Download Statistics-Chap2

Engineering Probability and
Statistics - SE-205 -Chap 2
By
S. O. Duffuaa
Lecture Objectives
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Present the following:
 Random experiment
 Sample space and event
 Relationships between event
• Disjoint events
• Intersection of events
• Union of events
Examples of a Random Experiment
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Measuring a current in a wire
Number of defective in a daily production
Time to do a task
Yearly rain fall in Dhahran
Throwing a coin
Number of accidents on campus per
month
Students must generate at least 5
examples
Outcome of a random experiments
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Every time the experiment is repeated a
different out come results.
The set of all possible outcomes is call
Sample Space denoted by S.
In the experiment of throwing the coin
the sample space S = { H, T}.
In the experiment on the number of
defective parts in three parts the sample
space S = { 0, 1, 2, 3}
Event
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An event E is a subset of the sample
space.
Example of Events in the experiment of the
number of defective in a sample of 3 parts
are:
E1 = { 0}, E2 = { 0,1}, E3 = { 1, 2}
Example of Events
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A sample of polycarbonate plastic is analyzed for
scratch resistance and shock resistance. The results
from 49 samples are:
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Shock resistance
H
H
40
L
4
Scratch Resistance
L
2
3
Let A denote the event a sample has high shock resistance
and B denote the event a sample has high scratch
resistance. Determine the the number of samples in
AB, AB and A`
Listing of Sample Spaces
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Tree Diagrams
Experience
Two events are mutually exclusive iff
E1  E2 = 
Lecture Objectives
Present the following:
 Types of sample spaces
 Concept of probability
 Probability of an event
 Axioms of probability
 Additive law of probability
Types of Sample Spaces
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A sample space is discrete if it consists of a
finite ( or countably infinite ) set of
outcomes. Examples are:
S = { H, T}, S = { 1, 2, 3, …}
Students should give more examples
Concepts of Probability
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Degree of belief
Relative frequency
Equally likely then generalize
** Whenever a sample space consists of N
equally likely outcomes then the probability
of each outcome is 1/N **
Probability of an Event
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For discrete a sample space, the
probability of an event denoted as P(E)
equals the sum of the probabilities of the
outcomes in E.
Example: S = { 1, 2, 3, 4, 5} each outcome
is equally likely. E is even numbers within
S. E = { 2, 4}, P(E) = 2/5.
Axioms of Probability
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If S is the sample space and E is any event
then the axioms of probability are:
1. P(S) = 1
2. 0  P(E)  1
3. If E1 and E2 are event such that E1  E2
= , then, P(E1  E2) = P(E1 ) + P(E2)
Addition Rules
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Addition Rule
P(AB) = P(A) + P(B) – P( AB)
If AB) = , then,
P(AB) = P(A) + P(B)
This rule can be generalized to k events
If Ei  Ej = , then
P( E1  E2  … Ek) = P(E1) + P(E2) + …
+ P(EK)
Conditional Probability
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Conditional Probability Concept
P(A B) = P(A B)/ P(B) for P(B) > 0
Give Examples
Solve problems
Multiplication Rule
P(A B) = P(AB) ) P(B) = P(BA) ) P(A)
 Example:
The probability that an automobile battery subject
to high engine compartment temperature suffer
low charging is 0.7. The probability a battery is
subject to high engine compartment temperature is
0.05.
What is the probability a battery is subject to low
charging current and high engine compartment
temperature?
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Solution of Example
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Let A denote the event a battery suffers low
charging current. Let B denote the event that a
battery is subject to high engine compartment
temperature. The probability the battery is subject
to both low charging current and high engine
compartment temperature is the intersection of A
and B.
 P(A B) = P(AB) ) P(A) = 0.7 x 0.05 = 0.035
Example On Conditional and
Multiplication ( Product) Rule
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Consider a town that has a population of
900 persons, out of which 600 are males.
The rest are females. A total of 600 are
employed, out of which 500 are males. Let
M denote male, F denote female and E
employed and NE not employed. A person
is picked at random. Find the following
probabilities. P(M), P(E), P(EF), P(EF),
P(E  F).
Solution of Example
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P(M) = 600/900 = 2/3
P(E) = 600/900 = 2/3
P(EF) = 100/300 = 1/3
P(EF) = P(EF) P(F) = (1/3) x (1/3) = 1/9
P(E  F) = P(E) + P(F) – P(EF)
= 2/3 + 1/3 – 1/9 = 8/9
Statistical Independence
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Two events are statistically independent if
the knowledge about one occurring does
not affect the probability of the other
happing. Mathematically expressed as:
P(AB) = P(A) 
P(A B) = P(A) P(B) Why ?
Example of Independence
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Let us consider the experiment of throwing
the coin twice. Let B denote the event of
having a head (H) in the first throw and A
denote having a tale (T) in the second
throw.
P(AB) ) = ½ = P(A)
P(A B) = ½ x ½ = ¼ = P(A) P(B)
Therefore A and B are independent
Example of Dependent
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A daily production of manufactured parts
contains 50 parts that do not meet specifications
while 800 meets specification. Two parts are
selected at random without replacement from the
batch. Let A denote the event the first part is
defective and B the event the second part is
defective.
 Are A and B independent?
 The answer is NO. Work it out before you see
the next slide
Example of Dependent
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P( BA ) = 49/849 why?
P(B) = P(B A )P(A) + P(B A)P(A)
= (49/849)(50/850) + (50/849)(800/850)
= 50/850
Therefore A and B are not independent.
Objective of Class
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Present Total Probability Rule
(Theorem)
Present Bayes Theorem ( Rule)
Total Probability Rule
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In a chip manufacturing process 20% of
the chips produced are subjected to a high
level of contamination. 0.1 of these chips
causes product failure. The probability is
0.005 that a chip that is not subjected to
high contamination levels during
manufacturing causes a product failure.
What is the probability that a product
using one of these chips fails?
Total Probability Rule
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Let B the event that a chip causes product
failure. We can write B as part of B in A and
part of B in A.
B = (B A)  (B  A)
P(B) = P(BA) ) P(A) + P(B A) ) P(A)
Graphically on next slide.
Graphical Representation
A
A
B
A
General Form of Total Probability Rule
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Assume E1, E2, … Ek are mutually
exclusive and exhaustive events. Then
 P(B) = P(B  E1) + P(B E2) + …+ P(B Ek) )
= P(B E1) P(E1) + P(B E2) P(E2) + …+ P(B Ek) P(Ek)
Bayes Rule
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P(A B) = P(AB) ) P(B) = P(BA) ) P(A)
Implies
P(AB) ) = P(BA) ) P(A)/ P(B) , P(B) > 0
OR Refer to the slide on about the general
total probability rule, we get
 P(Ei B) = P(Ei B)/ P(B) = P(B Ei )P(Ei)/ P(B)
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= P(B Ei )P(Ei)/
P(B E1) P(E1) + P(B E2) P(E2) + …+ P(B Ek) P(Ek)
Example on Bayes Theorem
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Refer to the example about the chip production. If
you know a chip caused failure what is the chance
that the chip is subjected to a high level of
contamination when its produced.
 We want P(A  B)
 P(A  B) = P(B  A) P(A)/ P(B) = (.1)(.2)/0.024
= 5/6 = 0.833
What is the probability of the chip is not subjected to a high
level of contamination when produced ?
Answer in two ways.
Examples on Bayes Theorem
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KFUPM students when driving to building 24 th
use two roads. The main road that passes in front
of gate 1 and the second road that passes in front
of gate 2. The students use the main road 80% of
the time because it is shorter. The radar is on 60%
of the time on the main road and 30% of the time
on the other road. The students are always
speeding. Find the chance a student will be caught
speeding. If you know student is caught speeding
what is the probability he is coming to building 24
by the main road. Answer the same question for
the other road.