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Basic Probability Concepts
Irrigation and Hydraulics Department
Faculty of Engineering – Cairo University
Academic Year 2011-2012
Dr. Mohamed Attia
Assistant Professor
by
Probability Experiments
� A random experiment is an experiment that
can result in different outcomes, even though
it is repeated in the same manner every time.
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Probability Experiments
� An experiment:
� Flipping a coin once.
� Rolling a die once.
� Pulling a card from a deck.
Sample Spaces
� The set of all possible outcomes of a random
experiment is called a sample space.
� You may think of a sample space as the set of
all values that a variable may assume.
� We are going to denote the sample space by S.
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Examples
� Experiment: Tossing a coin once.
� S = {H, T}
� Experiment: Rolling a die once.
� S = {1, 2, 3, 4, 5, 6}
Examples
� Experiment: Drawing a card.
S = 52 cards in a deck
� K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 (or Ace)
� 13 spades (♠), 13 hearts (♥), 13 diamonds (♦)
and 13 clubs (♣)
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Classification of Sample
Spaces
� We distinguish between discrete and continuous sample spaces.
� The outcomes in discrete sample spaces can be counted. The number of outcomes can be finite or infinite.
� In the case of continuous sample spaces, the outcomes fill an entire region in the space (interval on the line).
Events� An event, E, is a set of outcomes of a probability
experiment, i.e. a subset in the sample space.
� E ⊆ S.
� Simple event
� Outcome from a sample space with one
characteristic
� e.g.: A red card from a deck of cards
� Joint event
� Involves two outcomes simultaneously
� e.g.: An ace that is also red from a deck of cards
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Visualizing Events
� Contingency Tables
� Tree Diagrams
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Full Deck
of Cards
Red
Cards
Black
Cards
Not an Ace
Ace
Ace
Not an Ace
Visualizing Events
� Venn Diagram
� Sample space S
� Events A and B
A BS
Sometimes it is convenient to represent the
sample space by a rectangular region, with
events being circles within the region. Such a
representation is called Venn diagrams.
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Simple Events
The Event of a Triangle
There are 55 triangles in this collection of 18 objects
The event of a triangle AND red in color
Joint Events
Two triangles that are red
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Special Events
� Impossible event
� e.g.: Club & diamond on one card
draw
� Complement of event
� For event A, all events not in A
� Denoted as A’
� e.g.: A: queen of diamonds
A’: all cards in a deck that are
not queen of diamonds
♣♣
Null Event
Special Events
� Complement of an event
� Sample space S
� Event A
� Complement of A � A’
A
S
A’
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Mutually Exclusive Events
� Two events, E and F, are called mutually
exclusive, if
� That is, it is impossible for an outcome to be
an occurrence of both events.
Φ=FE I
e.g.: A: queen of diamonds; B: queen of
clubs
Events A and B are mutually exclusive
One of the events must occur
The set of events covers the whole sample space
e.g.: -- A: all the aces; B: all the black cards; C: all
the diamonds; D: all the hearts
Events A, B, C and D are collectively
exhaustive
Events B, C and D are also collectively
exhaustive
Collectively exhaustive events
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Special Events
� Collectively Exhaustive Events
� A, B, C, D are collectively exhaustive
� A and B are NOT mutually exclusive
� A and C are NOT mutually exclusive
� A and D are NOT mutually exclusive
� B, C, and D are also collectively exhaustive
� B, C, and D are mutually exclusive
B♦
C D♥
A
Special Events
A B
SBAU
BAIB
S
A
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Contingency Table
A Deck of 52 Cards
Ace Not anAce
Total
Red
Black
Total
2 24
2 24
26
26
4 48 52
Sample
Space
Red Ace
Full Deck of Cards
Tree Diagram
Event Possibilities
Red Cards
Black
Cards
Ace
Not an Ace
Ace
Not an Ace
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Probability
� Probability is the numerical
measure of the likelihood
that an event will occur
� The probability of an event E
is a number, P(E), such that
� Sum of the probabilities of
all mutually exclusive and
collective exhaustive events
is 1
1)(0 ≤≤ EP
Certain
Impossible
.5
1
01 )( =SP
(There are 2 ways to get one 6 and the other 4)
e.g. P( ) = 2/36
Computing Probabilities
� The probability of an event E:
� Each of the outcomes in the sample space is
equally likely to occur
number of event outcomes( )
total number of possible outcomes in the sample space
P E
X
T
=
=
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Computing Joint Probability
� The probability of a joint event, A and B:
( and ) = ( )
number of outcomes from both A and B
total number of possible outcomes in sample space
P A B P A B∩
=
E.g. (Red Card and Ace)
2 Red Aces 1
52 Total Number of Cards 26
P
= =
P(A1 and B2) P(A1)
TotalEvent
Joint Probability Using
Contingency Table
P(A2 and B1)
P(A1 and B1)
Event
Total 1
Joint Probability Marginal (Simple) Probability
A1
A2
B1 B2
P(B1) P(B2)
P(A2 and B2) P(A2)
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Computing Compound (or
Multiple) Probability
� Probability of a compound event, A or B:
( or ) ( )
number of outcomes from either A or B or both
total number of outcomes in sample space
P A B P A B= ∪
=
E.g. (Red Card or Ace)
4 Aces + 26 Red Cards - 2 Red Aces
52 total number of cards
28 7
52 13
P
=
= =
P(A1)
P(B2)
P(A1 and B1)
Compound Probability
(Addition Rule)
P(A1 or B1 ) = P(A1) + P(B1) - P(A1 and B1)
P(A1 and B2)
TotalEvent
P(A2 and B1)
Event
Total 1
A1
A2
B1 B2
P(B1)
P(A2 and B2) P(A2)
For Mutually Exclusive Events: P(A or B) = P(A) + P(B)
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Computing Conditional Probability
� The probability of event A given that event B has
occurred:
� It means the likelihood of realizing a sample point in
A assuming that it belongs to B
( and )( | )
( )
P A BP A B
P B=
E.g.
(Red Card given that it is an Ace)
2 Red Aces 1
4 Aces 2
P
= =/52
/52= 0.5
B
A&B
A
Conditional Probability Using
Contingency Table
Black
ColorType Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Revised Sample Space
(Ace and Red) 2 / 52 2(Ace | Red)
(Red) 26 / 52 26
PP
P= = =
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Conditional Probability and
Statistical Independence
� Conditional probability:
� Multiplication rule:
( and )( | )
( )
P A BP A B
P B=
( and ) ( | ) ( )
( | ) ( )
P A B P A B P B
P B A P A
=
=
Conditional Probability and
Statistical Independence
� Events A and B are independent if
� Events A and B are independent when the probability of one event, A, is not affected by another event, B
(continued)
( | ) ( )
or ( | ) ( )
or ( and ) ( ) ( )
P A B P A
P B A P B
P A B P A P B
=
=
=
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� B1, B2, …, Bk are mutually exclusive and
collectively exhaustive
� Suppose it is known that event A has
occurred
� What is the (conditional) probability that event
Bi has occurred?
Bayes’s Theorem
B1 B2 Bk
A
Bayes’s Theorem
( )( ) ( )
( ) ( ) ( ) ( )
( )( )
1 1
||
| |
and
i i
i
k k
i
P A B P BP B A
P A B P B P A B P B
P B A
P A
=+ •••+
=Adding up
the parts
of A in all
the B’sSame
Event
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Bayes’s Theorem Using Contingency
Table (Example 1)
Fifty percent of graduating engineers in a certain year worked as water resources (WR) managers. Out of those who worked as WR managers, 40% had a Masters degree. Ten percent of those graduating engineers who did not work as WR managers had a Masters degree. What is the probability that a randomly selected engineer who has a Masters degree is a WR manager?
?)/(
1.0)/(4.0)/(5.0)(
=
===
MWRP
RWMPWRMPWRP
Bayes’s Theorem
Using Contingency Table (continued)
WR WR
Masters
Masters
1.0.5 .5
.2
.3
.05
.45
.25
.75
Total
Total
8.025.0
2.0
)5.0)(1.0()5.0)(4.0(
)5.0)(4.0(
)()|()()|(
)()|()/(
==+
=
+=
RWPRWMPWRPWRMP
WRPWRMPMWRP
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Example 2
� A water treatment plant may fail for two reasons: inadequacy of materials (event A) or mechanical failure (event B).
If P(A) = 2 P(B), P(A | B) = 0.8 and the probability of failure of the treatment plant equals 0.001 , what is the probability that a mechanical failure occurs? What is the probability of a failure due to inadequate materials?
The probability of failure P (A U B) may be written as:
P (A U B) = P (A) + P (B) – P (A ∩ B)
= 2 P (B) + P (B) – P (B) P (A | B)
= 3 P (B) – 0.8P (B)
0.001 = 3 P (B) – 0.8P (B)
� P (B) = 0.00045
� P (A) = 0.0009
Example 3
� Each of two pumps Q1 and Q2 may not operate. On inspection
one may observe one of the outcomes of the following set: {(ƒ,
ƒ), (ƒ, o), (o, ƒ), (o, o)}. The notation (ƒ, o) means that pump Q1
fails and pump Q2 operates. The sample space has four
elements. From experience one knows that:
� P ((ƒ, ƒ)) = 0.1, P ((ƒ, o )) = 0.2
� P ((o, ƒ)) = 0.3, P ((o, o )) = 0.4
� Let
� A: Q1 operates
� B: Q2 operates
� C: at least one of the pumps operates
� Compute P(A), P(B), P(A ∩ B), P(C)
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Example 3 (Cont.)
� P ( A ) = 0.7
� P ( B ) = 0.6
� P ( A ∩ B ) = P (( o , o )) = 0.4
� P ( C ) = ( A U B )
= P ( A ) + P ( B ) – P ( A ∩ B ) = 0.9
Example 4
� The demand of a water supply system can be low (event L) , moderate (event M) or high (event H) with known probabilities P(L) = 0.1, P(M) = 0.7 and P(H) = 0.2. Failure of the system (event F) can only occur if a certain pump fails to function. From past experience the following conditional probabilities are known: P(F | L) = 0.05, P(F | M) = 0.10 and P(F | H) = 0.25. What is the probability of a pump failure P(F).
First we observe that the events L, M and H are clearly disjoint and that their union is just the sure event S. Hence
P ( F ) = P (( L ∩ F ) U ( M ∩ F ) U ( H ∩ F ))
= P (L ∩ F) + P (M ∩ F) + P (H ∩ F)
Using the definition of conditional probability, we get
P ( F ) = P (L) P (F | L) + P (M) P (F | M) + P (H) P (F | H)
= 0.1 × 0.05 + 0.7 × 0.10 + 0.2 × 0.25
= 0.125
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Example 5
� Weather forecast information is sent using one of four routes. Let Ri denote the event that route i is used for sending a message (i = 1, 2, 3, 4). The probabilities of using the four routes are 0.1, 0.2, 0.3 and 0.4, respectively. Further, it is known that the probabilities of an error being introduced while transmitting messages are 0.10, 0.15, 0.20 and 0.25 for routes 1, 2, 3 and 4, respectively. In sending a message an error occurred. What is the probability that route 2 was used for transmission?
Let E denote the event of having an erroneous message
� P(E | R1) = 0.10, P(E | R2) = 0.15,
� P(E | R3) = 0.20, P(E | R4) = 0.25.
Example 5 (Cont.)
� Using Bayes' rule one can compute the probability that the message was sent via route 2 given the event that an error has occurred. This probability is given by
� P(R2 | E) = )(
)( 2
EP
ERP I
∑ =
=4
1
22
)|()(
)|()(
i ii REPRP
REPRP
25.04.020.03.015.02.010.01.0
15.02.0
×+×+×+×
×=
.15.020.0
03.0==
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Example 6
� The probability of receiving more than 50 mm of rain in the months of the year is 0.25, 0.30, 0.35, 0.40, 0.20, 0.10, 0.05,0.05, 0.05, 0.05, 0.10 and 0.20 for January, February, ..., December, respectively. A monthly rainfall record selected at random is found to be more than 50 mm. What is the probability that this record belongs to month m (m = 1 for January, ..., m = 12 for December).
� Denote the probability of selecting month m as P(Mm).
� Denote the conditional probability of receiving more than 50 mm of rain, given that month m is selected as P(E | Mm).
� Then P(Mm | E) can be computed using Bayes' rule as indicated in the following equation and table
∑=
==12
1
)|()(
)|()(
)(
)|()()|(
m
mm
jjjj
j
MEPMP
MEPMP
EP
MEPMPEMP
Example 6 (Cont.)
Application of Bayes’ rule
m P(Mm) P(E | Mm) P(Mm) P(E | Mm) P(Mm | E)
1 1/12 0.25 0.02083 0.1191
2 1/12 0.30 0.02500 0.1429
3 1/12 0.35 0.02917 0.1667
4 1/12 0.40 0.03333 0.1905
5 1/12 0.20 0.01667 0.0952
6 1/12 0.10 0.00833 0.0476
7 1/12 0.05 0.00417 0.0238
8 1/12 0.05 0.00417 0.0238
9 1/12 0.05 0.00417 0.0238
10 1/12 0.05 0.00417 0.0238
11 1/12 0.10 0.00833 0.0476
12 1/12 0.20 0.01667 0.0952
12/12 0.17500 = P(E) 1.0000
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Thank you
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