Chapter 7 Probability 7.1 Experiments, Sample Spaces, and Events 7.2 Definition of Probability 7.3 Rules of Probability 7.4 Use of Counting Techniques.

Chapter 7 Probability

• 7.1 Experiments, Sample Spaces, and Events

• 7.2 Definition of Probability

• 7.3 Rules of Probability

• 7.4 Use of Counting Techniques in Probability

• 7.5 Conditional Probability and Independent Events

• 7.6 Bayes’ Theorem

An experiment is an activity with observable results (called outcomes).

A sample point is an outcome of an experiment. The sample space is the set of all possible sample points.

An event is a subset of a sample space.

Section 7.1 Experiments, Sample Spaces, and Events

Ex. Rolling a die

Outcomes: landing with a 1, 2, 3, 4, 5, or 6 face up.

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

Events: , {1}, {2}, {3}, {4}, {5}, {6}, S

Impossible event

Certain event

Ex. An experiment consists of spinning the hand on the disk below two times. Find the sample space.

C

W

P

S = {(P,C), (P,W), (P,P), (C,P), (C,W), (C,C), (W,P), (W,C), (W,W)}

Events

The union of events A and B is the event

The intersection of events A and B is the event

The complement of event A is the event AC.

.A B

.A B

Ex. Rolling a die. S = {1, 2, 3, 4, 5, 6}

Let A = {1, 2, 3} and B = {1, 3, 5}

{1,2,3,5}A B

{1,3}A B

{2}CA B

Events A and B are mutually exclusive if .A B

Ex. When rolling a die, if event A = {2, 4, 6} (evens) and event B = {1, 3, 5} (odds), then A and B are mutually exclusive.

Ex. When drawing a single card from a standard deck of cards, if event A = {heart, diamond} (red) and event B = {spade, club} (black), then A and B are mutually exclusive.

The probability of an event occurring is a measure of the proportion of the time that the event will occur in the long run.

Suppose that in n trials an event E occurs m times. The relative frequency of the event E is m/n.

If the relative frequency approaches some value P(E) as n becomes larger, then P(E) is called the empirical probability of E.

Section 7.2 Definition of Probability

Ex. The table below represents the frequency of certain types of license plates observed by a family on a recent trip. Find the probability distribution.

State Number

Wisconsin 45

Illinois 80

Iowa 20

Indiana 5

Probability

45/150 = 0.300

80/150 = 0.533

20/150 = 0.133

5/150 = 0.033

Notice 150 total observations

The function P, which assigns a probability to each simple event is called a probability function.

Let S = {s1, s2, s3,…,sn} where each si represents a simple event (all mutually exclusive) and let P(si) represent the probability of event si.

Also P(si) has the following properties:

1 2

0 ( ) 1

( ) ( ) ... ( ) 1

( ) ( )

i

n

i j i j

P s

P s P s P s

P s s P s P s

probabilities are between 0 and 1

Sum of the probabilities is 1

Probabilities of the union is the sum of their probabilities

Probability of an Event in a Uniform Sample Space

If

S = {s1, s2, … , sn}

is the sample space for an experiment in which the outcomes are equally likely, then we assign the probabilities

to each of the outcomes s1, s2, … , sn.

1 2

1( ) ( ) ( )nP s P s P s

n 1 2

1( ) ( ) ( )nP s P s P s

n

Ex. Assume that when rolling a die each face is equally likely to show up. If event E = {2} then since S = {1, 2, 3, 4, 5, 6}, we have P(E) = 1/6. That is, the probability of rolling a 2 is 1 in 6.

Similarly, the probability of rolling any face number is 1/6.

Finding the Probability of an Event E

1. Determine a sample space S associated with the experiment.

2. Assign probabilities to the simple events of S.

3. If E = {s1, s2, s3,…,sn} (each a simple event) then

P(E) = P(s1) + P(s2) +…+ P(sn).

If E is the empty set then P(E) = 0.

Ex. An experiment consists of spinning the hand on the disk below one time. Assume each outcome is equally likely.

C

W

A

Find P(C) and then find .P C W

( ) ( )P C W P C P W 1 1 2

3 3 3

Notice S = {C, A, W} each of which has a probability of 1/3.

1

3P C

Applied Example: Rolling Dice

• A pair of fair dice is rolled.

• Calculate the probability that the two dice show the same number.

• Calculate the probability that the sum of the numbers of the two dice is 6.

Applied Example 3, page 365

Applied Example: Rolling Dice

Solution• The sample space S of the experiment has 36 outcomes

S = {(1, 1), (1, 2), … , (6, 5), (6, 6)}

• Both dice are fair, making each of the 36 outcomes equally likely, so we assign the probability of 1/36 to each simple event.

• The event that the two dice show the same number is

E = {(1, 1), (2, 2) , (3, 3), (4, 4), (5, 5), (6, 6)}

• Therefore, the probability that the two dice show the same number is given by

( ) [(1,1)] [(2,2)] [(6,6)]

1 1 1 1

36 36 36 6

P E P P P

( ) [(1,1)] [(2,2)] [(6,6)]

1 1 1 1

36 36 36 6

P E P P P

Six terms

Applied Example 3, page 365

Applied Example: Rolling Dice

Solution

• The event that the sum of the numbers of the two dice is 6 is given by

E6 = {(1, 5), (2, 4) , (3, 3), (4, 2), (5, 1)}

• Therefore, the probability that the sum of the numbers on the two dice is 6 is given by

6( ) [(1,5)] [(2,4)] [(3,3)] [(4,2)] [(5,1)]

1 1 1 1 1 5

36 36 36 36 36 36

P E P P P P P

6( ) [(1,5)] [(2,4)] [(3,3)] [(4,2)] [(5,1)]

1 1 1 1 1 5

36 36 36 36 36 36

P E P P P P P

Applied Example 3, page 365

Properties of the Probability Function

Property 1 ( ) 0 for any P E E

Property 2 ( ) 1 P S

Property 3 ( ) ( ) ( )P E F P E P F

If E and F are mutually exclusive (E F = Ø), then

Section 7.3 Rules of Probability

Dollars spent Probability

0.05

0.10

0.15

0.25

0.4550x

200x

100 150x

150 200x

50 100x

Ex. A local grocery store has found kept track of the amount of money spent by its customers on a single visit. Find the probability that if a customer is selected at random, the amount spent by the customer will be

a. More than $150

b. More than $50 but less than or equal to $200

= 0.15

= 0.50

Property 4 Addition Rule

If E and F are any two events of an experiment, then

( ) ( ) ( ) ( )P E F P E P F P E F Subtract overlap

Note: If E and F are mutually exclusive, then ( ) 0P E F

E F

Ex. A card is drawn from a well-shuffled deck of 52 playing cards. What is the probability that it is a king or a heart?

4 13 1( ) , ( ) , ( )

52 52 52P K P H P K H

( ) ( ) ( ) ( )P K H P K P H P K H

4 13 1

52 52 52

16 4

52 13

K = King and H = Heart

Property 5 Rule of Complements

If E is an event of an experiment and EC denotes the complement of E, then

1 ( )CP E P E

Ex. A card is drawn from a well-shuffled deck of 52 playing cards. What is the probability that it is not a king?

K = pick a king,

1 ( )CP K P K 4

152

48 12

52 13

4( )

52P K

Computing the Probability of an Event in a Uniform Sample Space

Let S be a uniform sample space and let E be any event. Then

number of favorable outcomes in

number of possible outcomes in

n EEP E

S n S

Section 7.4 Use of Counting Techniques in Probability

Ex. Suppose that you reach into a box of 12 size AA batteries and you know that 4 of them are dead. Find the probability that

a. in one draw you get a good battery.

b. in two draws without replacement you get two good batteries.

good batteries

batteries

n

n8 2

12 3

ways to get 2 good

ways to draw 2 batteries

n

n

8,2

12,2

C

C

8,1

12,1

C

C

28 14

66 33

Ex. Three balls are selected at random without replacement from the jar below. Find the probability that

a. All 3 of the balls are green.

b. One ball is red and two are black.

3,3

8,3

C

C

draw 3 green

draw 3

n

n1

56

draw 1 red, 2 black

draw 3

n

n

2,1 3,2

8,3

C C

C

6 3

56 28

Ex. Refer to the jar of marbles below. Two marbles are drawn at random without replacement.

Find the probability that no yellow are drawn.

3,21

11,2

C

C

3 521

55 55

1 both yellowP

The probability of an event is affected by the knowledge of other information relevant to the event.

Ex. You roll a fair die. Find the probability that you roll a 2 given that your roll is an even.

Notation: P(A|B) is read “the probability of event A given that event B has occurred.”

2 | evenP 1

3

Knowing it is even restricts the sample space to {2, 4, 6}.

So

Section 7.5 Conditional Probability and Independent Events

Conditional Probability of an Event

If A and B are events in an experiment and then the conditional probability that the event B will occur given that A has already occurred is

|P A B

P B AP A

( ) 0,P A

Which can be written (the Product Rule):

|P A B P A P B A

Ex. In a box of 20 size AA batteries, 10 are brand X and 10 are brand Y. You also know that 3 of the brand X batteries are dead, while 2 of the brand Y are dead. Find the probability that in a (random) draw

a. you get a dead (D) brand X battery.

b. you get brand Y given that you drew a dead (D) battery.

|P X D P X P D X

|P Y D

P Y DP D

2 / 20

5 / 20 2

5

10 3

20 10 3

20

Independent Events

If A and B are independent events, then | ( )

( | ) ( )

P B A P B

P A B P A

Test for Independent Events

Events A and B are independent events if and only if ( ) ( )P A B P A P B

Note: this generalizes to more than two independent events.

Ex. If A die is rolled twice, show that rolling a 5 on the first roll and rolling a 4 on the second roll are independent events.

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)

S

(roll 1, roll2)

1

36P V R

V = 5 on first roll, R = 4 on second roll

P V P R

Therefore V and R are independent

6 6 1

36 36 36

Bayes’ Theorem

1 1 2 2

||

| | ... |i i

in n

P A P E AP A E

P A P E A P A P E A P A P E A

Let A1, A2, …, An be a partition of a sample space S and let E be an event of the experiment such that P(E) is not zero. Then the posteriori probability P(Ai|E) is given by

1 i n Where

Posteriori probability: probability is calculated after the outcomes of the experiment have occurred.

Section 7.6 Bayes’ Theorem

Ex. A store stocks light bulbs from three suppliers. Suppliers A, B, and C supply 10%, 20%, and 70% of the bulbs respectively. It has been determined that company A’s bulbs are 1% defective while company B’s are 3% defective and company C’s are 4% defective. If a bulb is selected at random and found to be defective, what is the probability that it came from supplier B?

||

| | |

P B P D BP B D

P A P D A P B P D B P C P D C

Let D = defective

0.2 0.03

0.1 0.01 0.2 0.03 0.7 0.04

0.1714

So about 0.17