3.1 Probability ExperimentsProbability experiment: An action, or trial, through which specific results

(counts, measurements, or responses) are obtained.

Roll a die

Outcome: The result of a single trial in a probability experiment.

{3}

Sample Space: The set of all possible outcomes of a probability experiment.

{1, 2, 3, 4, 5, 6}

Event: Consists of one or more outcomes and is a subset of the sample space.

{Die is even}={2, 4, 6}

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Example: A probability experiment consists of tossing a coin and then rolling a 6-sided die. Describe the sample space.

H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6

The sample space has 12 outcomes:{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Fundamental Counting PrincipleFundamental Counting Principle

• If one event can occur in m ways and a second event can occur in n ways, the number of ways the two events can occur in sequence is m*n.

• Can be extended for any number of events occurring in sequence.

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Example: You are purchasing a new car. Your choices are:Manufacturer: Ford, GM, HondaCar size: compact, midsizeColor: white (W), red (R), black (B), green (G)

How many different ways can you select 1 manufacturer, 1 size, & 1 color?

3 ∙ 2 ∙ 4 = 24 ways

Types of ProbabilityClassical (theoretical) Probability

• Each outcome in a sample space is equally likely.

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Number of outcomes in event E( )

Number of outcomes in sample spaceP E

Example: You roll a six-sided die. Find the probability of each event.

1. Event A: rolling a 3

2. Event B: rolling a 7

3. Event C: rolling a number less than 5

Sample space: {1, 2, 3, 4, 5, 6}1

( 3) 0.1676

P rolling a

0( 7) 0

6P rolling a

4( 5) 0.667

6P rolling a number less than

Part Whole

Types of ProbabilityEmpirical (statistical) Probability

• Based on observations obtained from probability experiments.

• Relative frequency of an event.

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Frequency of event E( )

Total frequency

fP E

n

Part Whole

Example: A company is conducting an online survey of randomly selected individuals to determine if traffic congestion is a problem in their community. So far, 320 people have responded to the survey. What is the probability that the next person that responds to the survey says that traffic congestion is a serious problem in their community? Response Number of times, f

Serious problem 123

Moderate problem 115

Not a problem 82

Σf = 320

123( ) 0.384

320

fP Serious problem

n

Law of Large Numbers

• As an experiment is repeated over and over, the statistical probability of an event approaches the theoretical (actual) probability of the event.

Range of Probabilities RuleRange of probabilities rule

• The probability of an event E is between 0 and 1, inclusive.

• 0 ≤ P(E) ≤ 1

[ ]0 0.5

1

Impossible UnlikelyEven

chance Likely Certain

Complement of event E

• The set of all outcomes in a sample space that are not included in event E.

• Denoted E ′ (E prime)

• P(E ′) + P(E) = 1• P(E) = 1 – P(E ′)• P(E ′) = 1 – P(E)

E ′E

Example: You survey a sample of 1000 employees at a company and record the age of each. Find the probability of randomly choosing an employee who is not between 25 and 34 years old.

Employee ages

Frequency, f

15 to 24 54

25 to 34 366

35 to 44 233

45 to 54 180

55 to 64 125

65 and over 42

Σf = 1000

366( 25 34) 0.366

1000

fP age to

n

366( 25 34) 1

1000634

0.6341000

P age is not to

Example: Probability Using the Fundamental Counting Principle

Your college identification number consists of 8 digits. Each digit can be 0 through 9 and each digit can be repeated. What is the probability of getting your college identification number when randomly generating eight digits?

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• Each digit can be repeated

• There are 10 choices for each of the 8 digits

• Using the Fundamental Counting Principle, there are

10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10

= 108 = 100,000,000 possible identification numbers

• Only one of those numbers corresponds to your ID number

P(your ID number) =1

100,000,000Part Whole

1 ID number

Total Possible

3.2 Conditional ProbabilityConditional Probability

• The probability an event occurs, given that another event already occurred

• Denoted P(B | A) (read “probability of B, given A”)

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Example1: Two cards are selected in sequence from a standard deck. Find the probability that the second card is a queen, given that the first card is a king. (Assume that the king is not replaced.)

Because the first card is a king and is not replaced, the remaining deck has 51 cards, 4 of which are queens.

4( | ) (2 |1 ) 0.078

51nd stP B A P card is a Queen card is a King

Example2: The table shows the results of a study in which researchers examined a child’s IQ and the presence of a specific gene in the child. Find the probability that a child has a high IQ, given that the child has the gene.

Gene Present

Gene not present Total

High IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102

33( | ) ( | ) 0.458

72P B A P high IQ gene present

The Multiplication RuleMultiplication rule for the probability of A and B

• The probability that two events A and B will occur in sequence is P(A and B) = P(A) ∙ P(B | A)

• For independent events the rule can be simplified to P(A and B) = P(A) ∙ P(B) Can be extended for any number of independent events

Example1: Two cards are selected, from a deck without replacing the 1st card. Find the probability of selecting a king and then selecting a queen.

( ) ( ) ( | )

4 4

52 5116

0.0062652

P K and Q P K P Q K

Events are dependent:

Example2: A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6.

Events are independent:( 6) ( ) (6)

1 1

2 61

0.08312

P H and P H P

Example3: The probability that a particular knee surgery is successful is 0.85. Find the probability that three knee surgeries are successful.

P(3 successful surgeries) = (.85)(.85)(.85) ≈ 0.614

P(3 unsuccessful surgeries) = (.15)(.15)(.15)≈ 0.003P (at least 1 successful surgery) = 1 - .003 = .997

Events are independent:

3.3 Mutually Exclusive EventsMutually exclusive

• Two events A and B cannot occur at the same time

9

AB A B

A and B are mutually exclusive A and B are not mutually exclusive

Example:Event A: Rolling a 3 on a die Event B: Rolling a 4 on a die

Example:Event A: Randomly select a male student. Event B: Randomly select a nursing major

Addition rule for the probability of A or BThe probability that events A or B will occur is

P(A or B) = P(A) + P(B) – P(A and B)

For mutually exclusive events A and B, the rule can be simplified toP(A or B) = P(A) + P(B)Can be extended to any number of mutually exclusive events

Example: Select a card from a deck. Find probability it is a 4 or an ace. P (4 or Ace) = P(4) + P(Ace) = 4/52 + 4/52 = 8/52 = 2/13 = .154

Example: Roll a die. Find the probability of rolling less than 3 OR an odd number P (<3) + P(odd) – P(<3 and odd) = 2/6 + 3/6 – 1/6 = 4/6 = 2/3 = .667

Using the Addition RuleExample1: The frequency distribution shows the volume of sales (in dollars) and the number of months a sales representative reached each sales level during the past three years. If this sales pattern continues, what is the probability that the sales representative will sell between $75,000 and $124,999 next month?P(A) or P(B) = P(A) + P(B) = 7/36 + 9/36 = 16/36 = .444

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Sales volume ($) Months

0–24,999 3

25,000–49,999 5

50,000–74,999 6

75,000–99,999 7

100,000–124,999 9

125,000–149,999 2

150,000–174,999 3

175,000–199,999 1

AB

Example2: Find the probability the donor has type B or is Rh-negative.Type O Type A Type B Type AB Total

Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65

Total 184 164 45 16 409

P(Type B or RH-) = P(B) + P(RH-) – P(B and RH-) = 45/409 + 65/409 – 8/409 = 102/409 = .249

3.4 Counting (Permutations)

Larson/Farber

Determine the number of ways a group of objects can be arranged in order

A Permutation is an ordered arrangement of objects. The number of different

Permutations of n distinct objects is n! (n factorial) n! = n∙(n – 1)∙(n – 2)∙(n – 3)∙ ∙ ∙3∙2 ∙1 0! = 1 Examples:

• 6! = 6∙5∙4∙3∙2∙1 = 720

• 4! = 4∙3∙2∙1 = 24

Permutation of n objects taken r at a time

• The number of different permutations of n distinct objects taken r at a time

!

( )!n r

nP

n r

where r ≤ n

Example: Find the number of ways of forming three-digit codes in which no digit is repeated. (Select 3 digits from a group of 10.)

10 3

10! 10!

(10 3)! 7!

10 9 8 7 6 5 4 3 2 1

7 6 5 4 3 2 1

720 ways

P

N = 10, r = 3

3.4 Counting (Combinations)

Determine the number of ways to choose several objects from a group without regard to order

Combination of n objects taken r at a time

• A selection of r objects from a group of n objects without regard to order !

( )! !n r

nC

n r r

Example: A state’s department of transportation plans to develop a new section of interstate highway and receives 16 bids for the project. The state plans to hire four of the bidding companies. How many different combinations of four companies can be selected from the 16 bidding companies?• Select 4 companies

from a group of 16• n = 16, r = 4• Order is not important

16 4

16!

(16 4)!4!

16!

12!4!16 15 14 13 12!

12! 4 3 2 11820 different combinations

C

Ti 83/84

16 C 4 = 16 [Math][PRB][nCr][Enter] 4