• 4-1 Introduction• 4-2 Sample Spaces and Probability• 4-3 The Addition Rules for

Probability• 4-4 The Multiplication Rules and

Conditional Probability

• Determine the sample spaces and find the probability of an event using classical probability.

• Find the probability of an event using empirical probability.

• Find the probability of compounds events using the addition rules.

• Find the probability of compounds events using the multiplication rules.

• Find the conditional probability of an event.

4-1 Introduction

• What is probability?

Probability is the chance of occurrence of a particular event.

Probability is measured on a scale from 0 to 1.

0 1

Event can never

happen

Event is absolutely certain

to happen

4-2 Sample Spaces and Probability

• A probability experiment is a process that leads to well-defined results called outcomes.

• An outcome is the result of a single trial of a probability experiment.

• Sample space is the set of all possible outcomes of a probability experiment.

4-2 Sample Spaces Example

Toss a coin

Experiment Sample space

Head, tail

Roll a die

1,2,3,4,5,6

4-2 Event

• Event consists of a set of outcomes of a probability experiment.

• An event with one outcome is called a simple event.

• Event that consists of more than one outcome is called a compound event.

4-2 Simple Event

• Experiment: Select a day of a week and getting Tuesday.

• Outcome: Tuesday (One outcome)

4-2 Compound Event

• Experiment: Roll a die and getting odd number.

• Outcomes: 1, 3, 5 (Three outcomes)

4-2 Tree diagram

• A tree diagram can be used to find all possible outcomes of a probability experiment.

H

T

H

H

T

TFirst Toss

Second Toss

4-2 Type of Probability

Classical Probability

Empirical Probability

VS

4-2 Formula for Classical Probability

• Classical probability assumes that all outcomes in the sample space are equally likely to occur.

• Equally likely events are events that have the same probability of occurring.

4-2 Formula for Classical Probability

Probability of an event E

= Total number of outcomes in the sample space

P (E) =

Number of outcomes in E

n(E)n(S)

Answer a multiple choice question with four choices (A, B, C and D).

The probability of occurrence for each outcome is the same which is 1/4.

Deliver a baby. The gender for a baby can be either male or female.

The probability of getting male or female will always be 1/2.

4-2 Classical Probability

Example

• An ordinary die is thrown. Find the probability that the number obtained

(a) is less than 5.(b) is a multiple of 3.

(c) is 9.

4-2 Classical Probability

Solutions:

(a) The outcomes are 1, 2, 3, and 4.P (less than 5) = 4/6 = 2/3

(b) The outcomes are 3 and 6. P (multiple of 3) = 2/6 = 1/3

(c) It is impossible to get a 9 when a die is rolled.P (9) = 0

4-2 Classical Probability

Question:

• A card is drawn at random from an ordinary pack containing 52 playing cards. Find the probability that the card drawn (a) is the diamond(b) is the four of spades

4-2 Empirical Probability

• Some of the outcomes are not equally likely thus their probabilities need to be determined through empirical method.

• Empirical probability estimate the probability of an outcome based on the actual experience, observation, or experiment.

4-2 Formula for Empirical Probability

Given a frequency distribution, the probability of an event being a given class is

P (E) =

=

Frequency for the class

Total frequencies in the distribution

f

n

4-2 Empirical Probability

Example:

• In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution.

4-2 Empirical Probability

Type FrequencyA 22B 5

AB 2O 21

n = 50

4-2 Empirical Probability

Find the following probabilities based on the frequency distribution :

(a) A person has type O blood.(b) A person has type A or type B blood.(c) A person has neither type A nor type O blood.(d) A person does not have type AB blood.

4-2 Empirical Probability

Solutions:

(a) P(O) = 21/50 (b) P(A or B) = 22/50 + 5/50 = 27/50(c) P(neither A nor O) = 5/50 + 2/50 = 7/50(d) P(not AB) = 1 - 2/50 = 48/50

4-2 Empirical Probability

Question

A ball is drawn from a box containing 10 red, 15 white, 5 green, and 5 black. Find the probability that the ball is (a)black(b)red or green(c)not white

4-2 Complementary Events

The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is denoted by E (read as “E bar”).

4-2 Complementary Events

P(E)

P(S) = 1

(a) Simple probability (b) P(E) = 1- P(E)

P(E)

P(E)

Venn Diagram

4-2 Rule for Complementary Events

P (E) = 1 - P (E) or P (E) = 1 - P (E) or P (E) + P (E) = 1

P (E) = 1 - P (E) or P (E) = 1 - P (E) or P (E) + P (E) = 1

4-2 Complementary Events

Examples:

Find the complement of each event.

(a) Rolling a die and getting 4.

(b) Selecting a month and getting a month that begins with a J.

(c) Selecting a day of the week and getting a weekday.

4-2 Complementary Events

Solutions:

(a) Getting 1, 2, 3, 5, or 6

(b) Getting February, March, April, May, August, September, October, November, or December.

(c) Getting Saturday or Sunday.

4-3 The Addition Rules for Probability

• Two events are mutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common).

4-3 Addition Rules 1

When two events A and B are mutually exclusive, the probability that A or B will occur is

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

4-3 The Addition Rules for Probability

P (A) P (B)

P (S) = 1

Mutually exclusive events

4-3 Addition Rules 1

Example 1:

A box contains 3 chocolate doughnuts, 4 jelly doughnuts, and 5 strawberry doughnuts. If a person select one doughnut randomly, find the probability that it is either a chocolate doughnut or a strawberry doughnuts. P (A or B) = P (A) + P (B)

4-3 Addition Rules 1

Solution: P (chocolate or strawberry) = P (chocolate) + P (strawberry) = 3/12 + 5/12 = 8/12 = 2/3

4-3 Addition Rules 1

Example 2:

A day of the week is selected at random. Find the probability that it is a weekend.

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

4-3 Addition Rules 1

Solution: P (Saturday or Sunday) = P (Saturday) + P (Sunday) = 1/7 + 1/7 = 2/7

In a conference, there are 12 researchers, 10 scientists, and 8 educators. If an attendant is selected, find the probability of getting a researcher or an educator.

4-3 Addition Rules 2

When two events A and B are not mutually exclusive, the probability that A or B will occur is

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

or

4-3 Addition Rules 2

P (A) P (B)

P (A and B)

P (S) = 1

Non-mutually exclusive events

4-3 Addition Rules 2

Example:

In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.

4-3 Addition Rules 2

STAFF FEMALES MALES TOTAL

NURSES 7 1 8

PHYSICIANS 3 2 5

TOTAL 10 3 13

4-3 Addition Rules 2

Solution:

P (nurse or male) = P (nurse) + P (male) - P (male nurse)= 8/13 + 3/13 - 1/13= 10/13

In a group of 30 students all study at least one of the subjects physics and biology, 20 attend the physics class and 21 attend the biology class. Find the probability that a student chosen at random studies both physics and biology.

In a statistics class there are 18 juniors and 10 seniors; 6 of the seniors are females, and 12 of the juniors are males. If a student is selected at random, find the probability of selecting:

(a)A junior or a female(b)A senior or a female(c)A junior or a male

4-4 The Multiplication Rules and Conditional Probability

• Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring.

• Example: Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events.

When two events A and B are independent, the probability of both occurring is

P (A and B) = P (A) • P (B)

or

4-4 Multiplication Rules 1

Example 1:

• A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die.

4-4 Multiplication Rules 1

Solution:P (head and 4)= P(head) • P(4)= 1/2 • 1/6= 1/12

or

4-4 Multiplication Rules 1

Example 2:

A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.

or

4-4 Multiplication Rules 1

Solution:Let S denote stress. ThenP (S and S and S) = P (S) • P (S) • P (S) = (0.46) (0.46) (0.46) = 0.097

or

4-4 Multiplication Rules 1

Example 3:

The probability that a specific medical test will show positive is 0.32. If four person are tested, find the probability that all four will show positive.

4-4 Multiplication Rules 1

Solution:Let T denote a positive test result. ThenP (T and T and T and T) = P (T) • P (T) • P (T) • P (T)= (0.32)4

= 0.010.

or

4-4 Multiplication Rules 1

Two types of metal A and B, which have been treated with a special coating of paint have probabilities of 1/4 and 1/3 respectively of lasting four years without rusting. Find the probability that

(a)Both last four years without rusting(b)At least one of them lasts four years without rusting.

At a local university 54% of incoming first-year student have computers. If three students are selected at random, find the probability that

(a)All have computers(b) None have computers(c) At least one has computer

• When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent.

• Example: Having high grades and getting a scholarship are dependent events.

or

4-4 The Multiplication Rules and Conditional Probability

• The conditional probability of an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred.

• The notation for the conditional probability of B given A is P(B|A).

• Note: This does not mean B divided by A.

r

4-4 The Multiplication Rules and Conditional Probability

When two events A and B are dependent, the probability of both occurring is

P (A and B) = P (A) • P (B|A)

or

4-4 Multiplication Rules 2

Example 1:

In a shipment of 25 refrigerators, two are defective. If two refrigerators are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested.

P (A and B) = P (A) • P (B|A)

4-4 The Multiplication Rules and Conditional Probability

Solution:Let D denote defective. Since the events are dependent, P (D1 and D2) = P (D1) • P (D2|D1)

= (2/25) (1/24) = 2/600 = 1/300

P (A and B) = P (A) • P (B|A)

or

4-4 The Multiplication Rules and Conditional Probability

Example 2:

A person owned a collection of 30 CDs, of which 5 are country music. If 2 CDs are selected at random, one by one without replacement, find the probability that both are country music.

4-4 The Multiplication Rules and Conditional Probability

Solution:Let C denote country music. Since the events are dependent, P (C1 and C2) = P (C1) • P (C2|C1)

= (5/30) (4/29) = 20/870 = 2/87

P (A and B) = P (A) • P (B|A)

or

4-4 The Multiplication Rules and Conditional Probability

Example 3:Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tail up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball.

or

4-4 The Multiplication Rules and Conditional Probability

4-4 The Multiplication Rules and Conditional Probability

P(B1)

P(B2)

1/2

1/2

Box 1

Box 2

P(R|B1)

P(B|B1)

P(R|B2)

P(R|B2)

2/3

1/3

1/4

3/4

BOX BALL

Red

Blue

Blue

Red

1/2 • 2/3 = 1/3

1/2 • 1/3 = 1/6

1/2 • 1/4 = 1/8

1/2 • 3/4= 3/8

Solution:Since red ball can be obtained from box 1 and box 2.

P(red) = 1/3 + 1/8 = 8/24 + 3/24 = 11/24

4-4 The Multiplication Rules and Conditional Probability

A box contains 6 red pens and 3 blue pens. A pen is selected at random, the colour is noted and the pen is retained. After this, a second pen is selected and the colour is noted. Find the probabilities of obtaining

i) both red pens ii) two pens of different colours

The probability that second event B occurs given that the first event A has occurred can be found by dividing the probability that both events occurred by the probability that the first event has occurred. The formula is

4-4 Conditional Probability - Formula

P (B|A) =P (A and B)

P (A) P (B|A) =

P (A and B)

P (A)

Example:

For married couples the probability that the husband has passed his driving test is 7/10 and the probability that the wife has passed her driving test is 1/2. The probability that both of them passed the driving test is 7/15. Find the probability that the husband has passed, given that the wife has passed.

` 4-4 Conditional Probability

Solution

P(H) = 7/10; P(W) =1/2; P (H and W) = 7/15

P (H|W) = P (H and W) / P(W) = 7/15 / 1/2 = 14/15

4-4 Conditional Probability