1 / 71

# Probability & Statistics for P-8 Teachers - PowerPoint PPT Presentation

Probability & Statistics for P-8 Teachers. Chapter 4 Probability. Probability. Probability can be defined as the chance of an event occurring. It can be used to quantify what the “odds” are that a specific event will occur.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Probability & Statistics for P-8 Teachers' - hamish-medina

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Probability & Statistics for P-8 Teachers

Chapter 4

Probability

• Probability can be defined as the chance of an event occurring.

• It can be used to quantify what the “odds” are that a specific event will occur.

• Some examples of how probability is used everyday would be weather forecasting, “75% chance of snow” or for setting insurance rates.

• Probabilities must be between 0 and 1, inclusive.

• A probability of 0 indicates impossibility.

• A probability of 1 indicates certainty.

• A probability experiment is a chance process that leads to well-defined results called outcomes.

• An outcome is the result of a single trial of a probability experiment.

• A sample space is the set of all possible outcomes of a probability experiment.

• An event consists of outcomes.

Experiment

Sample Space

Toss a coin

Tail

Roll a die

1, 2, 3, 4, 5, 6

True, False

question

Toss two coins

HH,

HT, TH, TT

Find the sample space for rolling two dice.

36 different possible outcomes

Find the sample space for the outcomes of tossing 3 coins

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

8 different possible outcomes

Use a tree diagram to find the sample space for tossing three coins.

HHH

HHT

HTH

HTT

THH

THTTTHTTT

H

H

T

H

H

T

T

H

H

T

T

H

T

T

P- denotes a probability.

A, B, andC- denote specific events.

P(A)- denotes the probability of event A occurring.

Notation

P (E) =

number of possible outcomes

Sample Spaces and Probability

Probability of an event is the ratio of the outcome of the event to the sample space

Determine the probability of getting exactly two heads when tossing a coin three times

Sample Space

HHH THH

HHT THT

HTH THT

HTT TTT

8 possible outcomes

The probability of exactly two heads is 3/8.

If two dice are rolled one time, find the probability of getting a sum of 7.

P(sum of 7) = 6/36

P(A or B) The event “either A or B or both occur”

P(A & B) The event “both A and B occur”

P(not E) The event “E does not occur”

Union

Intersection

Complement

The complement of event Eis the set of outcomes in the sample space that are not included in the outcomes of event

P (not E) = 1 – P(E)

Find the complement of each event.

Event

Complement

of the Event

Rolling a die and getting a 4

Getting a 1, 2, 3, 5, or 6

Selecting a letter of the alphabet

Getting a consonant (assume y is a

and getting a vowel

consonant)

Selecting a month and getting a

Getting February, March, April, May,

month that begins with a J

August, September, October,

November, or December

Selecting a day of the week and

Getting Saturday or Sunday

getting a weekday

Use the complement rule to find the probability of rolling a dice and getting a number larger than 1.

P(1) =

1/6

P(>1) =

5/6

Complement Rule

P(>1) = 1 – P(1)

= 1 – 1/6

= 5/6

• The rule of additionapplies to the following situation:

• We have two events from the same sample space, and we want to know the probability that either event occurs.

P (A or B) = P(A) + P(B) – P(A and B)

A card is drawn randomly from a deck of ordinary playing cards. Find the probability of picking a heart or an ace.

We know the following:

• There are 52 cards in the deck.

• There are 13 hearts, so P(H) = 13/52.

• There are 4 aces, so P(A) = 4/52.

• There is 1 ace that is also a heart, so P(H and A) = 1/52.

• Therefore, based on the rule of addition:

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

P(H or A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13

• Two events are mutually exclusive events if they cannot occur at the same time

• no outcomes in common

• P(A and B) = 0

Determine which events are mutually exclusive and which are not, when a single die is rolled.

a. Getting an odd number and getting an even number

Getting an odd number: 1, 3, or 5

Getting an even number: 2, 4, or 6

Mutually Exclusive

Determine which events are mutually exclusive and which are not, when a single die is rolled.

b. Getting an odd number and getting a number less than 4

Getting an odd number: 1, 3, or 5

Getting a number less than 4: 1, 2, or 3

Not Mutually Exclusive

If two events are mutually exclusivethen the addition rule can be simplified since no intersection occurs:

0

P (A or B) = P(A) + P(B) – P(A and B)

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

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities.

a. A person has type O blood.

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities.

b. A person has type A or type B blood.

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities.

c. A person does not have type AB blood.

• Additional information or other events occurring may have an impact on the probability of an event.

• Conditional probability is the probability that the second event B occurs given that the first event A has occurred.

• Additional information may have an impact on the probability of an event.

• Example:

• Consider rolling a dice and getting a 6

• Sample space is {1, 2, 3, 4, 5, 6}

• P(rolling a 6) = 1/6

• Suppose we know an even number was rolled

• New sample space is {2, 4, 6}

• P(rolling a 6) = 1/3

P(B|A)

represents the probability of event B occurring after it is assumed that event A has already occurred. (read P(B|A) as “B given A”)

Conditional Probability

Let A = drawing a card less than five (2, 3, or 4)

Find P(B|A)

Conditional Probability

Given a card less than five

New sample space (12 cards)

P(B|A) = 3/12

Let A = drawing a card less than five (2, 3, or 4)

Find P(B|A) using the formula

Conditional Probability

P(A) = 12/52

P(A and B) = 3/52

3/52

=

= 3/12

12/52

The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket.

N= parking in a no-parking zone, T= getting a ticket

A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown.

Find the probability that the respondent answered yes (Y), given that the respondent was a female (F).

• We can create a multiplication rule for the intersection of two events A and B by rewriting the conditional probability formula

• Solving for the intersection yields:

P(A and B)

P(B|A) =

P(A)

P(A and B) = P(A) P(B|A)

or

P(A and B) = P(B) P(A|B)

• From the bag of marbles, find the probability of picking two red marbles from a bag with 4 red and 3 blue marbles.

• Pick a red marble from the first draw is 4/7

• Since a red has already been drawn, then 3 red marbles are left from the remaining 6 marbles

P(red and red) = P(red) P(red|red)

= (4/7) (3/6)

= 6/21

At a university in western Pennsylvania, there were 5 burglaries reported in 2003, 16 in 2004, and 32 in 2005. If a researcher wishes to select at random two burglaries to further investigate, find the probability that both will have occurred in 2004.

• Two events A and B are independent events if the fact that A occurs does not affect the probability of B occurring.

• Examples:

• Tossing a coin and rolling a die

• Choosing a 3 from a deck of cards, replacing it, and then choosing an ace as the second card

• If two events are independent, then the events do not affect the probability outcomes

• One way to represent this is by:

• since event A has no affect on event B regardless of whether A was given to us or not

P(B|A) = P(B)

If two events are independentthen the multiplication rule can be simplified since the conditional probability has no affect on the initial probability

P(B)

P (A and B) = P(A) P(B|A)

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

A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die.

This problem could be solved using sample space.

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

A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.

Determine if the following events, based on a standard deck of cards, are independent of each other

A = Drawing an ace

B = Drawing a heart

Show P(B|A) = P(B)

P(B) = 13/52 = 1/4

P(B|A) = 1/4

Since P(B|A) = P(B) = 1/4

then the events are independent

• Counting problems are of the following kind:

• “How many different 8-letter passwords are there?”

• “How many possible ways are there to pick 11 soccer players out of a 20-player team?”

• “What is the probability of winning the lottery?”

• Stan is about to order dinner at a restaurant. He has a choice of:

• two appetizers (soup or salad)

• three main courses (pasta, steak, or fish)

• two desserts (cake or pie)

• How many different meal combinations can Stan choose?

Appetizers

Main Courses

Desserts

cake

pie

pasta

steak

fish

cake

pie

soup

cake

pie

12

Combinations

cake

pie

pasta

steak

fish

cake

pie

cake

pie

• The fundamental counting rule is also called the multiplication of choices.

• If a task is made up of many stages, the total number of possibilities for the task is given by

• m × n× p× . . .

• where m is the number of choices for the first stage, and n is the number of choices for the second stage, p is the number of choices for the third stage, and so on.

A paint manufacturer wishes to manufacture several different paints. The categories include

Color: red, blue, white, black, green, brown, yellow

Type: latex, oil

Texture: flat, semigloss, high gloss

Use: outdoor, indoor

How many different kinds of paint can be made if you can select one color, one type, one texture, and one use?

# of

Colors

# of

Types

# of

Textures

# of

Uses

×

×

×

7

×

2

×

3

×

2

= 84

84 different kinds of paint

Colleen has six blouses, four skirts and four sweaters.

How many different outfits can she choose from, assuming she wears three different items at once?

Blouses

Skirts

Sweaters

= 96 outfits

4

×

6

4

×

The final score in a soccer game is 5 to 4 for Team A.

How many different half-time scores are possible?

Team B

(0 - 4)

Team A

(0 - 5)

= 30 different possible half-time scores.

6

×

5

How many four-letter arrangements are possible?

use any

of the 26

letters:

A - Z

26

1st

26

2nd

26

3rd

26

4th

= 456,976

×

×

×

How many four-letter arrangements are possible if a letter cannot be repeated?

use any

of the 26

letters:

A - Z

26

1st

25

2nd

24

3rd

23

4th

= 358,800

×

×

×

only 25 letters left

only 24 letters left

only 23 letters left

Apermutation is an arrangement of a set of objects

for which the order of the objects is important.

Show the number of ways of arranging the letters of the word CAT:

CAT

CTA

ACT

ATC

TCA

TAC

There are six arrangements or permutations of the word CAT. By changing the order of the letters, you have a different permutation.

Using the Fundamental Counting Principle, we would have 3 x 2 x 1 number of distinct arrangements or permutations.

In calculating permutations, we often encounter expressions such as 3 x 2 x 1. The product of consecutive natural numbers in decreasing order down to the number one can be represented using factorial notation:

• Factorial is the product of all the positive numbers from 1 to a number.

3 x 2 x 1 = 3!

This is read as “three factorial”.

4! = 4 × 3 × 2 × 1 = 24

10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 3,628,800

= 720

• Permutation is an arrangement of objects in a specific order. Order matters.

In general, if we have n objects but only want to select r objects at a time

Suppose a business owner has a choice of 5 locations in which to establish her business. She decides to rank each location according to certain criteria, such as price of the store and parking facilities. How many different ways can she rank the 5 locations?

Using factorials, 5! = 120.

Using permutations, 5P5 = 120.

Suppose the business owner wishes to rank only the top 3 of the 5 locations. How many different ways can she rank them?

Using permutations, 5P3 = 60.

How many three-letter permutations can be formed from the letters of the word DINOSAUR?

the number of permutations of eight objects

taken three at a time.

This may be written as 8P3.

8!

(8–3)!

8!

5!

8P3 =

=

=

8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

5 × 4 × 3 × 2 × 1

= 336

Find the following permutations

6!

4!

6P2 =

= 6 × 5 = 30

7!

3!

7P4 =

= 7 × 6 × 5 × 4 = 840

3!

0!

3P3 =

= 3 × 2 × 1 = 6

Note:nPn= n!

A school musical director can select 2 musical plays to present next year. One will be presented in the fall, and one will be presented in the spring. If she has 9 to pick from, how many different possibilities are there?

Order matters, so we will use permutations.

How many four-letter words can be made using

the letters of PEER?

4P4 = 4!

Note that PEER and PEER are not distinguishable,

therefore, they count as one permutation.

When this happens, you must divide by the factorial of

the number of repeated terms.

For PEER, the number of arrangements is:

4!

2!

4P4

2!

= 4 × 3 = 12

“E” occurs twice

=

• Consider selecting two people out of three individuals (A, B, C) for President and Vice-President of a committee

• 3P2 = 6

• Now consider selecting two people out of three individuals (A, B, C) for committee membership

• Here choice A,B is the same committee as choice B,A

A B

B A

A C

C A

B C

C B

6

choices

A B

A C

B C

3

choices

• Combination is a grouping of objects. Order does not matter.

When order matters, you have permutations.

When order does not matter, you have combinations

A committee of four students is to be selected from a group of ten students. In how many ways can this be done?

10C4

= 210

The committee of four can be

selected in 210 ways.

A newspaper editor has received 8 books to review. He decides that he can use 3 reviews in his newspaper. How many different ways can these 3 reviews be selected?

The placement in the newspaper is not mentioned, so order does not matter. We will use combinations.

A company is hiring people to fill five identical positions.

There are 12 applicants. How many ways can the five positions be filled?

12C5

The company can fill the five

positions 792 ways.

= 792

The company wants to hire Applicant A and any four of the others. How many ways can the five positions be filled now?

With Applicant A, there are 330 ways to fill the position

1C1 x 11C4

= 330

There are seven books to choose from.

How many ways can five or more books be selected?

Select 5 or 6 or 7:

7C5 + 7C6 + 7C7

= 21 + 7 + 1

= 29

There are 29 ways to select 5 or more books.

In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?

There are not separate roles listed for each committee member, so order does not matter. We will use combinations.

There are 35·10=350 different possibilities.

How many five-card hands can be dealt from a standard deck of 52 cards if:

a) each hand must contain two aces?

b) each hand must contain three red cards?

4C2 x 48C3 = 103 776

26C3 x 26C2 = 845 000

The counting rules can be combined with the probability rules in this chapter to solve

many types of probability problems.

By using the fundamental counting rule, the permutation rules, and the combination rule, you can compute the probability of outcomes of many experiments, such as getting a full house when 5 cards are dealt or selecting a committee of 3 women and 2 men from a club consisting of 10 women and 10 men.

Two cards are picked at random from a deck of playing cards.

What is the probability that they are both kings?

Method 1:Using the fundamental counting principle

P(2 kings) =

Method 2:Using combinations

Number of ways of choosing 2 kings from 4 kings is 4C2.

Number of ways of choosing 2 cards from 52 cards is 52C2.

P(2 kings) =

A store has 6 TV Graphic magazines and 8 Newstimemagazines on the counter. If two customers purchased a magazine, find the probability that one of each magazine was purchased.

TV Graphic: One magazine of the 6 magazines

Newstime: One magazine of the 8 magazines

Total: Two magazines of the 14 magazines

A combination lock consists of 50 numbers (1 through 50). If a 3-number combination is needed, find the probability that the combination is the number 1-2-3 in that order. The same number can be used more than once. (Note: A combination lock is really a permutation lock.)

There are 50·50·50 = 125,000 possible combinations.

The number 1-2-3 is one combination.

P(1-2-3) = 1/125,000