Engineering Probability and Statistics - SE-205 -Chap 2

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Engineering Probability and Statistics - SE-205
-Chap 2

• By
• S. O. Duffuaa

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Lecture Objectives

• Present the following
• Random experiment
• Sample space and event
• Relationships between event
• Disjoint events
• Intersection of events
• Union of events

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Examples of a Random Experiment

• 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

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Outcome of a random experiments

• 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

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Event

• 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

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Example of Events

• A sample of polycarbonate plastic is analyzed for
  scratch resistance and shock resistance. The
  results from 49 samples are
• Shock
  resistance
• 
  H L
• H
  40 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

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Listing of Sample Spaces

• Tree Diagrams
• Experience
• Two events are mutually exclusive iff
• E1 ? E2 ?

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Lecture Objectives

• Present the following
• Types of sample spaces
• Concept of probability
• Probability of an event
• Axioms of probability
• Additive law of probability

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Types of Sample Spaces

• 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

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Concepts of Probability

• 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

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Probability of an Event

• 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.

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Axioms of Probability

• 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)

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Addition Rules

• 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)

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Conditional Probability

• Conditional Probability Concept
• P(A ?B) P(A ?B)/ P(A) for P(A) gt 0
• Give Examples
• Solve problems

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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

• 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

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Example On Conditional and Multiplication (
Product) Rule

• 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).

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Solution of Example

• 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

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Statistical Independence

• 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 ?

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Example of Independence

• 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

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Example of Dependent

• 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

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Example of Dependent

• 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.

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Objective of Class

• Present Total Probability Rule (Theorem)
• Present Bayes Theorem ( Rule)

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Total Probability Rule

• 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?

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Total Probability Rule

• 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.

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Graphical Representation

A?
A
A?
B
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General Form of Total Probability Rule

• 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)

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Bayes Rule

• 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) gt 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)
  
• P(B? Ei )P(Ei)/ P(B? E1) P(E1) P(B?
  E2) P(E2) P(B? Ek) P(Ek)

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Example on Bayes Theorem

• 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.

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Examples on Bayes Theorem

• 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.