DR KOAY probability

8/7/2019 DR KOAY probability

1/25

Larson/Farber Ch. 3

pPPpPROBABILITYBy

Dr. Koay Chen Yong

Email: [email protected]


2/25

Larson/Farber Ch. 3

Weather forecast

Games

Sports

3

Application of Probability

& Its Uses

Business

Medicine

Probability


3/25


4/25

Larson/Farber Ch. 3

{1 2 3 4 5 6 }

{ Die is even } = { 2 4 6 }

{4}

Roll a dieProbability experiment:An action through which counts, measurements orresponses are obtained

Sample space:The set of all possible outcomes

Event:

A subset of the sample space.

Outcome:

Important Terms

The result of a single trial


5/25

Larson/Farber Ch. 3

Probability Experiment: An action throughwhich counts, measurements, or responses areobtained

SampleSpace: The set of all possible outcomes

Event: A subset of the sample space.

Outcome: The result of a single trial

Choose a car from production line

Another Experiment


6/25

Larson/Farber Ch. 3

Classical (equally probable outcomes)

Probability blood pressure will decrease

after medication

Probability the line will be busy

Empirical

Intuition

Types of Probability


7/25

Larson/Farber Ch. 3

Two dice are rolled.

Describe thesample space.

1st roll

36 outcomes

2nd roll

Start

1 2 3 4 5 6

1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6

Tree Diagrams


8/25

Larson/Farber Ch. 3

1,1

1,2

1,31,4

1,5

1,6

2,1

2,2

2,32,4

2,5

2,6

3,1

3,2

3,33,4

3,5

3,6

4,1

4,2

4,34,4

4,5

4,6

5,1

5,2

5,35,4

5,5

5,6

6,1

6,2

6,36,4

6,5

6,6

Find the probability the sum is 4.

Find the probability the sum is 11.

Find the probability the sum is 4 or 11.

Two dice are rolled and the sum is noted.

SampleSpace and Probabilities

P(4) =3/36 = 1/12 = 0.083

P(11) = 2/36 = 1/18= 0.056

P(4 or 11)= 5/36

= 0.139


9/25

Larson/Farber Ch. 3

Complementary Events

The complement of event E is event E. E consistsof all the events in the sample space that are notin

event E.

The days production consists of12 cars, 5 of

which are defective. If onecar is selected at

random, find theprobability it is not defective.

E E

Solution:

P(defective) = 5/12

P(not defective) = 1 - 5/12 = 7/12 = 0.583

P(E) = 1 - P(E)


10/25

Larson/Farber Ch. 3

Section 3.2

Independent Events

and theMultiplication Rule


11/25

Larson/Farber Ch. 3

Two dice are rolled. Find the probability

the second die is a 4 given the first was a 4.

Original sample space: {1, 2, 3, 4, 5, 6}

Given the first die was a 4, the conditional

sample space is: {1, 2, 3, 4, 5, 6}

The conditional probability, P(B|A) = 1/6Note: For independent events,

P(A\B) = P(A) or P(B\A) = P(B)

Independent Events


12/25

Larson/Farber Ch. 3

Two dice are rolled. Find the probability both are 4s.

Let A = first die is a 4 and B = second die is a 4.

P(A) = 1/6 P(B) = 1/6 and P(B\A) = 1/6

P(A and B) = 1/6 x 1/6 = 1/36 = 0.028

When two events A and B are independent, then

P (A and B) = P(A) x P(B)

Note: For independent events P(B) and P(B|A)

are the same.

Multiplication Rule


13/25

Larson/Farber Ch. 3

Section 3.3

The Addition Rule


14/25

Larson/Farber Ch. 3

Compare A and B to A or B

Thecompound event A and B means that Aand B both occur in the same trial. Use the

multiplication rule to find P(A and B).

Thecompound event A or B means either Acan occur without B, B can occur without A or

both A and B can occur. Use theaddition rule

to find P(A or B).

A B

A or BA and B

A B


15/25

Larson/Farber Ch. 3

Mutually Exclusive Events

Two events, A and B, are mutually exclusive if

they cannot occur in the same trial.

A = a person is under 16 years old

B = a person is having driving license

A = a person was born in Kuala Lumpur

B = a person was born in Penang

A BMutually exclusive

P(A and B) = 0

When event A occurs it excludes event B in the same trial.


16/25

Larson/Farber Ch. 3

Non-Mutually Exclusive Events

If two events can occur in the same trial, they arenon-mutually exclusive.

A = a person is under 25 years old

B = a person is a lawyer

A = a person was born in Alor Setar

B = a person watches football on TV

A B

Non-mutually exclusive

P(A and B) 0

A and B


17/25

Larson/Farber Ch. 3

The Addition Rule

The probability that one or the other of two events willoccur is: P(A) + P(B) P(A and B)

A card is drawn from a deck. Find the

probability it is a king or it is red.

A = the card is a kingB = the card is red.

P(A) = 4/52 and P(B) = 26/52but P(A and B) = 2/52

P(A orB) = 4/52 + 26/52 2/52

= 28/52 = 0.538


18/25

Larson/Farber Ch. 3

The Addition Rule

A card is drawn from a deck. Find the

probability the card is a king or a 10.

Let A = the card is a king B = the card is a 10.

P(A) = 4/52 and P(B) = 4/52 and P(A B) = 0

P(A orB) = 4/52 + 4/52 0 = 8/52 = 0.054

When events are mutually exclusive,

P(A or B) = P(A) + P(B)


19/25

Larson/Farber Ch. 3

3. Probability atleast one of two events occur

P(A or B) = P(A U B)= P(A) + P(B) - P(A B)

Use Addition Rule

Addthe simple probabilities, but to prevent double counting, dont forget

to subtract the probability of both occurring.

4. Ifmutually exclusive, then P(A or B) = P(A U B)= P(A) +P(B)

as P(A B) = 0

1. For complementary events P(E') = 1 - P(E)

Subtract the probability of the event from one.

2. The probability both independentevents A and B occur

P(A and B) = P (A B) = P(A) X P(B)

UseMultiplication Rule

Summary


20/25

Larson/Farber Ch. 3

Section 3.4

Using Tree Diagram


21/25

Larson/Farber Ch. 3

If one event can occurm ways and a second event can occur

n ways, the number of ways the two events can occur insequence is m n. This rule can be extended forany number

of events occurring in a sequence.

If a meal consists of 2 choices of soup, 3 main dishes and 2 desserts,

how many different meals can be selected?

= 12 meals

Start

2

Soup

3

Main

2

Dessert

Fundamental Counting Principle


22/25

Larson/Farber Ch. 3

Example:

ETheprobabilities that two students, A and B willpass the driving test are1/3 and 2/5 respectively.Calculate theprobability that

(a) both A and B pass the driving test,

(b) either one of them passes the driving test,

(c) at least one of them passes the driving test.B 2/5 pass Let A = A pass

pass A = A fail

1/3 3/5 fail B = B pass

A B = B fail

2/5 pass

fail

2/3 B

3/5 fail


23/25

Larson/Farber Ch. 3

Continue:

(a) P (both A and B pass) = P(A B)

= P(A) X P(B)

= 1/3 X 2/5

= 2/15(b) P (either one of them passes)

= P(A B) + P(A B)

= (P(A) X P(B)) + (P(A) X P(B))

= (1/3 X 3/5) + (2/3 X 2/5)

= 7/15


24/25

Larson/Farber Ch. 3

Continue:

(c) P(at least one of them passes the driving test)

= 1 P(both of them fail)

= 1 (P(A) X P(B))

= 1 ( 2/3 x 3/5)

= 1 2/5

= 3/5


25/25

Larson/Farber Ch. 3

Group Activities

Have FUN with probabilities!

Documents

DR KOAY probability