+… P(Ak)
 For any two events
P(A or B)=P(A∪B)=P(A)+P(B)–P(A∩B)
 P(A|B)=P(A∩ B)/P(B)
More
 P(A and B)=P(A∩B)=P(A)P(B|A) or P(B)P(A|B)
 For independent events
P(A|B)=P(A);
P(B|A)=P(B);
P(A∩B)=P(A)P(B)
 For independent events A1, A2,…, Ak
P(A1∩ A2 ∩ …∩ Ak)=P(A1)…P(Ak)
More
 For any events
P(A1∩ A2 ∩ …∩ Ak)
=P(A1)P(A2|A1)…P(Ak|A1∩…∩Ak-1)
Example 6.11
Toss a coin 3 times. Find
P(2 heads)=?
HHT (½)(½)(½)=1/8
HTH (½)(½)(½)=1/8
THH (½)(½)(½)=1/8
P(2 heads)=P(HHT∪HTH∪THH)=P(HHT)+P(HTH)+P(THH)=3/8
P(at least one head)=?
=1–P(no head)=1–P(TTT)
=1-1/2*1/2*1/2
=1-1/8
=7/8
Example 6.12
Pick 5 cards
P(2 aces|1 king)= P(2 aces and 1 king)/P(1 king)
#of ways to pick up 2 aces and 1 king
#of ways to pick up 1 king
 4  4  44 
   
 2 1  2 
 4  48 

1 

4 

 

 4   4   44 
  

2
1
2

probability      
 4   48 
 

1
4
 

Example 6.13
 Deal 5 cards. What is the probability that you get
5 face cards? (J,Q, or K)
12 
Method 1:
 
# of hands with all face cards  5 

# of hands
 52 
 
5 
Method 2: multiply conditional probability
12 11 10 9 8
   
52 51 50 49 48
Example 6.14
 Deal 5 cards, what is the probability of getting
4 aces?
# of hands with 4 aces
=(# of ways to pick 4 aces)*(# of ways to pick other card)
 4  48 
=     48
 4 1 
48
P (4 aces)=
 52 
 
5 
Exercise
 Suppose that the probability is 0.45 that a
rare tropical disease is diagnosed correctly
and, if diagnosed correctly, the probability
is 0.60 that the patient will be cured. What
is the probability that a person who has the
disease will be diagnosed correctly and
cured?
Exercise
 The probability that Henry will like a new
movie is 0.70 and the probability that Jane,
his girlfriend, will like it is 0.60. If the
probability is 0.28 that he will like it and
she will dislike it, what is the probability
that he will like it given that she is not
going to like it?
 Are the two events (Henry likes the movie
and Jane likes it) independent?
6.7 Bayes Theorem
 A test for cancer has the following properties:
C = a patient has cancer
+ = test is positive, claiming patient has cancer
P(C)=0.001
P(+|C) =0.98
P(+|C’)= 0.01
Question 1: what % of patients test positive?
Venn diagram

P(C)=0.001 P(+|C) =0.98 P(+|C’)= 0.01
C
+
P(C’∩+)
=P(C’)P(+|C’)
=0.999*0.01
=0.00999
C∩+
C∩+
0.001*0.98=0.00098
P(+)=P(C∩+)+P(C’∩+)
=P(C)P(+|C)+P(C’)P(+|C’)
=0.001*0.98+0.999*0.01
=0.01097
Tree diagram
+
C
_
C’
+
_
P()  weighted average of P(+|C) and P(+|C')
=P(C)P(+|C)+P(C')P(+|C')
Q2: What is the probability a patient has
cancer given that patient tests positive?
P(C∩+)
0.00098
———— = ———— =0.089
P(+)
0.01097
Most of + cases are from C’ !!!
A tree diagram helps!
Rule of total probability
P(A)=P(B)P(A|B)+P(B')P(A|B')
Weighted average of P(A|B) & P(A|B’)
P(B)P(A|B)
A
P(A|B)
B
P( B)
B’
P( B’)
A
P(B')P(A|B')
P(A|B’)
Generalization: B1, B2, B3
Use this formula when
Given: P(A|B)
P(A|B’)
Need: P(A)
P(B)
Bayes’ Theorem
P (A and B)
P ( B | A) 
P(A)
P( B) P( A | B)
=
P ( B ) P ( A | B )  P ( B ') P ( A | B ')
Use this when
Given: P(A|B) P(A|B’)
Need: P(B|A)
P(B)
Example: Insurance Company figures
A=accident in a year
AP=accident prone
Given:
P(A|AP)=0.10
P(A|AP’)=0.05
P(AP)=0.2
Find P(A)
P( A)  P( AP) P( A | AP)  P( AP ') P( A | AP ')
= 0.20*0.10 +0.80*0.05
=0.02+0.04
=0.06
Find P(AP|A)
P( AP | A) 
P( AP A) P( AP) P( A | AP) 0.20*0.10 0.02



P( A)
P( A)
0.06
0.06
Bayes’ Theorem
If B1, B2,…, and Bk are mutually excusive
events of which one must occur, then
P( Bi | A)=
P( Bi ) P( A | Bi )
P( B1 ) P( A | B1 )  P( B2 ) P( A | B2 )  ...  P( Bk ) P( A | Bk )
for i=1, 2,..., or k.
Exercise
 In a cannery, assembly lines I, II, and III account
for 50%, 30%, and 20% of total output. If 0.4% of
the cans from assembly line I are improperly
sealed, and the corresponding percentages for
assembly lines II and III are 0.6% and 1.2%.
Q1: Probability of a can being improperly sealed.
Q2: Probability of a can being sealed from line I if it
is found to be improperly sealed
Tree diagram
 A =event that a can is improperly sealed
 Bi=event that a can is sealed from line i
0.5
0.3
0.2
I
0.004
A
0.0020
II
0.006
A
0.0018
III 0.012
A
0.0024
Solutions:
 Answer #1:
P ( A)  P ( A | B1 ) P ( B1 )  P ( A | B2 ) P ( B2 )  P ( A | B3 ) P ( B3 )
 0.004*0.5  0.006*0.3  0.012*0.2
 0.002  0.0018  0.0024
 0.0062
 Answer #2:
P( B1 | A) 
P( B1 ) P( A | B1 ) 0.004*0.5

 0.32
P( A)
0.0062
Exercise
 Imagine that the probability is 0.95 that a
certain test will correctly diagnose a person
with diabetes as being diabetic, and that it is
0.05 that the test will incorrectly diagnose a
person without diabetes as being diabetic.
Given that roughly 10% of the population is
diabetic, what is the probability that a
person diagnosed as being diabetic actually
has diabetes?
Homework on 6.7
(not to be turned in)
6.70*, 6.71*, 6.72*, 6.73*, 6.78
For the problems marked *, write down
what is given and what you need to find in
symbolic notation.