chapter 8 sets and probabilities. 8.1 sets the use of braces: { } element (member) of a set, , ...

Chapter 8

Sets and Probabilities

8.1 SETS

• The use of braces: { }• Element (member) of a set, , • Empty set • Distinguish: 0, and {0}• Equality of sets• Set-builder notation:

{ x x has property P }• Universal set U

Subset• A set A is a SUBSET of a set B

(written A B) if every element of A is also an element of B.

• Proper subset

• For any set A: A and A A.

• {a, b} has 4 subsets, {a, b, c} has 8 subsets.

• Number of subsets: If set A has n elements, then A has 2n subsets.

VENN diagrams

A

B C

U

COMPLEMENT

• Let A and B be any sets with U the universal set.

• Then:The complement of A, written A’, is

A’ = { x x A and x U }

• Example: U = {1, 2, 3, 4, 5}, A = {1, 3, 5}A’ = ?

A

A’

INTERSECTION

• The intersection of A and B is:

A B = { x x A and x B }

• Example:

A = {1, 2, 3, 4, 5}

B = {1, 3, 5}

A B = ?

U

A B

DISJOINT SETS

• A and B are DISJOINT sets if A B =

A BU

UNION

• The union of two sets A and B is:

A B = { x x A or x B or both}

A B

U

How to read set expressionsx A

x A

A B

A B

A B

A B

A \ B

A B

A’

x belongs to A / x is an element of A

x does not belong to A / x is not an element of A

A is contained in B / A is a subset of B

A contains B / B is a subset of A

A cap B / A meet B / A intersection B

A cup B / A join B / A union B

A minus B / the difference between A and B

A cross B / the cartesian product of A and B

A prime

8.2 APPLICATIONS OF VENN DIAGRAMS

Example 1 (p. 452).

Shade the region representing the sets:

• A’ B

• A’ B’

8.2 APPLICATIONS OF VENN DIAGRAMS

Example 2 (p. 453).

Shade the region A’ (B C’)

8.2 APPLICATIONS OF VENN DIAGRAMS

Example 3 (p. 454): A group of 60 business students was surveyed with the following results:

19 of the students read Business Week

18 read the Wall Street Journal

50 read Fortune

13 read Business Week and the Journal

11 read the and Fortune

13 read Business Week and Fortune

9 read all three magazines.

8.2 APPLICATIONS OF VENN DIAGRAMS

Example 3 (p. 454): (cont.)

(a) How many students read none of the publications?

(b) How many read only Fortune?

(c) How many read Business Week and the Journal, but not Fortune?

ADDITION RULE FOR COUNTING

Denote n(X): number of elements in X

n(A B) = n(A) + n(B) – n(A B)

Example 5 (p. 457):

A group of 10 students meets to plan a school function. All are majoring in accounting or economics or both. Five of the students are economics major and 7 are in accounting major. How many students major in both subjects?

ADDITION RULE FOR COUNTING

Example 6 (p. 457): Below is the result of American reading habit:

(a)Find n(G B)

(b)Find n(G B)

(c)Find n( (A C) H’)

ANone

B1-5

C6-10

D11-50

E51+

FNo answer

Total

GH

20052002

1618

3831

1415

2527

68

11

100100

8.3 PROBABILITY

• Experiment: activity or occurrence with an observable result

• Trial: each repetition of an experiment.

• Outcomes: possible results of each trial• Sample space: set of all possible

outcomes for an experiment.– Example: tossing a coin, rolling a die.

Sample space and event• Event: any subset of a sample space• Example: rolling a die,

Sample space S = {1, 2, 3, 4, 5, 6}

Determine the following events:– The die shows an even number: E1

– The die shows a 1: E2

– The die shows a number less than 5: E3

– The die shows a multiple of 3: E4

Sample space and event• Event: any subset of a sample space

– Simple event: event with only one possible outcome

– Certain event: event that equals the sample space

– Impossible event: empty set

• Example: rolling a die, S = {1, 2, 3, 4, 5, 6}– Simple event: The die shows a 4, E = {4}– Certain event: The die shows a number less

than 10, E = S– Impossible event: The die shows a 7, E =

Set operations for eventsLet E and F be events for a sample space S. Then:

– Event E F occurs when both E and F occur;– Event E F occurs when E or F or both occur– Event E’ occurs when E does not occur– If E F = , E and F are disjoint events (or

mutually exclusive events)

•Example: rolling a die, S = {1, 2, 3, 4, 5, 6}– E: The die shows an even number– F: The die shows a number greater than 2– Determine the following events:

E F, E F, E’, E F’

BASIC PROBABILITY PRINCIPLE(for spaces with equally likely outcomes)

• Suppose event E is a subset of a sample space S. Then the probability that event E occurs, written P(E), is:

• For any event E, 0 P(E) 1.• For any sample space S,

P(S) = 1 and P() = 0

Example: find P(E), P(F) in example in previous slide

)(

)()(

Sn

EnEP

Standard 52-card Deck

Spade

Club

Heart

Diamond

DiamondKing

Heart Queen

DiamondJack

Heart Ace

ExamplesA single card is drawn from a standard 52-card deck, find the probability of the following events:

(a) Drawing an ace

(b) Drawing a face card

(c) Drawing a spade

(d) Drawing a spade or a heart

(e) Drawing a queen

(f) Drawing a diamond

(g) Drawing a red card

ExampleThe table below lists the smoking habits of a group of college students:

If a student is chosen at random, find the probability of getting someone who is a woman and a heavy smoker.

Sex Non-smoker Regular Smoker Heavy Smoker Total

Man 135 70 5 210

Woman 187 21 15 223

Total 322 91 20 433

ADDITION RULE & COMPLEMENT RULE

ADDITION RULE• For any events E and F from a sample space S,

P(E F) = P(E) + P(F) – P(E F)• For mutually exclusive events E and F,

P(E F) = P(E) + P(F)

COMPLEMENT RULE• For any event E,

P(E’) = 1 – P(E), P(E) = 1 – P(E’).

8.4 BASIC CONCEPTS OF PROBABILITY

Examples

Example 1: A single card is drawn from a standard 52-card deck, find the probability that it will be a red or a face card.

Example 2: Two dice are rolled. Find the probability of the following events:

(a) The 1st die shows a 2 or the sum is 6 or 7

(b) The sum is 11 or the 2nd die shows a 5

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

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

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

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

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

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

ExamplesExample 2: Two dice are rolled. Find the probability of the following events:

(a) The 1st die shows a 2 or the sum is 6 or 7

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

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

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

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

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

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

ExamplesExample 2: Two dice are rolled. Find the probability of the following events:

(b) The sum is 11 or the 2nd die shows a 5

Examples

Example 4: A die is rolled, what is the probability that any number but 5 will come up?

Example 5: Two dice are rolled. Find the probability that the sum of the numbers showing is greater than 3.

ODDS• The odds in favor of an event E are defined as

the ratio of P(E) to P(E’), or , , P(E’) 0.

• If the odds favoring event E are m to n, then

P(E) = , and P(E’) =

Example 6 (p. 472)

Example 7 (p. 473)

Example 8 (p. 473)

)'(

)(

EP

EP

nm

m

nm

n

ODDSExample 6 (p. 472)

The probability of rain tomorrow is 1/3. Find the

odds in favor of rain tomorrow.

Example 7 (p. 473)

There is a 40% chance that it will snow tomrrow.

Find the odds in favor of snow tomorrow.

Example 8 (p. 473)

The odds that a particular bid will be low bid are 4

to 5. Find the probability that the bid will be the

low bid.

Relative Frequency ProbabilityExample 9: the table lists the number of siblings indicated by respondents in a survey:

(a)Find the relative frequency probability of having 0, 1, 2, …, 10 or more siblings.

(b)Find the probability that a randomly chosen American has 1 or 2 siblings.

Number of siblings Frequency0 1401 5052 5833 4574 3145 2246 1577 1158 779 49

10 or more 137Total 2758

Number of siblings Frequency Probability0 140 0.0511 505 0.1832 583 0.2113 457 0.1664 314 0.1145 224 0.0816 157 0.0577 115 0.0428 77 0.0289 49 0.018

10 or more 137 0.050Total 2758 1.000

PROPERTIES OF PROBABILITY

Let S be a sample space consisting of n distinct outcomes s1, s2, …, sn. An acceptable probability

assignment consists of assigning to each outcome si a number pi (the probability of si) according to

these rules:

1.The probability of each outcome is a number between 0 and 1 (0 pi 1, i = 1..n)

2.The sum of the probabilities of all possible outcomes is 1. (p1 + p2 + … + pn = 1)

Relative Frequency ProbabilityExample: two dice are rolled, and the sum is calculated. Make the probability distribution for the sum:

Find the probability that the sum is at least 10.

The sum Probability23456789

101112

Example 10Let L indicate the event that the respondent had a “liberal” political tendency, and let M indicate that the respondent believes that marijuana use should be legal. Below are the survey estimates:

P(L) = .27, P(M) = .37, P(L M) = .15

(a)Find the probability that a respondent does not have a liberal tendency and does not support legalizing the use of marijuana.

(b)Find the probability that a respondent does not have a liberal tendency or does not support legalizing the use of marijuana.

8.5 CONDITIONAL PROBABILITY

• The conditional probability of an event E given event F,

written P(E|F), is:

, P(F) 0

• Example 2 (p. 483): given P(E) = .4, P(F) = .5 and

P(EF)=.7. Find P(E|F).

• Example 3 (p.483): 2 coins were tossed, find the

probability that both were heads, if it is known that at

least one was head.

)()(

)|(FPFEP

FEP

PRODUCT RULE OF PROBABILITY

If E and F are events, then P(E F) may be found by either of these formulas:

P(E F) = P(F)P(E|F) or P(E F) = P(E)P(F|E)

Example 4. (p. 484)In a class with 2/5 women and 3/5 men, 25% of the women are business majors. Find the probability that a student at random from the class is a female business major.

Example 5. (p. 485)A company needs to decide between person A and B to be a new director of advertising. The research shows that A is in charge of twice as many advertising campaigns as B. And A’s campaigns have satisfactory results three out of four times, while B’s campaigns have satisfactory results only two out of five times.a)Find the probability that A runs a campaign that produces satisfactory results.b)Find the probability that B runs a campaign that produces satisfactory results.c)Find the probability that a campaign is satisfactory, unsatisfactoryd)Find the probability that either A runs the campain or the results are satisfactory

Example 6. (p. 486)

From a box containing 1 red, 3 white and 2 green marbles, two marbles are drawn one at a time without replacing the first before the second is drawn. Find the probability that one white and one green marble are drawn.

Example 7. (p. 488)

Two cards are drawn without replacement from a deck. Find the probability that the first card is a heart and the second card is red.

Example 8. (p. 488)

Three cards are drawn without replacement from a deck. Find the probability that exactly 2 of the cards are red.

INDEPENDENT EVENTS

• E and F are independent events if

P(F|E) = P(F) or P(E|F) = P(E)

PRODUCT RULE FOR INDEPENDENT EVENTS

• E and F are independent events if and only if P(EF) = P(E)P(F)

Example 9. (p. 490)

A calculator requires a key-stroke assembly and a logic circuit. Assume that 99% of the key-stroke assemblies are satisfactory and 97% of the circuits are satisfactory. Find the probability that a finished calculator will be satisfactory. Suppose that the failure of key-stroke assemblies and the failure of logic circuits are independent.

Example 10. (p. 490)

On a typical day in Saigon the probability of very hot weather is 0.10, the probability of a traffic jam is 0.80, and the probability of very hot weather or a traffic jam is 0.82. Are the event “very hot weather” and event “a traffic jam” independent?

8.6 BAYES’ FORMULA• BAYES’ FORMULA (Special case)

• Example 1 (p. 495)For a fixed length of time, the probability of worker error on a production line is 0.1, the probability that an accident will occur when there is a worker error is 0.3, and the probability that an accident will occur when there is no worker error is 0.2. Find the probability of a worker error if there is an accident.

)'|()'()|()()|()(

)|(FEPFPFEPFP

FEPFPEFP

• BAYES’ FORMULA (General case)

• Example 2 (p. 497)A survey indicated that 87% of married women have one or more children, 40% of never-married women have one or more children, and 88% of women who are divorced, separated or widowed have one or more children. The survey also indicated that 43% of women were currently married, 24% had never been married, and 33% were divorced, separated or widowed. Find the probability that a woman who have one or more children is married.

)|()()|()()|()(

)|(11 nn

iii FEPFPFEPFP

FEPFPEFP

USING BAYES’ FORMULA

1. Start a probability tree with branches representing events F1, F2, …, Fn. Label each branch with its corresponding probability.

2. From the end of each of these branches, draw a branch for event E. Label this branch with probability of getting to it, or P(E|Fi).

USING BAYES’ FORMULA3. There are now n defferent paths that result in

event E. Next to each path, put its probability – the product of the probabilities that the first branch occurs, P(Fi), and that the second branch occurs, P(E|Fi): that is P(Fi) P(E|Fi).

4. P(Fi|E) is found by dividing the probability of the branch for Fi by the sum of the probabilities of all the branches producing event E.

Example 3 (p. 497)A manufacturer buys items from 6 different suplliers. The fraction of the total number of items obtained from each supplier, along with the probability that an item purchased from that supplier is defective, is shown in the table below. Find the probability that a defective item came from supplier 5.

Supplier Fraction of total supplied

Probability of Defective

1 .05 .04

2 .12 .02

3 .16 .07

4 .23 .01

5 .35 .03

6 .09 .05