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PROBABILITY MODELS. 1.1 Probability Models and Engineering. Probability models are applied in all aspects of Engineering. Traffic engineering, reliability, manufacturing process control, design of industrial experiments, signal processing, decision analysis and risk analysis. - PowerPoint PPT Presentation
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PROBABILITY
MODELS
1.1 Probability Models and Engineering
Probability models are applied in all aspects of Engineering
• Traffic engineering, reliability, manufacturing process control, design of industrial experiments, signal processing, decision analysis and risk analysis
• Measurements in every aspect of engineering are subject to variability. Engineers deal with measurement data every day and it is vital for them to understand the relevance of measurement error.
• Probability methods are used to model variability and uncertainty
• Quality in manufacturing is inversely proportional to the variability of the manufacturing process
Example 1: Let’s Make a Deal
At the end of a TV game show (Let’s Make a Deal) thewinning contestant selects one of 3 curtains. Behind onecurtain are the keys to a new car. The spaces behind theother curtains are empty. When the choice is made it hasalways been the practice of the host to open one of theother curtains to reveal an empty space. The contestant isthen offered the chance to change their mind.
Does opening the other curtain make a difference?
Should the contestant change their selection?
Opened Selected
A B C
You choose A
Host opens C
A B
C
A B COne of these closed curtains hides the car keys
Should you switch your choice to B? YES
Example 2: Reliability of a Network
A 2 terminal network has 6 components A, B, C, D and E connected asfollows:
Each component has reliability 0.9 over the time period.
What is the reliability of the connection between 1 & 2?
To solve this example we will need to consider conditional probabilities
1 2
B
E
D
A C
For a simple situation involving a Random Experiment
A probability model is built up by following the steps:
1. List all the possible elementary outcomes
This list is called the sample space of the experiment.
2. Assign probability weights to each elementary outcome.
3. Identify events, as subsets of the sample space, for which probabilities are to be calculated.
The probability that an event A occurs is the sum of the probability weights of the elementary outcomes comprising A.
This is denoted by P(A).
1.3 Building a Probability Model
P(A) is a measure of the likelihood that an event A will occur when the experiment is performed.
Building a probability model this way guarantees that the probabilities of all events conform to the 3 basic laws of probability:
• P(A) 0, for every event A.• P(S) = 1, where S is the sample space. Equivalent to saying the list is complete• P(either A or B occurs) = P(A) P(B), for any pair of disjoint events A and B.
Events A and B disjoint means that they cannot occur together for the same experiment.
Two dice are rolled. (As in Backgammon, Monopoly or Craps)
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)}
Sample space is:
A = total score is 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}.
B = both dice even = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2),(6,4), (6,6)}
C = at least one 6 = {(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2),
(6,3), (6,4), (6,5)}
Example 3: Two Dice
D = doubles = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
1.4 Combining Events
S
AB
S
A B
A B is the event that both A and B occur.
NOTE: If A B = then A and B cannot occur together for the same experiment.
A and B are mutually exclusive or disjoint events.
IF A and B are events then:
A B
C
VENN DIAGRAMS
S
CBA A B
C
S
CBA A B
C
BA BandAIn
CBA A B
C
CBA CandBandAIn
CBA A B
C
CB)(A A B
C
CB)(A A B
C
BA BorAIn
CB)(A A B
C
CandB)or(AIn CB)(A
CBA A B
C
BA A B
C
BA A B
C
A A of Outside
BA A B
C
BA Bin andA of Outside
BA A B
C
CBA A B
C
CBA A B
C
A AIn
CBA A B
C
BA B outside andA In
A B
C
CBA
BA B outside andA In
A B
C
CBA
Cin or B outside andA In
A B
C
CBA
Cin or B outside andA In
A B
C
C)B(A
C)B(A
Do the ( ) first
B B Outside
A B
C
C)B(A A B
C
CB Cin or B Outside
C)B(A
CB Cin or B Outside
A B
C
C)B(A
C)in or B (outside andA In
A B
C
C)B(A
C)in or B (outside andA In
A B
C
A B
C
PROBABILITY
A probability is a number assigned to an event representing the chance or likelihood that the event occurs when the random experiment is performed.
The probability of an event A is denoted P(A)
Probabilities have to be assigned in a consistent way. The probabilities of all events of a random experiment must satisfy the three rules
1 P(A)0 for any event A2 P(S) = 13 P(AB) = P(A) + P(B) for any pair of disjoint events
Complement: P(A ) = 1 - P(A)
P(A)=p
A
S
P(A)=1-p
A
Result:
P(S)=1
Union: P(AB ) = P(A) + P(B) - P(A B)
Result:
S
A
P(A)
B
P(B)
Notice that when we add the probabilities, this region, is added in twice - once from A and once from B
So we subtract P (A B) to correct the double overlap
THREE IMPORTANT RULES
Intersection P(AB) = P(A).P(B) If A and B are independent events.
(More on independence later)
Complement: P(A ) = 1 - P(A)
Union: P(AB ) = P(A) + P(B) - P(A B)
Note: and multiply
or add
Example
A and B are independent events with P(A) = 0.7 P(B) = 0.6
Find P(A B)
P(A B) = P(A) + P(B) - P(AB)
so we need P(AB)
As A,B are independent P(AB) = P(A).P(B) = 0.42
P(A B) = 0.7 + 0.6 - 0.42 = 0.88
MORE COMPLICTED EVENTS
If P(A) = 0.7, P(B) = 0.6 and P(A B) = 0.9,
determine P(A B)
Example
0.3 0.2
P(AB ) = P(A) + P(B) - P(A B)
0.9 = 0.7 + 0.6 - P(A B)
P(A B) = 0.4
0.4
Region outside A,B has probability 1 - (0.3 +0.4 +0.2) = 0.1
0.1
Strategy: Find the probability of each of the disjoint areas in the Venn Diagram
ie outside A or in B
P(A B)
S
A
0.3
B
0.20.4
0.1
ie outside A or in B
P(A B)
S
A
0.3
B
0.20.4
0.1
P(A B) = 0.1 + 0.4 + 0.2
Example
A cup coffee which is supposed to contain milk and sugar obtained from the coffee dispensing machine in an engineering school cafeteria is likely to have a number of different short-comings. They are represented as the events: A - coffee burnt, B - no sugar, C - no milk.
It is known that:
P(A) = 0.7, P(B) = 0.4 and P(A B) = 0.2, P(C) = 0.3
P(A C) = 0.2, P(B C) = 0.2 , P(AB C) = 0.1
Calculate the probability that the coffee:
(1) is burnt but has sugar and milk,
(2) is not burnt and either has no sugar or has milk.
A Good Strategy: Work out from the centre
P(A) = 0.7, P(B) = 0.4 and P(A B) = 0.2, P(C) = 0.3
P(A C) = 0.2, P(B C) = 0.2 , P(AB C) = 0.1
A B
C
0.1
0.1
0.10.1
0.4 0.1
0.10.0
A B
C
0.1
0.1
0.10.1
0.4 0.1
0.10.0
A - coffee burnt, B - no sugar, C - no milk.
PROBLEM (1) What corresponds to: is burnt but has sugar and milk,
Burnt and Sugar and Milk
AB C
In A and not in B and not in C
Translate:
0.4
Answer = 0.4
A (B C )
Not in A and (in B or outside C)
A - coffee burnt, B - no sugar, C - no milk.
PROBLEM 2: What corresponds to: is not burnt and either has no sugar or has milk
Not Burnt and (No Sugar or Milk)Translate:
A B
C
0.1
0.1
0.10.1
0.4 0.1
0.10.0
0.4
A B
C
0.1
0.1
0.10.1
0.4 0.1
0.10.0
A (B C )
Not in A and (in B or outside C)
A - coffee burnt, B - no sugar, C - no milk.
What corresponds to:
(2) is not burnt and either has no sugar or has milk
Not Burnt and (No Sugar or Milk)Translate:
Answer = 0.3
TREE DIAGRAMS
Example Let’s Make a Deal
At the end of a TV game show (Let’s Make a Deal) thewinning contestant selects one of 3 curtains. Behind onecurtain are the keys to a new car. The spaces behind theother curtains are empty. When the choice is made it hasalways been the practice of the host to open one of theother curtains to reveal an empty space. The contestant isthen offered the chance to change their mind.
Does opening the other curtain make a difference?
Should the contestant change their selection?
Opened Selected
A B C
You choose A
Host opens C
A B
C
A B COne of these closed curtains hides the car keys
Should you switch your choice to B? YES
32
31
Wrong
Right
Whatever host does makes no difference to you
NON SWITCHING STRATEGY
31
P(Right)
32
31
Wrong
Right
0
1
Wrong
Right
1
0
Wrong
Right
SWITCHING STRATEGY
1st choice wrong:Host opens other wrong boxOnly box to choose is right
1st choice right:Can only choose wrong box
32
P(Right)
The actual effect of the swapping strategy is to swap your first choice over.
If you chose wrongly you end up choosing correctly.
If you chose correctly you end up choosing wrongly.
The probability of choosing wrongly on the first step is 2/3 and so the swapping strategy gives a probability of 2/3 of winning.
CONDITIONAL PROBABILITY
For a joke the entire MM1 class closes its eyes and walks across Symonds St.
S = being a MM1 student
M = being a male MM1 student
F = being a female MM1 student
L = being a live MM1 student after crossing the road
LM = being a live male MM1 student after crossing the road
L F = being a live female MM1 student after crossing the road
The probability of a student living = SinstudentsofnumberLinstudentsofnumber
SL FM
L M L F
The probability of a student living given they were male restricts the set under consideration to males Minstudentsofnumber
MLinstudentsofnumber
ie the conditional probability is the probability re-calculated for a restricted set
P(M)M)P(L
M)|P(L
Example
Three urns contain various numbers of red and blue marbles.
A B C
An urn is chosen at random and a marble drawn
Calculate the probability that Urn A was chosen given the marble was red
Want P(A|R)
C
A
B
3/5
2/3
2/5
1/3
1/4
3/4
R)|P(AR)P(A
R)(P
1/3
1/3
1/3
B
R
BR
B
R
1/5
P(R) = 1/5 + 2/9 + 1/12
= 91/180
3956.0180/915/1
R)|P(A
P(AR) =
R)|P(AR)P(A
R)(P
C
A
B
3/5
2/3
2/5
1/3
1/4
3/4
1/3
1/3
1/3
B
R
BR
B
R
3956.0180/915/1
R)|P(A
Note: We recalculate the probability restricting ourselves to the red set
BA C P(Colour)
2/91/5 1/12 91/180
1/92/15 1/4 89/180
1/31/3 1/3 1
R
B
P(Urn)
Alternatively the information could be displayed in a table of probabilities
P(R)
P(AR)
3956.0180/915/1
R)|P(A
Another Example
An urn contains various numbers of red and blue marbles.
A
Two marble are drawn from an urn.
Calculate the probability that the first marble was red given the second marble was red
Want P(R1|R2)
2/4
2/4
3/4
1/4
)R|P(R 21)RP(R 21
)R(P 2
= 6/20
= 6/20 + 6/20
= 12/20
5.020/1220/6
)R|P(R 21
R1
B1
R2
R2
B2
B2
3/5
2/5
P(R1 R2)
P(R2)
)R|P(R 21)RP(R 21
)R(P 2
5.020/1220/6
)R|P(R 21
2/4
2/4
3/4
1/4
R1
B1
R2
R2
B1
B2
3/5
2/5 Note: We recalculate the probability restricting ourselves to the red set
P(Colour2)
3/103/10 6/10
1/103/10 4/10
4/106/10 1
R2
B2
P(Colour1)
Alternatively the information could be displayed in a table of probabilities
P(R2)
P(R1R2 )
5.010/610/3
)R|P(R 21
R1 B1
Note: Since the denominators cancel we could get the same results from a table of frequencies
5.063
)R|P(R 21
Total for Colour2)
33 6
13 4
46 10
R2
B2
Total for Colour1
R1 B1
eg in 10 trials of the last urn experiment
R2
R1R2
Should really be10/610/3
But the 10s cancel
Example
One person in 100 000 suffers from a condition that will prove fatal. A test exists which has a reliability of 99% ie if you have the condition it will report it correctly 99% of the time. It also gives false positives 3% of the time ie it will report you having the condition when you don’t.
You have the test and are diagnosed as having this condition.
What is the probability of you having this condition given that you test positive?
H
D
0.00001
0.99999
)|P(H)P(H
)(P
000033.0
0.99
0.01
+
-
0.03
0.97
+
-
Note: We recalculate the probability restricting ourselves to the red set
H (D): you have (don’t have) the condition
+(-) : test is positive(negative)
03.099999.099.000001.099.000001.0
)|P(H
99.000001.0
03.099999.099.000001.0
INDEPENDENCE
We are now in a position to understand
P(AB) = P(A).P(B) for independent events A,B
If A and B are independent then
P(A|B) = P(A) (since event B cannot affect event A)
P(A)P(B)
B)P(A
P(A)P(B)B)P(A
