Probabilistic and Statistical Techniques 1 Lecture 10 Eng. Ismail Zakaria El Daour 2011.

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Probabilistic and Probabilistic and Statistical Techniques Statistical Techniques 1 Lecture 10 Eng. Ismail Zakaria El Daour 2011

Transcript of Probabilistic and Statistical Techniques 1 Lecture 10 Eng. Ismail Zakaria El Daour 2011.

Page 1: Probabilistic and Statistical Techniques 1 Lecture 10 Eng. Ismail Zakaria El Daour 2011.

Probabilistic and Probabilistic and Statistical TechniquesStatistical Techniques

1

Lecture 10

Eng. Ismail Zakaria El Daour

2011

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Example

=

( )( | )

( )

P ABP A B

P B =

Probabilistic and Statistical TechniquesProbabilistic and Statistical Techniques

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ExampleProbabilistic and Statistical TechniquesProbabilistic and Statistical Techniques

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Probabilistic and Statistical Techniques

ExampleExample

Let the event O be an on time repair and let the event S be a satisfactory repair.It is known that P(S | O) = 0.85 and P(O) = 0.77.The question asks for P(O S) which isP(O S) = P(S | O) × P(O) = 0.85 × 0.77 = 0.6545.

A car repair is either on time or late and either satisfactory or unsatisfactory .If a repair is made on time, then there is a probability of 0.85 that it is satisfactory. There is a probability of 0.77 that a repair will be made on time . What is the probability that a repair is made on time and

is satisfactory?

SolutionSolution

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Probabilistic and Statistical Techniques

ExampleExample

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1. P(A) =

2. P(D) =

3. P(C B) =

4. P(A D) =

5. P(B D) =

Example

EventEvent C D Total

A 4 2 6

B 1 3 4

Total 5 5 10

What’s the Probability?

Probabilistic and Statistical Techniques

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SolutionThe Probabilities Are:

1. P(A) = 6/10

2. P(D) = 5/10

3. P(C B) = 1/10

4. P(A D) = 9/10

5. P(B D) = 3/10

EventEvent C D Total

A 4 2 6

B 1 3 4

Total 5 5 10

Probabilistic and Statistical Techniques

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EventEvent B1 B2 Total

A1 P(A 1 B1) P(A1 B2) P(A1)

A2 P(A 2 B1) P(A2 B2) P(A2)

P(B1) P(B2) 1

Event Probability Using Two–Way Table

Total

Probabilistic and Statistical Techniques

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Conditional Probability Using Two–Way Table

Experiment: Draw 1 Card. Note Kind, Color & Suit.

ColorType Red Black Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

P(Ace Black) 2 / 52 2P(Ace | Black) =

P(Black) 26 / 52 26

Probabilistic and Statistical Techniques

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Using the table then the formula, what’s the probability?

Example

1. P(A|D) =

2. P(C|B) =

3. Are C & B Independent?

EventEvent C D Total

A 4 2 6

B 1 3 4

Total 5 5 10

Probabilistic and Statistical Techniques

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SolutionUsing the formula, the probabilities Are:

Dependent

P(A | D) = P(A D)

P(D)

= =2 105 10

25

//

P(C | B) = P(C B)

P(B)

P(C) =

510

= =

1 104 10

14

14

//

= P(C | B)

Probabilistic and Statistical Techniques

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Example

Suppose that we have two events, A and B with P(A)=0.5, P(B)=0.6 and P(A &B)=0.4

Find:

P(A/B)

P(B/A)

Are A and B independent events? Why?

Probabilistic and Statistical Techniques

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Example

Assume that we have two events, A and B that are disjoints, assume also that P(A)=0.3, P(B)=0.4

What is (A & B)

What is P(A & B)

What is P(A/B)

Probabilistic and Statistical Techniques

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From a very large sample

Relative Frequencies

Died of

Cancer

Did Not Die

of Cancer

Totals

Never Smoke Cigars.00570.880.886

Former Cigar Smoker.00066.057.057

Current Cigar Smoker.00103.056.057

Totals.00739.9931.000Is Died of Cancer independent of cigar smoking?

Probabilistic and Statistical Techniques

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P (Died of Cancer) = 0.00739P (Died of Cancer/ Current Cigar Smoker)

= .00103/ .057= 0.018Since P (Died of Cancer) does not equal P

(Died of Cancer/ Current Cigar Smoker)So the events are dependent

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From a very large sample

Probabilistic and Statistical Techniques

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Example

In a survey of MBA students, the following data were obtained on ‘students’ first reason for application to the school in which they joined.

Probabilistic and Statistical Techniques

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Probabilistic and Statistical Techniques

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What can you say about: - P(A B) + P(A B’) = - P(A B) + P(B A’) =

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A bag contains 200 balls that are red or blue and either dull or shiny. There are 55 shiny red balls, 91 shiny balls, and 70 red balls. If a ball is chosen at random.

1- Find P( either shing or red )? 2- Find P( dull and bule )? 3- What is the probability of the chosen ball

being shiny conditional on it being red ? 4-What is the probability of the chosen ball

being dull conditional on it being red ?

Probabilistic and Statistical Techniques

Example

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A bag contains 150 balls that are red or blue and either dull or shiny. There are 36 shiny red balls, 54 blue balls. If a ball is chosen at random.

What is the probability of the chosen ball being shiny conditional on it being red ?

What is the probability of the chosen ball being dull conditional on it being red ?

Probabilistic and Statistical Techniques

Example

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