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Chapter 6 Section 2

Vocabulary Of Probability

Sample Space, the set of all possible outcomes

Identify the sample space for flipping a coin, rolling two dice, and

rolling one die then flipping a coin

Event, any outcome or set of outcomes of a random

phenomenon. An event is a subset of the sample space.  Example: Heads (event) then roll a number on a die…

Probability model, mathematical description of a random

phenomenon consisting of two parts:  Sample space

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Counting Principle/Multiplication Principle

Helping to determine sample space

Suppose that two events occur in order. If the first can occur in

“m” ways and the second in “n” ways (after the first has

occurred), then the two events can occur in order in m·n ways.

(3)

Tree Diagrams/Listed

Helping to determine sample space

Tree Diagram: Listed:

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Example:

At Brusters Ice-Cream you can get a regular

cone, a sugar cone, or a waffle cone. If

Brusters has 25 flavors how many one scoop ice-cream cones can you get?

How many two scoop cones can you get

(assume that the second scoop must be different)

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Example:

In NC, automobile license plates display 3 letters followed by

four digits. How many license plates can be produced if repetition of letters and numbers are allowed? – known as replacement

___ ___ ___ - ____ ____ ___ ___

What if repetition is not allowed? – known as without

replacement

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Example:

At the Olympics 8 runners race in the 100 m dash. In how

many ways can they finish?

___ ___ ___ ___ ___ ___ ___ ___8·7·6·5·4·3·2·1

How else could we write this?8! Factorial

Where is Factorial on your calculator??

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Counting Principle Examples:

A restaurant offers 6 main course, eight

beverages and 3 desserts, how many different

dinner options do they have?

A red die, a blue die and a white die are rolled,

how many different outcomes are possible?

A company has 2844 employees. Each employee

is to be given an ID number that consist of one

letter followed by two digits. Is it possible that

each employee has an individual ID?

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Deck Of Cards

4 suits

2 Black Suits (Clubs and Spades)2 Red Suits (Diamonds and Hearts)52 Cards

Each Suit (13 cards): Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, King,

and Queen.

Face Cards: Jack, King and Queen

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Deck Of Cards Example

Two cards are chosen in order from a deck of cards. In how

many ways can this be done if the first card is red and the second card is a club?

What if both cards are red?

What if both cards are Kings?

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Probability Rules

All probabilities are numbers between 0 and 1

0 < P(A) < 1

All possible outcomes together must have a probability of 1

P(S) = 1

The probability that an event does not occur (called the

complement) is 1 minus the probability that the event does occur

P(Ac) = 1 – P(A)

If two events have no outcomes in common, the probability that

one or the other occurs is the sum of their individual probabilities

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

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Probability Rules

Mutually Exclusive (disjoint), If two events have no

outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities

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

Mutually Inclusive, If two events have an outcome(s) in common, the probability that one or the other occurs is the sum of their individual probabilities minus the overlapping probability.

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

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Venn Diagrams and Probability

The complement AC contains exactly the outcomes that are not in A.

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Suppose we choose a student at random. Find the probability that the student

(a) has pierced ears.

(b) is a male with pierced ears.

(c) is a male or has pierced ears.

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Chapter 6 Section 2

(b) We want to find P(male and pierced ears), that is, P(A and B). Look at the intersection of the “Male” row and “Yes” column. There are 19 males with pierced ears. So, P(A and B) = 19/178.

(c) We want to find P(male or pierced ears), that is, P(A or B). There are 90 males in the class and 103 individuals with pierced ears. However, 19

males have pierced ears – don’t count them twice!

P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178

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Two-Way Tables and Probability

Note, the fact that we can’t use the addition rule for mutually exclusive events unless the events have no outcomes in common.

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Venn Diagrams and Probability

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Example

A person is selected at random

What is probability that they got a C?

What is the probability that they didn’t get a B?

What is the probability that they got an A or a B?

Grade F D C B A

Probabilit y

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Roll dice

P(roll a sum of 4)

P(roll two identical numbers)P(two and three)

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Benford’s Law

 P(1st digit is a 2)

 P(1st digit is 7 or greater)

 P(1st digit is a 4 and second digit is less than 3)

 P(1st digit is not a 2)

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Multiplicative Rule For Independent Events

Two events A and B are independent if one event does not

change the probability that the other will occur.

If A and B are independent then:P(A and B) = P(A)P(B)

P(A∩B) = P(A)P(B)

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Picking Cards

If two cards are drawn, what is the probability of drawing two

Aces?

If two cards are drawn, what is the probability of drawing a

heart and a spade?

A five-card hand is drawn from a standard deck of cards.

(22)

Flash Drive

Problems

A collection of 38 flash drives contains 4 that have a

virus. If 2 are selected at random, what is the probability that both are good?

References

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