Aim: Fundamental Counting Principle Course: Alg. 2 & Trig. Do Now: Choose one item from each...

Aim: Fundamental Counting Principle Course: Alg. 2 & Trig.

Do Now: Choose one item from each category to make an entire meal.

Aim: Probability! Probability! - How do I count the ways?

Main Course Drink Dessert

Spaghetti Milk Ice Cream

Hamburger Coke Apple Pie

Hotdog Chocolate Cake


Dinner is Served!

Spaghetti

Hamburger

Hotdog

Main Course Drink

Milk

Coke

DessertIce CreamApple PieChocolate Cake

Ice CreamApple PieChocolate Cake

S M IS M AS M C

S C IS C AS C C





Milk

Coke

Milk

Coke

H M A

Ht M IHt M AHt M C

H M I

H M C

H C IH C AH C C

Ht C IHt C AHt C C

Sample Space

Three consecutive Events


Am I lost?

MJ Petrides

Outerbridge Crossing

Great Adventure

How many different ways will get us from MJ Petrides to Great Adventure?

Tracing the different routes we find there are 6 different routes.

Is there a shortcut method for finding how many different routes there are?

3

2


The Fundamental Counting Principle

To find the total number of possible outcomes in a sample space, multiply the number of choices for each stage or event...

in other words...

If event M can occur in m ways and is followed by event N that can occur in n ways, then the event M followed by event N can occur in m · n ways.

Main Idea

Counting Principle 2 events: m · n 3 events: m · n · o 4 events: m · n · o · p 5 events: etc.

P En En S

( )( )( )


18x x =

Spaghetti

Hamburger

Hotdog

Main Course Drink

Milk

Coke

DessertIce CreamApple PieChocolate Cake


S M IS M AS M C

S C IS C AS C C





Milk

Coke

Milk

Coke

H M A

Ht M IHt M AHt M C

H M I

H M C

H C IH C AH C C

Ht C IHt C AHt C C

Sample Space

3 2 3


Model Problem

Jamie has 3 skirts - 1 blue, 1 yellow, and 1 red. She has 4 blouses - 1 yellow, 1 white, 1 tan and 1 striped. How many skirt-blouse outfits can she choose? What is the probability she will chose the blue skirt and white blouse?

Skirt blouse3 4

Blue

Yellow

Red

yellow white tan stripedyellow white tan striped

yellow white tan striped

12 outcomes in sample

spaceb y b w b t b sy y y w y t b y

r y r w r t r y

P En En S

( )( )( )

1( , )

12P b w


Regents Question

A four-digit serial number is to be created from the digits 0 through 9. How many of these serial numbers can be created if 0 can not be the first digit, no digit may be repeated, and the last digit must be 5?

1) 448 2) 2240 3) 504 4) 2,520

possible outcomes

E1 E2 E3 E4

8 8 7 1 = 448

0, 1, 2, 3, 4, 5, 6, 7, 8, 9 10 possible outcomes to start


Types of Events

Compound Event - two or more activities

Ex. Rolling a pair of dice What is the probability of rolling a pair of dice and getting a total of four?

6

5

4

3 2 1

Die 1

2 3 4 5 6 73 4 5 6 7 84 5 6 7 8 95 6 7 8 9 106 7 8 9 10 117 8 9 10 11 12

1 2 3 4 5 6Die 2

P(4) = 3/36 = 1/12


Types of Events

Mutually Exclusive Events – Two or more events that can not happen at the same time

Ex. mutually exclusive?

rolling a 2 and a 3 on a die

rolling an even number or a multiple of 3 on a die

Yes

No

Independent Event – when the outcome of one event does not affect the outcome of a second event.

Dependent Event - when the outcome of one event affects the outcome of a second event.


Probability of Two Independent Events

The probability of two independent events can be found by multiplying the probability of the first event by the probability of the second event.

(The Counting Principle w/Probabilities)

The probability of two independent events can be found by multiplying the probability of the first event by the probability of the second event.

(The Counting Principle w/Probabilities)

P(A and B) = P(A) · P(B) P(A and B) = P(A) · P(B)

Ex: A die is tossed and a spinner is spun.

What’s the probability of throwing a 5 and spinning red? P(5 and R)?

P ( )516

P red( ) 14

Event A Event B

P(5 and Red) =16

14

124

Faster than drawing a tree diagram!! Independent Event

AND Probabilities AND Probabilities


Not So Independent!

There are 4 red, 3 pink, 2 green and 1 blue chips in a bag. What is P(pink)? 3/10What is the probability of picking a pink and then reaching in and picking a second pink without replacing the first picked pink?

ONLY IF THE FIRST PINK CHIP WAS NOT RETURNED TO THE BAG.

The selection of the second event was affectedby the selection of the first.

Dependent Event

P(pink)

Event A Event B

P(pink)

3/10 2/9

P(pink and pink) =3 2 6 1

10 9 90 15


Model Problem

P(pink)

Event A

3/10

Event B

P(pink)

2/9

Dependent Events

P(pink, pink) =

Find the probability of choosing two pink chips without replacement.

3/10 • 2/9 = 6/90 or 1/15

Counting Principle w/Probabilities


Model Problem

P(blue)

Event A

1/10

Event B

P(red)

4/9 1/10 • 4/9 = 4/90 or 2/45

Find the probability of choosing blue and then a red chip without replacement.

Dependent Events

P(blue, red) =



Dependent Events

Two events are dependent events if the occurrence of one of them has an effect on the probability of the other.

AND Probabilities with Dependent EventsIf A and B are dependent events, thenP(A and B) = P(A) · P(B given that A has

occurred)extends to multiple dependent events

You are dealt three cards from a 52-card deck. Find the probability of getting 3 hearts.

P(1st heart) = 13/52

P(2nd heart) = 12/51

P(3rd heart) = 11/50

P(hearts) = 13/52 · 12/51 · 11/50 = 1716/162600 0.0129


Probability of Dependent Events

1. Calculate the probability of the first event. 2. Calculate the probability of the second event, etc. ... but NOTE:

The sample space for the probability of the subsequent event is reduced because of the previous

events.3. Multiply the the probabilities.

Ex. A bag contains 3 marbles, 2 black and one white. Select one marble and then, without replacing it in the bag, select a second marble. What is the probability of selecting first a black and then a white marble?

Event A Event B

P(A) = 23 P(B) =

12

P(Black 1st, White 2nd = 23

12

26

13

Key words - without replacement


Model Problem

20

20

10 10

10

50

10

From a deck of 10 cards (5 ten-point cards, 3 twenty-point cards, and 2 fifty-point cards), Ronnie can only pick 2 cards. In order to win the game, he must pick the 2 fifty-point cards. What is the probability that he will win?

10 10 10 10 10 20 20 20 50 50


Model Problem

20

20

10 10

10

50

10

From a deck of 10 cards Ronnie can only pick 2 cards. In order to win the game, he must pick the 2 fifty-point cards. What is the probability that he will win?

10 10 10 10 10 20 20 20

Dependent

P(50, 50) =

2/10 • 1/9 = 2/90 = 1/45


Event A Event B

P(50) =

2/10

P(50) =

1/9

5050


Regents Question

Penny has 3 boxes, each containing 10 colored balls. The first box contains 1 red ball and 9 white balls, the second box contains 3 red balls and 7 white balls, and the third box contains 7 red balls and 3 white balls. Penny pulls 1 ball out of each box.

Box 1 Box 2 Box 3

A. What is the probability that Penny pulled 3 red balls?

P(r,r,r) = 1/10 • 3/10 • 7/10 = 21/1000

B. If Penny pulled 3 white balls and did not replace them, what is the probability that she will now pull 3 red balls?

P(r,r,r) = 1/9 • 3/9 • 7/9 = 21/729


Venn Diagrams

Find the probability of rolling a die and getting a number that is both odd and greater than 2.

24

6

1 3

5

4

6

3

5

1 3

5

odd > 2

P(odd) = 3/6 P(> 2) = 4/6

P(odd > 2) = 2/6

In logic, a sentence p and q, written p q, is true only when p is true and q is true.


Probability of A and B

Probability of (A B)

(A and B are separate events)

In probability, an outcome is in event (A and B) only when the outcome is in event A and the outcome is also in event B.

)(

)()(

Sn

BandAnBandAP

)(

)()(

Sn

BAnBAP

Example: Find the probability of rolling a die and getting a number that is both odd and greater than 2.

}5,3,1{,6

3

)(

)()(

Sn

EnoddP

}6,5,4,3{,6

4

)(

)()2(

Sn

EnP

6

2)2( oddP

{3, 5}


Mutually Exclusive Events

Mutually exclusive – two events A & B are mutually exclusive if they can not occur at the same time. That is, A and B are mutually exclusive when A B =

An outcome for A or B is in one or the other. If the events are mutually exclusive then

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

If one card is randomly selected from a deck of cards, what is the probability of selecting a king or a queen? mutually

exclusive?yes

( ) ( ) ( )

4 4 8 2

52 52 52 13

P king or queen P k P q


‘Or’ Probabilities Not Mutually Exclusive

From a standard deck you randomly select one card. What is the probability of selecting a diamond or a face card? mutually

exclusive?no

13( )

52P diamonds

12( )

52P facecards

common elements A B

n(A B) = 3 {K, Q, J}

P(or fcd) = 13 12 25

52 52 52

13 12 3 22

52 52 52 52


Mutually Exclusive or Not

1. A card is drawn from a standard deck of 52. Find P(king or queen)

P(king) = P(queen) =

P(king or queen) =

2. A card is drawn from a standard deck of 52. Find P(king or face card)

P(king) = P(face) =

P(king or face) =

In #1 the 2 events have no common elements. They are mutually exclusive. In #2 a card can be both face and king. They are not mutually exclusive.

4

52

4

524 4 8 2

52 52 52 13

mutually exclusive

not mutually exclusive

4

52

12

524 12 4 8 2

52 52 52 52 13


Probability of A or B

What is the probability of spinning a number greater than 8 or an odd number?

1211

10

9

8

7 6

5

4

3

21 Count the number of

successes for

n > 8

n - odd

9, 10, 11, 12 4

4 6 2 8 2( 8 or odd)

12 12 12 12 3P

61, 3, 5, 7, 9, 11

10(greater than 8 or odd) ( 8) (odd) =

12P P P



Probability of (A or B)

Example: Find the probability of rolling a die and getting a number that is odd or greater than 2.

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

( ) 3(odd)

( ) 6

n EP

n S

( ) 4( 2)

( ) 6

n EP

n S

successes

(odd) ( 2) (odd >2)(odd >2)

( ) ( ) ( )

n n nP

n S n S n S

3 4 2 5(odd >2)

6 6 6 6P

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

P(A B) = P(A) + P(B) - P(A B)

P(A B) = n(A) + n(B) - n(A B) n(S) n(S) n(S)

If A and B are not mutually exclusive events, then


Probability Rules1. The probability of an impossible event is 0.

2. The probability of an event that is certain to occur is 1.

3. The probability of an event E must be greater than or equal to 0 and less that or equal to 1.

4. P(A and B) = n(A B) n(S)

5. P(A or B) = P(A) + P(B) - P(A B)

6. P(Not A) = 1 - P(A)

7. The probability of any even is equal to the sum of the probabilities of the singleton outcomes in the event.

8. The sum of the probabilities of all possible singleton outcomes for any sample space must always equal 1.


Model Problems

In drawing a card from the deck at random, find the probability that the card is:

A. A red king

B. A 10 or an ace

C. A jack or a club

A red king must be red and a king

P(red and king) = 252

There are 4 jacks and 13 clubs, but one of the cards is both (jack of clubs)

P(jacks or clubs) = 452

1352

+ 152

_ 1652=

10’s and aces have no common outcomes

P(10’s or aces) = 452

452

+ 052

_ = 852

P(A and B) = P(A) · P(B)

26 4 1

52 52 26 g

P(A B) = P(A) + P(B) - P(A B)

P(A B) = P(A) + P(B) - P(A B)

mutually exclusive



Model Problems

Based on the table below, if one person is randomly selected from the US military, find the probability that this person is in the Army or is a woman.

Air Force

Army Marines Navy Total

Male 290 400 160 320 1170

Female 70 70 10 50 200

Total 360 470 170 370 1370

Active Duty US Military Personnel, in 000’s

P(A B) = P(A) + P(B) - P(A B)

P(Army Female) = P(A) + P(F) - P(A F)470

1370

200

1370

70

1370

60

137



Model Problems

Five more men than women are riding a bus as passengers. The probability that a man will be the first passenger to leave the bus is 2/3. How many passengers on the bus are men, and how many are women?

x = number of womenx + 5 = number of men

2x + 5 = number of passengers

P(man) = Number of menNumber of passengers

x + 52x + 5

23

= 4x + 10 = 3x + 15x = 5

x + 5 = 10

510


Model Problem

A special family has had nine girls in a row. Find the probability of this occurrence.

9

1 1 1 1 1 1 1 1 1(nine girls in a row) =

2 2 2 2 2 2 2 2 2

1 1

2 512

P

g g g g g g g g

Having a girl is an independent event with P(1 girl) = 1/2

P(A and B) = P(A) · P(B)

Probability of two Independent Events

extends to multiple independent events


Model Problem

If the probability that South Florida will be hit by a hurricane in any single year is 5/19

a) What is the probability that S. Florida will be hit by a hurricane in three consecutive years?

b) What is the probability that S. Florida will not be hit by a hurricane in the next ten years?

35 5 5 5 125

(hurricane - 3) = 0.01819 19 19 19 6859

P

(no hurricane) = 1 (hurricane)

5 141 0.737

19 19

P P

10

1014(no hurricane 10 yrs) = 0.737 0.047

19P

• The probability of event (A) plus the probability of "not A” or ~A, equals 1:

P(A) + P(~A) = 1; P(A) = 1 – P(~A); P(~A) = 1 – P(A)


Model Problem

Three people are randomly selected, one person at a time, from 5 freshman, two sophomores, and four juniors. Find the probability that the first two people selected are freshmen and the third is a junior.

P(1st selection is freshman) = 5/11

P(2nd selection is freshman) = 4/10

P(3rd selection is junior) = 4/9

P(F, F, J) = 5/11 · 4/10 · 4/9 = 8/99


Model Problems

A sack contains red marbles and green marbles. If one marble is drawn at random, the probability that it is red is 3/4. Five red marbles are removed from the sack. Now, if one marble is drawn, the probability that it is red is 2/3. How many red and how many green marbles were in the sack at the start?

x = original red marblesy = original number of green marbles

x__ x + y

34 =

515

x - 5 x + y - 5

23 =

3x + 3y = 4x 2x + 2y - 10 = 3x - 15 3y = x 2y + 5 = x

3y = 2y + 5 y = 5

3y = x = 15


Several players start playing a game with a full deck of 52 cards. Each player draws two cards at random, one at a time, from the full deck. Find the probability that a player does not draw a pair.

1

1

1st card48

51

2nd card48

51g

Model Problem


Model Problem

Find the probability of rolling a die and getting a number that is odd or greater than 2.

24

6

1 3

5

4

6

3

5

1 3

5

odd > 2

P(odd) = 3/6 P(> 2) = 4/6

A B = {1, 3, 4, 5, 6}

n(A B) = 5

( ) 5(> 2 or odd)

( ) 6

n A BP

n U

n(U) = 6


Model Problem

In a group of 50 students, 23 take math, 11 take psychology, and 7 take both. If one student is

selected at random, find the probability that the student takes math or psychology

23( )

50P M

11( )

50P Psy

7( )

50P M Psy

P(A B) = P(A) + P(B) - P(A B)

23 11 7 27( )

50 50 50 50P M Psy

23416 7

M Psy


Determine the number of outcomes:

4 coins are tossed

A die is rolled and a coin is tossed

A tennis club has 15 members: 8 women and seven men. How many different teams may be formed consisting of one woman and one man on each team?

A state issues license plates consisting of letters and numbers. There are 26 letters, and the letters may be repeated on a plate; there are 10 digits, and the digits may be repeated. The how many possible license plates the state may issue when a license consists of: 2 letters, followed by 3 numbers, 2 numbers followed by 3 letters.


One bag contains 3 red and 4 white balls. A 2nd bag contains 6 yellow and 3 green balls. One ball is drawn from each bag. Find the probability of choosing a red and yellow ball.

Model Problem


The Product Rule

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Aim: Fundamental Counting Principle Course: Alg. 2 & Trig. Do Now: Choose one item from each...

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