1 1 presented by e. g. gascon introduction to probability section 7.3, 7.4, 7.5

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PRESENTED BY E. G. GASCON

Introduction to Probability Section 7.3, 7.4, 7.5

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VOCABULARY & EXAMPLES

Introduction to Probability Section 7.3

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Sample Space

The set of ALL possible outcomes for an experiment is the sample space. (This will make up the denominator of a probability)

Probability range is 0 ≤ P(E) ≤ 1

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Event

An event is a subset of a sample space defined by the problem (the numerator)

Complement of an Event• Is the set of ALL outcomes in a sample

space that are not included in the event. Denoted by E’

Example: If the event is throwing an 6 on a dice, then the complement is throwing 1,2,3,4,or 5.

So… P(less than 6) = 1 – P(6)

This is sometimes an easier way to find a difficult probability

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Conditional Probability

The Probability of an event occurring, given that another event has already occurred. Denoted by

P(B|A), read Probability of B given A. Ex P( Jack of Diamonds) P(Jack | Diamonds) There 1 jack of diamonds P(Jack of Diamonds) = 1/52

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Independent Events

Two event are independent if the occurrence of one of the events does NOT affect the probability of the occurrence of the other

Use Conditional probability to determineP(B|A) = P(B) or P(A|B) = P(A) then they are

independent

Example of Independent Events: Draw two cards WITH replacement.If the first card is a heart, and one puts it back in the deck, then if the second card drawn is red. The probability of drawing a red card does not change because a heart was drawn first .

Example of Dependent Events: Draw two cards WITHOUT replacement.If the first card is a heart, then if the second card drawn is red. The probability of drawing a red card changes because a heart was drawn first, therefore there is one less card in the deck, and that heart might have been red.

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Mutually Exclusive Events

Events E and F are mutually exclusive events if E F = , meaning that they have no elements in common.

One of these statements will be true:

•Event A and B cannot occur at the same time

•Event A and B have zero outcome in common

•P(A AND B) = 0

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Basic Probability Principle

Let S be a sample space of equally likely outcomes, and let event E be a subset of S. Then the probability that event E occurs is

( )( )

( )

n EP E

n S note: 0 P(E) 1

Sample space

Event

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Intersection Rule for Probability (Multiplication Rule)

Two events A AND B occur in sequence P(A and B) = P(A) * P(B|A)

SPECIAL CASE: If the events are INDEPENDENT P(A and B) = P(A) * P(B)

Take an extra minute to study this slideIt is important when faced with an AND problem to decide if the two events are

independent (slide # 6).

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How to Solve an AND problem

1. Identify the events A and B in sequence2. Decide whether the events are independent

or dependent3. Find P(A), P(B), or if necessary P(B|A)4. Use the appropriate Intersection rule

(Multiplication rule)

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Example AND Problem

Two cards are selected at random “without replacement,” find the probability that a King , AND then a Queen is selected.

The events are a King then a Queen (Dependent Events)

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52P(K) =

P(Q |K) =4

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16.006

2652

4

514

52 *

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Example Special Case

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

The events are a Heads then a “2” (Independent Events)

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2P(H) =

P(“2”) =1

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1.083

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1

61

2 *

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OR

For Mathematical purposes OR meansINCLUSIVE “OR”Three ways for event A OR B to occur

A occurs and B does not B occurs and A does not A and B both occur

Ex: In a group, the number wearing Red shirts OR Green Pants will be : # of Red shirts with not green pants + # of green

pants with not red shirts + number of Red shirts and Green pants.

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Union Rule for Probability (Addition Rule)

For ANY events E OR F from a sample space SP(E F) = P(E) + P(F) – P(E F)

Special Case:For mutually exclusive events E AND F from a

sample space S, P(E F) = P(E) + P(F) (as in examples of section

7.3)Take an extra minute to study this slideIt is important when faced with an OR problem to decide if the two events are

mutually exclusive. (slide # 7)

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How to Solve an OR problem

1. Identify the events A and B2. Decide whether the events are mutually

exclusive3. Find P(A), P(B), and if necessary P(A and B)4. Use the appropriate Union rule (ADDITION

rule)

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Example: Special Case

Probability of drawing a spade OR heart.

Probability of drawing a spade.( ) 13

( )( ) 52

n SpadesP Spade

n Cards in Deck

Probability of drawing a Heart.( ) 13

( )( ) 52

n HeartP Heart

n Cards in Deck

Probability of drawing a spade OR heart.( )

( ) ( )

( ) ( )

13 13 26 1

52 52 52 2

P Spade OR Heart

n Spade n Hearts

n Cards in Deck n Cards in Deck

Note: These are mutually Exclusive events.

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Example Probability of drawing a red face card.

Probability of drawing a red card.(Re ) 26(Re )

( ) 52

n dCardP dCard

n Cards in Deck

Probability of drawing a face card.( ) 12

( )( ) 52

n FaceCardP FaceCard

n Cards in Deck

BUT, these two events are NOT mutually exclusive...so… there is an extra step

Probability of drawing a red face card.

(Re | ) 6(Re | )

( ) 52

n d FaceCardP d FaceCard

n Cards in Deck

There are 26 red cards

There are 12 face cards

There are 6 red –face cards

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Solution to Events NOT mutually exclusive

Probability of drawing a red OR face card.( )

(Re ) ( ) ( )

( ) ( ) ( )

26 12 6 32 8

52 52 52 52 13

P red OR FaceCard

n d n FaceCard n redFaceCards

n Cards in Deck n Cards in Deck n Cards in Deck

Note: The difference between this problem and the example on slide 13

On the previous slide #13 was a problem that were mutually exclusive, and this one they were not.

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Complement Rule

IF you know the probability that event E occurs, then the probability that event E does NOT occur is:

P(E’) =1 − P(E)

Use this rule to save steps.

To find the probability of the roll of two fair dice yields a sum > 3, one could find the P(4) OR P(5) OR P(6) OR P(7) OR P(8) OR P(9) OR P(10) OR P(11) OR P(12)

BUT it is easierFind the P( sum 3) = P(1) OR P(2) OR P(3) = 0 + (1/36) + (2/36) = (3/36) = 1/12so… P( sum >3) = 1 − P(sum 3) = 1 − (1/12) = 11/12

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Odds

The odds in favor of an event E are defined as the ratio of P(E) to P(E’)

The odds that the roll of two fair dice yields a sum > 3 is…P(sum >3) : P(sum 3) =

112 1:111112

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Sometimes it is not easy to

logically figure the

conditional probability,

so we have a formula

Conditional Probability

P(E F)P E | F

( )P F

The conditional probability of event E given event F:

Also it can be stated that by re-writing the equation:

P(E F) ( ) P E | F

P(E F) ( ) P F | E

P F

or

P E

Notice that this

formula is a

derivation from the

AND rule.

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Example 7.5 # 17The Problem

Numbers 1,2,3,4,and 5 written on slips of paper, and 2 slips are drawn at random one at a time WITHOUT replacement.

Find the probability that the sum is 8, given the first number is 5.

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

P("5" "sum 8")P sum 8 | 5

("5")P

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Solution

Each number as a 1/5 chance of being drawn for the first slip of paper .

Each number that is left has a ¼ chance of being drawn for the second slip of paper.

There is only 1 way to get a “5” AND “sum of 8”, (5 + 3).

therefore…1 1 1

("5" " 8") 15 4 20

P AND sum

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Solutions cont.

1(5)

5P

Putting it all together:

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1 5 5 120P sum 8 | 5 1 20 1 20 45

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Visual Solutions

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3

4

5

First Slip drawnSecond Slip drawn

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243

5

5

3

1 245

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1

4

5

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132

This is the only scenario which has a sum of 8, and there is only one way to get it, therefore the probability of a sum of 8 given 5 on the first draw is 1/4

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Questions / Comments / Suggestions

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1 1 presented by e. g. gascon introduction to probability section 7.3, 7.4, 7.5

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