Engineering Probability and Statistics - SE-205 -Chap 2

By

S. O. Duffuaa

Lecture Objectives

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

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

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

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 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 4Scratch 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 AB, AB and A`

Listing of Sample Spaces

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

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

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

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

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

Addition Rule

P(AB) = P(A) + P(B) – P( AB) If AB) = , then, P(AB) = 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

Conditional Probability Concept P(A B) = P(A B)/ P(A) for P(A) > 0 Give Examples Solve problems

Multiplication Rule

P(A B) = P(AB) ) P(B) = P(BA) ) 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?

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(AB) ) P(A) = 0.7 x 0.05 = 0.035

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(EF), P(EF), P(E F).

Solution of Example

P(M) = 600/900 = 2/3 P(E) = 600/900 = 2/3 P(EF) = 100/300 = 1/3 P(EF) = P(EF) P(F) = (1/3) x (1/3) = 1/9 P(E F) = P(E) + P(F) – P(EF)

= 2/3 + 1/3 – 1/9 = 8/9

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(AB) = P(A) P(A B) = P(A) P(B) Why ?

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(AB) ) = ½ = P(A) P(A B) = ½ x ½ = ¼ = P(A) P(B) Therefore A and B are independent

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

Example of Dependent

P( BA ) = 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

Present Total Probability Rule (Theorem)

Present Bayes Theorem ( Rule)

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?

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(BA) ) P(A) + P(B A) ) P(A) Graphically on next slide.

Graphical Representation

A AB

A

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)

Bayes Rule

P(A B) = P(AB) ) P(B) = P(BA) ) P(A) Implies P(AB) ) = P(BA) ) 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) = P(B Ei )P(Ei)/ P(B E1) P(E1) + P(B E2) P(E2) + …+ P(B Ek)

P(Ek)

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.

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.