2. Basic Concepts
• Random Experiment is an experiment or a
process for which the outcome cannot be
predicted with certainty.
• Sample space (S) The set of all possible
outcomes of a random experiment
• Sample space: The collection of all possible
outcomes for an experiment.
• Event: is any subset of the sample space.
• Subset : a set A is called a subset of a set B if
each element of set A belongs to the set B.
A B
2
Dr. Yehya Mesalam
4. Solution
Draw the Venn and tree diagrams for the experiment
of tossing a coin once.
Example
a) Venn Diagram and
b) Tree diagram
4
Dr. Yehya Mesalam
5. Draw the Venn and tree diagrams for the experiment of
tossing a coin twice.
Example
Solution
a) Venn Diagram and b) Tree diagram
5
Dr. Yehya Mesalam
6. Example
Suppose we randomly select two workers from a company and
observe whether the worker selected each time is a man or a woman.
Write all the outcomes for this experiment. Draw the Venn and tree
diagrams for this experiment.
Solution
a) Venn Diagram and
b) Tree diagram
6
Dr. Yehya Mesalam
7. Example
• State the sample space for the experiment of
selecting two items from the batch consists of
three items {a, b, c}
1. without replacement
2. with replacement
1- S={{a, b}, {a, c}, {b, c}}.
2- S ={{a, a}, {a, b}, {a, c}, {b, b}, {b, c}, {c, c}}.
Solution
7
Dr. Yehya Mesalam
10. • Union of Events – If A and B are two events in
a sample space S, then the union, A U B, is the
set of all outcomes in S that belong to either
A or B
A B
The entire shaded area
represents
A U B
S
Axioms of sets
10
Dr. Yehya Mesalam
11. S
Event Relations - Union
The union of two events, A and B, is the event that
either A or B or both occur when the experiment
is performed. We write
A B
A B
A B
11
Dr. Yehya Mesalam
12. S
A B
Event Relations-Intersection
The intersection of two events, A and B, is
the event that both A and B occur.
We write A B.
A B
• If A and B are mutually exclusive, then P(A B) = 0.
12
Dr. Yehya Mesalam
13. S
Event Relations - Complement
The complement of an event A consists of
all outcomes of the experiment that do not
result in event A. We write AC ( The event
that event A doesn’t occur).
A
AC
13
Dr. Yehya Mesalam
14. Example
List the elements of each of the following sample
spaces:
1. The set of integers between 1 and 50 divisible by 8;
2. The set S=
3. The set outcomes when a coin is tossed until a tail
or three heads appear;
Solution
1- S={8, 16, 24, 32, 40, 48, }
2- S={-5, 1}
3- S={T, HT, HHT, HHH}
}
0
5
4
{ 2
X
X
X
14
Dr. Yehya Mesalam
15. Example
Let the Sample Space be the collection of all
possible outcomes of rolling one die:
S = [1, 2, 3, 4, 5, 6]
Let A be the event “Number rolled is even”
Let B be the event “Number rolled is at least 4”
Then
A = [2, 4, 6] and B = [4, 5, 6]
15
Dr. Yehya Mesalam
16. Example
S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]
5]
3,
[1,
A
6]
[4,
B
A
6]
5,
4,
[2,
B
A
S
6]
5,
4,
3,
2,
[1,
A
A
Complements:
Intersections:
Unions:
[5]
B
A
3]
2,
[1,
B
16
Dr. Yehya Mesalam
17. Example
• The die toss:
• Simple events: Sample space:
1
2
3
4
5
6
E1
E2
E3
E4
E5
E6
S ={E1, E2, E3, E4, E5, E6}
S
•E1
•E6
•E2
•E3
•E4
•E5
(or S ={1, 2, 3, 4, 5, 6})
Venn Diagram
Tree diagram
17
Dr. Yehya Mesalam
18. Example
18
• Represent the sample space of three items are
selected at random from a manufacturing
process. Each item is inspected and classified
defective, D, or nondefective, N.
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}.
Solution
Dr. Yehya Mesalam
20. Basic Concepts
• An event is a collection of one or more simple
events.
•The die toss:
–A: an even number
–B: a number less than 5
S
A ={2, 4, 6}
B ={1, 2, 3, 4}
B
A
•6
•3
•5
•4
•1
•2
A ∩ B ={2, 4} 20
Dr. Yehya Mesalam
21. If M = {x | 3 < x < 9} and N = {y | 5 < y < 12},
Find M ∪ N
21
Example
Solution
M ∪ N = {z | 3 < z < 12}.
Dr. Yehya Mesalam
22. An experiment involves tossing a pair of dice, one
blue and one black, and recording the numbers that
come up. If x equals the outcome on the blue die
and y the outcome on the black die, describe the sample
space S
(a) by listing the elements (x, y);
(b) by using the rule method.
22
Example
Solution
(a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2),
(2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4),
(5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
Dr. Yehya Mesalam
23. Possible outcomes for rolling a pair of dice
23
Solution
(b) S = {(x, y) | 1 ≤ x, y ≤ 6}.
Dr. Yehya Mesalam
24. • For a two-stage experiment,
m ways to accomplish the first stage
n ways to accomplish the second stage
then there are mn ways to accomplish the whole
experiment.
• For a k-stage experiment, number of ways equal to
n1 n2 n3 … nk
Example: Toss two coins. The total number of
simple events is:
2 2 = 4
Counting Rules
24
Dr. Yehya Mesalam
25. Examples
Example: Toss three coins. The total number of
simple events is:
2 2 2 = 8
Example: Two balls are drawn in order from a dish
containing four candies. The total number of simple
events is:
6 6 = 36
Example: Toss two dice. The total number of
simple events is:
4 3 = 12
25
Dr. Yehya Mesalam
26. A prospective car buyer can choose between a fixed and
a variable interest rate and can also choose a payment
period of 36 months, 48 months, or 60 months. How
many total outcomes are possible?
Example
There are two outcomes (a fixed or a variable interest
rate) for the first step and
three outcomes (a payment period of 36 months, 48
months, or 60 months) for the second step.
Hence,
Total outcomes = 2 x 3 = 6
Solution
26
Dr. Yehya Mesalam
27. A National Football League team will play 16 games
during a regular season. Each game can result in one of
three outcomes: a win, a loss, or a tie. The total possible
outcomes for 16 games are calculated as follows:
Example
Total outcomes = 3·3·3·3·3·3·3·3·3·3·3·3 ·3·3·3·3
= 316 = 43,046,721
One of the 43,046,721 possible outcomes is all 16 plays.
Solution
27
Dr. Yehya Mesalam
28. 60
)
1
(
2
)
1
)(
2
)(
3
)(
4
(
5
)!
3
5
(
!
5
5
3
P
Permutations
• n distinct objects, take r objects at a time and
arrange them in order. The number of different
ways you can take and arrange is
60
)
3
)(
4
(
5
.
1
!
0
and
)
1
)(
2
)...(
2
)(
1
(
!
where
)!
(
!
n
n
n
n
r
n
n
Pn
r
Example: How many 3-digit lock passwords can
we make by using 3 different numbers among 1,
2, 3, 4 and 5?
28
Dr. Yehya Mesalam
29. Example
• How many ways to select a student
committee of 3 members: chair, vice chair,
and secretary out of 8 students?
336
)
6
)(
7
(
8
)
1
)(
2
)(
3
)(
4
(
5
)
1
)(
2
)(
3
)(
4
)(
5
)(
6
)(
7
)(
8
(
)!
3
8
(
!
8
8
3
P
The order of the choice is
important! ---- Permutation
29
Dr. Yehya Mesalam
30. Permutations ofnobjects ofsamekinds
• Permutations of n objects of which n1 are alike, n2 are
alike, . . . , nr are alike is
60
)
!
2
!*
3
/(
!
6
The order of the choice is
.
1
!
0
and
)
1
)(
2
)...(
2
)(
1
(
!
where
!
!.....
!
!
..
or
!
!.....
!
!
2
1
2
1
2
1
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
r
r
r
Example: How many different letter arrangements can
be formed from the letters PEPPER?
30
Dr. Yehya Mesalam
32. 32
How many different letter arrangements can be made
from the letters in the word
STATISTICS?
Example
Solution:
Dr. Yehya Mesalam
33. Combinations
• n distinct objects, select r objects at a time without
regard to the order. The number of different ways
you can select is
Example: Three members of a 5-person committee must
be chosen to form a subcommittee. How many different
subcommittees could be formed?
)!
(
!
!
r
n
r
n
Cn
r
10
1
)
2
(
)
4
(
5
1
)
2
)(
1
)(
2
(
3
1
)
2
)(
3
)(
4
(
5
)!
3
5
(
!
3
!
5
5
3
C
The order of
the choice is
not important!
33
Dr. Yehya Mesalam
34. Example
• How many ways to select a student
committee of 3 members out of 8 students?
• (Don’t assign chair, vice chair and secretary).
56
)
1
)(
2
(
3
)
6
)(
7
(
8
)]
1
)(
2
)(
3
)(
4
(
5
)][
1
)(
2
(
3
[
)
1
)(
2
)(
3
)(
4
)(
5
)(
6
)(
7
(
8
)!
3
8
(
!
3
!
8
8
3
C
The order of the choice is
NOT important!
Combination
34
Dr. Yehya Mesalam
Solution:
35. An ice cream parlor has six flavors of ice cream. Ahmed
wants to buy two flavors of ice cream. If he randomly
selects two flavors , how many ways of his selection?
Example
n = total number of ice cream flavors = 6
r = # of ice cream flavors to be selected = 2
Thus, there are 15 ways to select two ice cream flavors
out of six.
15
1
2
3
4
1
2
1
2
3
4
5
6
!
4
!
2
!
6
)!
2
6
(
!
2
!
6
2
6
C
Solution:
35
36. Example
Permutations of the First 4 letters 4p4=24
abcd
abdc
acbd
acdb
adbc
adcb
bacd
badc
bcad
bcda
bdac
bdca
cabd
cadd
cbad
cbda
cdab
cdba
dabc
dacb
dbac
dbca
dcab
dcba
How many ways can arrangement the first 4 letters of
the alphabet taken many ways all at a time. (a, b, c, d)
The # of ways = n! = 4! = 24
36
Dr. Yehya Mesalam
37. The Probability of an Event
• The probability of an event A measures “how
often” we think A will occur. We write P(A).
• Suppose that an experiment is performed n
times. The relative frequency for an event A is
Number of times A occurs f
n n
events
simple
of
number
total
A
in
events
simple
of
number
)
(
N
n
A
P A
n
N
N
37
Dr. Yehya Mesalam
38. General Laws of Probability
For any two events, A and B, the probability of
their union, P(A B), is
)
(
)
(
)
(
)
( B
A
P
B
P
A
P
B
A
P
A B
38
Dr. Yehya Mesalam
39. Intersection of Events
• Intersection of Events – If A and B are two
events in a sample space S, then the
intersection, A ∩ B, is the set of all outcomes
in S that belong to both A and B
A B
AB
S
39
Dr. Yehya Mesalam
40. • A and B are Mutually Exclusive Events if they
have no basic outcomes in common
– i.e., the set A ∩ B is empty (A ∩ B =Ø )
A B
S
Mutually Exclusive Events
40
الحادثين عن يقال
A
و
B
معا حدوثهما استحال إذا متنافيان أنهما
.
Dr. Yehya Mesalam
41. Laws of Probability
For any two events, A and B, the probability of
their union, P(A B), is
)
(
)
(
)
( B
P
A
P
B
A
P
A B
The event A and B are mutually exclusive
A and B are mutually exclusive
0
)
(
B
A
P
)
( B
A
41
Dr. Yehya Mesalam
42. • A and B are Independent Events if their joint
probability equals the product of their probabilities
• If the occurrence (or non occurrence ) of one does
not affect the probability of the occurrence
A B
S
Independent Events
42
الحادثين يعتبر
A
أو
B
عدم أو إحداهما ع
وقو كان إذا مستقمين حادثين
اآلخر ع
وقو في يؤثر ال وقوعه
.
A
∩
B
P(A ∩ B) =P(A) .P(B)
Dr. Yehya Mesalam
43. Laws of Probability
For any two events, A and B, the probability of
their union, P(A B), is
)
(
).
(
)
(
)
(
)
( B
P
A
P
B
P
A
P
B
A
P
A B
The event A and B are
independent
A and B independent events
)
(
).
(
)
( B
P
A
P
B
A
P
)
( B
A
43
Dr. Yehya Mesalam
44. Probabilities for Complements
• We know that for any event A:
P(A AC) = 0
• Since either A or AC must occur,
P(A AC) =1
• so that P(A AC) = P(A)+ P(AC) = 1
P(AC) = 1 – P(A)
A
AC
44
Dr. Yehya Mesalam
45. Relation Between Events
A B
A ∩ B = Ø
Mutually Exclusive Events
P(A U B) =P(A) +P(B)
P(A ∩ B) = 0
Dr. Yehya Mesalam 45
46. Relation Between Events
A B
Independent Events
P(A ∩ B) =P(A) .P(B)
A ∩ B
P(A U B) =P(A) +P(B)- P(A).P(B)
Dr. Yehya Mesalam 46
47. Relation Between Events
A B
General Events
A ∩ B
P(A U B) =P(A) +P(B)- P(A ∩ B)
Dr. Yehya Mesalam 47
50. Relation Between Events
A
A ∩ B
A = 1+ 3
1 3
Event of A only A only = 1
Event of A
P(A only )= P(1)= P(A) – P(A ∩ B)
Dr. Yehya Mesalam 50
51. Relation Between Events
B
A ∩ B
B = 2+ 3 2
3
Event of B only
B only = 2
Event of B
P(B only )= P(2)= P(B) – P(A ∩ B)
Dr. Yehya Mesalam 51
52. Relation Between Events
A B
A ∩ B
1
2
3
Event of at least one event happened
Means A or B or A and B together
P(A U B) =P(A) +P(B)- P(A ∩ B)
(A U B) = 1 +2 + 3
Dr. Yehya Mesalam 52
53. Relation Between Events
A B
A ∩ B
1
2
3
Event of at Most one event happened
Means A only B only or A and B Not occur together
P(A ∩ B) / = 1- P(A ∩ B)
(A ∩ B)/ = 1 +2 + 4
S
4
4
4
White
Area
White
Area
Dr. Yehya Mesalam 53
54. Relation Between Events
A ∩ B
3
Event of at Most one event happened
Means A only B only or A and B Not occur together
P(A ∩ B) / = 1- P(A ∩ B)
(A ∩ B)/ = S - 3
S
Dr. Yehya Mesalam 54
55. Relation Between Events
A ∩ B
3
Event of at Most one event happened
Means A only B only or A and B Not occur together
P(A ∩ B) / = 1- P(A ∩ B)
(A ∩ B)/ = S - 3
S
Dr. Yehya Mesalam 55
A B
1 2
4
3
56. Relation Between Events
A B
A ∩ B
A only = 1
1
2
3
B only = 2
Dr. Yehya Mesalam 56
Only Event A or Event B
Means A only or B only is occur together
P(A only U B only )= 1 + 2
57. Relation Between Events
A ∩ B
3
Event of only event A and B event happened
Means A and B occur together
= P(A ∩ B)
(A ∩ B) = 3
S
Dr. Yehya Mesalam 57
A B
58. A Probability Table
B
A
A
B
)
B
P(A
)
B
A
P(
B)
A
P(
P(A)
B)
P(A
)
A
P(
)
B
P(
P(B) 1.0
P(S)
Probabilities and joint probabilities for two events A and
B are summarized in this table:
58
Dr. Yehya Mesalam
59. Conditional Probabilities
The probability that A occurs, given that event
B has occurred is called the conditional
probability of A given B and is defined as
0
)
(
,
)
(
)
(
)
|
(
B
P
B
P
B
A
P
B
A
P
“given”
0
)
(
,
)
(
)
(
)
|
(
A
P
A
P
B
A
P
A
B
P
59
Dr. Yehya Mesalam
60. Defining Independence
• We can redefine independence in terms of
conditional probabilities:
Two events A and B are independent if and
only if
P(A|B) = P(A) or P(B|A) = P(B)
Otherwise, they are dependent.
• Once you’ve decided whether or not two
events are independent, you can use the
following rule to calculate their intersection.
60
Dr. Yehya Mesalam
61. Multiplicative Rule for Intersections
• For any two events, A and B, the probability that
both A and B occur is
P(A B) = P(A) P(B given that A occurred)
= P(A)P(B|A)
• If the events A and B are independent,
then the probability that both A and B
occur is
P(A B) = P(A) P(B)
61
Dr. Yehya Mesalam
62. Example
• Toss a fair coin twice. What is the probability of
observing at least one head (event A)? Exactly
one Head (event B)?
H
1st Coin 2nd Coin Ei P(Ei)
H
T
T
H
T
HH
HT
TH
TT
1/4
1/4
1/4
1/4
P(at least 1 head)= P(A)
= P(E1) + P(E2) + P(E3)
= 1/4 + 1/4 + 1/4 = 3/4
Tree Diagram
P(exactly 1 head)=P(B)
= P(E2) + P(E3)
= 1/4 + 1/4 = 1/2
62
Dr. Yehya Mesalam
63. Example
• A bowl contains three balls, two reds, one blue. A
child selects two balls at random. What is the
Probability of observing exactly two reds?
1st M&M 2nd M&M Ei P(Ei)
r1b
r1r2
r2b
r2r1
1/6
1/6
1/6
1/6
1/6
1/6
P(exactly two reds)
= P(r1r2) + P(r2r1)
= 1/6 +1/6
= 1/3
r1
b
b
r1
r1
b
br1
br2
r2
r2
r2
r2
r1 b
63
64. Example
Suppose that there were 1000 students in a
college, and that they could be classified as
follows:
Male (B) Female
Colorblind (A) 40 2
Not Colorblind 470 488
A: Colorblind
P(A) = 42/1000=.042
B: Male
P(B) = 510/1000=.51
P(AB) = P(A) + P(B) – P(AB)
= 42/1000 + 510/1000 - 40/1000
= 512/1000 = .512
Check: P(AB)
= (40 + 2 + 470)/1000 =.512
Dr. Yehya Mesalam
65. A Special Case
Male Female
Colorblind 40 2
Not Colorblind 470 488
When two events A and B are
mutually exclusive, P(AB) = 0
and P(AB) = P(A) + P(B).
A: male and colorblind
P(A) = 40/1000
B: female and colorblind
P(B) = 2/1000
P(AB) = P(A) + P(B)
= 40/1000 + 2/1000
= 42/1000=.042
A and B are mutually
exclusive, so that
66. Example
Male Female
Colorblind 40 2
Not Colorblind 470 488
A: male
P(A) = 510/1000=.51
B: female
P(B) = 1- P(A)
= 1- .51
=.49
A and B are
complementary, so that
Select a student at random from
the college. Define:
66
Dr. Yehya Mesalam
67. Example
Male (A) Female
Colorblind (B) 40 2
Not Colorblind 470 488
A: Male
B: Colorblind
Find P(A), P(A|B)
Are A and B independent?
P(A) = 510/1000=.51
Select a student at random from
the college. Define:
P(A|B) = P(AB)/P(B)=.040/.042=.95
P(B) = 42/1000=.042 P(AB) = 40/1000=.040
P(A|B) and P(A) are not equal. A, B are dependent
67
68. Example
• If P (A/B) =0.4 and P (B) =0.8 and P (A) =0.5
are the events A and B independent?
Solution
If A and B are independent, then
But P (A/B) =0.4 and P (A) =0.5
then A and B not independent
P(A)
)
(
)
(
)
|
(
B
P
B
A
P
B
A
P
68
Dr. Yehya Mesalam
69. Example:
A: Red die show 1
B: Green die show 1
P(AB) = P(A) + P(B) – P(AB) = 6/36 + 6/36 – 1/36
= 11/36
P(A|B) = P(A and B)/P(B)
=1/36/1/6=1/6=P(A)
P(A) does not
change, whether B
happens or not…
A and B are
independent!
A:
B
69
70. The Law of Total Probability
P(A) = P(A S1) + P(A S2) + … + P(A Sk)
= P(S1)P(A|S1) + P(S2)P(A|S2) + … + P(Sk)P(A|Sk)
Let S1 , S2 , S3 ,..., Sk be mutually exclusive and
exhaustive events (that is, one and only one
must happen). Then the probability of any event
A can be written as
70
Dr. Yehya Mesalam
71. The Law of Total Probability
A
A Sk
A S1
S2….
S1
Sk
P(A) = P(A S1) + P(A S2) + … + P(A Sk)
= P(S1)P(A|S1) + P(S2)P(A|S2) + … + P(Sk)P(A|Sk)
71
Dr. Yehya Mesalam
72. Bayes’ Rule
Let S1 , S2 , S3 ,..., Sk be mutually exclusive and
exhaustive events with prior probabilities P(S1),
P(S2),…,P(Sk). If an event A occurs, the posterior
probability of Si, given that A occurred is
,...k
,
i
S
A
P
S
P
S
A
P
S
P
A
S
P
i
i
i
i
i 2
1
for
)
|
(
)
(
)
|
(
)
(
)
|
(
)
|
(
)
(
)
|
(
)
(
)
(
)
(
)
|
(
)
|
(
)
(
)
(
)
(
)
(
)
|
(
Proof
i
i
i
i
i
i
i
i
i
i
i
i
S
A
P
S
P
S
A
P
S
P
A
P
AS
P
A
S
P
S
A
P
S
P
AS
P
S
P
AS
P
S
A
P
72
Dr. Yehya Mesalam
73. Example
• Urn A1 contains 7 black and 3 white balls. A2
contains 3 black and 7 white balls, and urn A3
contains 5 white and 5 black balls. A fair die is to be
cast. If the die turns up 1, or 3, then a ball will be
selected from A1. If the die turns up 2, 4 or 6, a ball
will be selected from A2. Finally, a ball will be
selected from A3 otherwise.
a- Given that the ball selected is black, what is
probability that it was chosen from A2?
b- Given that the ball selected is white, what is
probability that it not chosen from A1?
73
Dr. Yehya Mesalam
74. Solution
• First calculate the P(B)
A1
A2
A3
P(W/A1)=0.3
P(W/A2)=0.7
P(W/A3)=0.5
P(B/A1)=0.7
P(B/A2)=0.3
P(B/A3)=0.5
P(A2)=3/6
P(B) = (2/6)*0.7 + 0.5*0.3 + (1/6)*0.5 = 0.466
Given that the ball selected is black, what is probability that it was chosen from
A2?
P(A2/B) = (0.5*0.3)/ 0.466 = 0.321
74
Dr. Yehya Mesalam
75. • b- Given that the ball selected is white, what is
probability that it not chosen from A1?
P(W) = (2/6)*0.3 + 0.5*0.7 + (1/6)*0.5 = 0.534
Or
P(W) = 1-P(B) =1 -0.466 = 0.534
• P(A1/w) = ((2/6)*0.3))/ 0.534 = 0.187
75
Solution
Dr. Yehya Mesalam
812
.
0
187
.
0
1
)
(
1
)
( 1
/
1
W
A
P
W
A
P
76. Example
One bag contains 4 white balls and 3 black balls, and a
second bag contains 3 white balls and 5 black balls.
One ball is drawn from the first bag and placed
unseen in the second bag. What is the probability
that a ball now drawn from the second bag is black?
76
Solution
Let B1 represent a black ball from bag 1
W1 represent, a white ball from bag 1
B2 represent a black ball from bag 2.
Dr. Yehya Mesalam
78. Example
0.90
0.90
a
b
0.95
0.80
0.90
0.95
The following circuit operates only if there is a path of
functional devices from left to right. The probability
that each device functions is shown on the graph.
Assume that devices fail independently. What is the
probability that the circuit operates
Solution
The probability that the circuit operates =
= (1- 0.13) (1- 0.052)(0.8) = 0.797202 78
79. Example
2 green and 4 red balls are in a box; Two of
them are selected at random.
A: First is green;
B: Second is red.
• Find P(AB).
79
Dr. Yehya Mesalam
80. Method 1
• Choose 2 balls out of 6. Order is recorded. (Total
number of ways, i.e. size of sample space S)
The order of
the choice is
important!
Permutation 30
)
5
(
6
!
4
!
6
)!
2
6
(
!
6
6
2
P
4
4
1
C
2
2
1
C
2 4 = 8 ways to
choose first green and
second red
( mn Rule)
• Event AB: First green, second red
First green
Second Red
30
8
#
#
)
(
S
B
A
B
A
P
80
Dr. Yehya Mesalam
81. Method 2
A: First is green;
B: Second is red;
AB: First green, second red
P(A)
P(B|A)
P(A B) = P(A)P(B|A)
P(A B) = (2/6)(4/5)=8/30
2/6
P(Second red | First green)=4/5
81
Dr. Yehya Mesalam
82. Probability Rules & Relations of Events
Complement Event
Additive Rule
Multiplicative Rule
Conditional probability
Mutually Exclusive Events
Independent Events
)
(
1
)
( A
P
A
P c
0
)
(
B
A
P
)
(
)
(
)
|
(
B
P
B
A
P
B
A
P
)
(
)
(
)
(
)
( B
A
P
B
P
A
P
B
A
P
)
|
(
)
(
)
( A
B
P
A
P
B
A
P
)
(
)
(
)
( B
P
A
P
B
A
P
)
(
)
(
)
( B
P
A
P
B
A
P
)
(
)
|
( A
P
B
A
P
)
(
)
(
)
(
)
( C
P
B
P
A
P
C
B
A
P
82
83. Key Concepts
III. Counting Rules
1. mn Rule, extended mn Rule
2. Permutations:
3. Combinations:
IV. Event Relations
1. Unions and intersections
2. Events
a. Disjoint or mutually exclusive:
b. Complementary:
)
(
1
)
( A
P
A
P c
0
)
(
B
A
P
)!
(
!
!
)!
(
!
r
n
r
n
C
r
n
n
P
n
r
n
r
)
(
)
(
)
( B
P
A
P
B
A
P
83
Dr. Yehya Mesalam
84. Key Concepts
3. Conditional probability:
4. Independent events
5. Additive Rule of Probability:
6. Multiplicative Rule of Probability:
)
(
)
(
)
( B
P
A
P
B
A
P
)
(
)
|
( A
P
B
A
P
)
(
)
(
)
|
(
B
P
B
A
P
B
A
P
)
(
)
(
)
(
)
( B
A
P
B
P
A
P
B
A
P
)
|
(
)
(
)
( A
B
P
A
P
B
A
P
)
(
)
(
)
(
)
( C
P
B
P
A
P
C
B
A
P
84
Dr. Yehya Mesalam
86. Example
One card is selected from an ordinary deck of playing cards. What is the
probability of getting?
1- The event (A) the king of hearts is selected
2- The event (B) a king is selected
3- The event (C) a heart is selected
4- The event (D) a face card is selected
5- The event (E) not a face card is selected.
86
87. 1- The event (A) the king of hearts is selected
1/52
2- The event (B) a king is selected
1/13 = 4/52
Solution
87
Dr. Yehya Mesalam
88. 3- The event (C ) a heart is selected
1/4 = 13/52
Solution
4- The event (D)a face card is selected
3/13=12/52
88
Dr. Yehya Mesalam
89. 5- The event (E) not a face card is selected (D )
40/52=10/13
Solution
-
89
Dr. Yehya Mesalam
90. Event (B & C) King and heart
1/13 X 1/4 = 1/52
King
heart
King and heart (B∩C)
90
Dr. Yehya Mesalam
91. Event (B or C) King or heart
King
heart
King or heart (BUC)
16/52 = 4/52 + 13/52 - 1/52
91
Dr. Yehya Mesalam
93. Spade or Face Card
P (spade) + P (face card) – P (spade & face card) = 1/4 + 3/13 – 3/52
= 22/52
Spade
Face
Spade U Face
94. Example
One card is selected from an ordinary deck of playing cards.
What is the probability of getting?
(a) A queen? (b) A Jack? (c) Either a queen or a jack?
(d) A queen or a red card? (e) A face card?
94
96. Example
• A bowl contains three balls, one red, one blue and
one green. A child takes two balls randomly one at a
time. What is the probability that at least one is red?
1st 2nd Ei P(Ei)
1/6
1/6
1/6
1/6
1/6
1/6
P(at least 1 red) =
P(E1) +P(E2) +
P(E3) + P(E6) =4/6
= 2/3
RB
RG
BR
BG
m
m
m
m
m
m
m
m
m
GB
GR
96
97. Example
Simple Events Probabilities
HHH
HHT
HTH
HTT
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
P(Exactly 2 heads)= P(B)
= P(HHT) + P(HTH) + P(THH)
= 1/8 + 1/8 + 1/8 = 3/8
P(at least 2 heads)=P(A)
= P(HHH)+P(HHT)+P(HTH)+P(THH)
= 1/8 + 1/8 +1/8 + 1/8 = 1/2
THH
THT
TTH
TTT
A={HHH, HHT, HTH, THH}
B={HHT,HTH,THH}
• Toss a fair coin 3 times. What is the probability of
observing at least two heads (event A)? Exactly two
Heads (event B)?
97
99. Example
• Toss a fair coin twice. What is the
probability of observing at least one
head (Event A)?
S
A
A
P
#
#
)
(
H
1st Coin 2nd Coin Ei P(Ei)
H
T
T
H
T
HH
HT
TH
TT
1/4
1/4
1/4
1/4
P(at least 1 head)
= P(A)
= P(HH) + P(HT) + P(TH)
= 1/4 + 1/4 + 1/4 = 3/4
A={HH, HT, TH}
S={HH, HT, TH, TT}
99
Dr. Yehya Mesalam
100. Example
• A bowl contains three M&Ms®, two reds, one blue.
A child selects two M&Ms at random. What is the
probability that exactly two reds (Event A)?
S
A
A
P
#
#
)
(
1st M&M 2nd M&M Ei P(Ei)
r1b
r1r2
r2b
r2r1
1/6
1/6
1/6
1/6
1/6
1/6
P(A)
= P(r1r2) + P(r2r1)
= 1/6 +1/6 = 2/6=1/3
r1
b
b
r1
r1
b
br1
br2
r2
r2
r2
r2
r1 b
A={r1r2, r2r1}
10
101. Tree Diagram
Experiment: Select 2 balls from 20 balls :
14 Blue & 6 Red. Don’t Replace.
Dependent!
B
R
B
R
B
R
P(R) = 6/20
P(R|R) = 5/19
P(B|R) = 14/19
P(B) = 14/20
P(R|B) = 6/19
P(B|B) = 13/19
10
Dr. Yehya Mesalam
102. Tree Diagram
Experiment: Select 2 balls from 20 balls :
14 Blue & 6 Red. With Replacement.
Independent!
B
R
B
R
B
R
P(R) = 6/20
P(R|R) = 6/20
P(B|R) = 14/20
P(B) = 14/20
P(R|B) = 6/20
P(B|B) = 14/19
10
Dr. Yehya Mesalam
103. Example
A survey of job satisfaction of teachers was
taken, giving the following results
Satisfied Unsatisfied Total
College 74 43 117
High School 224 171 395
Elementary 126 140 266
Total 424 354 778
Job Satisfaction
L
E
V
E
L
10
Dr. Yehya Mesalam
104. Example
If all the cells are divided by the total number
surveyed, 778, the resulting table is a table of
empirically derived probabilities.
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
10
Dr. Yehya Mesalam
105. Example
For convenience, let C stand for the event that the
teacher teaches college, S stand for the teacher
being satisfied and so on. Let’s look at some
probabilities and what they mean.
is the proportion of teachers who are
college teachers.
P(C) 0.150
is the proportion of teachers who are
satisfied with their job.
P(S) 0.545
is the proportion of teachers who are
college teachers and who are satisfied
with their job.
P(C S) 0.095
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
10
106. Example
is the proportion of teachers who
are college teachers given they are
satisfied. Restated: This is the
proportion of satisfied that are
college teachers.
P(C S)
P(C | S)
P(S)
0.095
0.175
0.545
is the proportion of teachers who
are satisfied given they are
college teachers. Restated: This is
the proportion of college teachers
that are satisfied.
P(S C)
P(S | C)
P(C)
P(C S) 0.095
P(C) 0.150
0.632
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
106
107. Example
P(C S) 0.095
P(C) 0.150 and P(C| S) 0.175
P(S) 0.545
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C|S) P(C) so C and S are dependent events.
Are C and S independent events?
107
Dr. Yehya Mesalam
108. Example
Satisfied Unsatisfied Total
College 0.095 0.055 0.150
High School 0.288 0.220 0.508
Elementary 0.162 0.180 0.658
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C) = 0.150, P(S) = 0.545 and
P(CS) = 0.095, so
P(CS) = P(C)+P(S) - P(CS)
= 0.150 + 0.545 - 0.095
= 0.600
P(CS)?
108
Dr. Yehya Mesalam
109. We know:
P(F) =
P(M) =
P(H|F) =
P(H|M) =
Example
From a previous example, we know that 49% of the
population are female. Of the female patients, 8% are high
risk for heart attack, while 12% of the male patients are
high risk. A single person is selected at random and found
to be high risk. What is the probability that it is a male?
Define H: high risk F: female M: male
61
.
)
08
(.
49
.
)
12
(.
51
.
)
12
(.
51
.
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
F
H
P
F
P
M
H
P
M
P
M
H
P
M
P
H
M
P
.12
.08
.51
.49
109
110. Example
Suppose a rare disease infects one out of
every 1000 people in a population. And
suppose that there is a good, but not perfect,
test for this disease: if a person has the
disease, the test comes back positive 99% of
the time. On the other hand, the test also
produces some false positives: 2% of
uninfected people are also test positive. And
someone just tested positive. What are his
chances of having this disease?
110
Dr. Yehya Mesalam
111. We know:
P(A) = .001 P(Ac) =.999
P(B|A) = .99 P(B|Ac) =.02
Example
Define A: has the disease B: test positive
0472
.
02
.
999
.
99
.
001
.
99
.
001
.
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
c
A
B
P
c
A
P
A
B
P
A
P
A
B
P
A
P
B
A
P
We want to know P(A|B)=?
111
112. Tom and Dick are going to take
a driver's test at the nearest DMV office. Tom
estimates that his chances to pass the test are
70% and Dick estimates his as 80%. Tom and
Dick take their tests independently.
Define D = {Dick passes the driving test}
T = {Tom passes the driving test}
T and D are independent.
P (T) = 0.7, P (D) = 0.8
Example
112
Dr. Yehya Mesalam
113. What is the probability that at most one of the
two friends will pass the test?
Example
P(At most one person pass)
= P(Dc Tc) + P(Dc T) + P(D Tc)
= (1 - 0.8) (1 – 0.7) + (0.7) (1 – 0.8) + (0.8) (1 – 0.7)
= .44
P(At most one person pass)
= 1-P(both pass) = 1- 0.8 x 0.7 = .44
113
Dr. Yehya Mesalam
114. What is the probability that at least one of the
two friends will pass the test?
Example
P(At least one person pass)
= P(D T)
= 0.8 + 0.7 - 0.8 x 0.7
= .94
P(At least one person pass)
= 1-P(neither passes) = 1- (1-0.8) x (1-0.7) = .94
114
115. Suppose we know that only one of the two
friends passed the test. What is the probability
that it was Dick?
Example
P(D | exactly one person passed)
= P(D exactly one person passed) / P(exactly one
person passed)
= P(D Tc) / (P(D Tc) + P(Dc T) )
= 0.8 x (1-0.7)/(0.8 x (1-0.7)+(1-.8) x 0.7)
= .63
115
