Chapter 6 Section 3.pptx

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

26 x 26 x 26 – 10 x 10 x10 x 10 = 175,760,000

 _{What if repetition is not allowed? – known as without}

replacement

26 x 25 x 24 – 10 x 9 x 8 x 7 = 78,624,000

Basic

Probability

 _{Sample Space: A list of all possible outcomes of a given experiment.}

a. Tossing a coin b. Rolling a six sided die

c. Drawing a marble from a bag containing two red, three blue, and one white marble

Heads,

Tails

1, 2, 3,

4, 5, 6

R, R,

B, B, B,

Intersection

of two sets

(A



B):

all the elements that appear in both sets

Example:

Given set A: {3,4,5,6,7}, and set

B: {5,6,7,8,9,10}, find (A



B).

(A



B) =

{5,6,7}

Union

of two sets

(A



B):

Everything

in both sets

Example:

Given set A: {3,4,5} and set

B: {5,6,7}, find (A



B).

(A



B) =

{3,4,5,6,7}

You TRY

Example:

Given the following sets, find

A



B and A



B.

A = {1, 3, 5, 7, 9, 11, 13, 15}

B = {0,3,6,9,12,15}

A



B =

A



B =

{3, 9, 15}

Example: Use the Venn Diagram to answer the following questions. Let A = Factors of 12 and B = Factors of 16:

1.What are the elements of set A?

2.What are the elements of set B?

3.Why are 1, 2, and 4 in both sets?

4. What is A  B?

5. What is A  B? Factors of 12

1 4 3 6 12

2 Factors of 16 8 16

A = {1, 2, 4, 3, 6, 12}

B = {1, 2, 4, 8, 16}

They are factors of 12

and 16.

A



B= {1, 2, 4}

Compliment of a set:

all elements in the universal set that are NOT in the initial set

Ex: S = {…-3, -2, -1, 0, 1, 2, 3, 4,…} and A = {…-2, 0, 2, 4,…}

If A is a subset of S, what is AC_?

Note that P(AC_{) is every outcome except (or not) A, so we can find P(A}C_{) by}

finding:

P(A

C

) = 1 - P(A)

 _{Why do you think this works?}

A and A

are the only options, so the sum of their

probabilities should be 1 for 100%.

P(A

) + P(A) = 1

Example: Use the Venn Diagram to find the following:

What is AC?

What is BC?

What is (A  B)C?

What is (A  B)C?

A:Chorus

B:Band

S

15

5

24

16

A

= 39

(A



B)

= 55

B

= 31

Example 3:

A pair of dice is rolled. What is the probability of NOT rolling doubles?

For a complex problem like this we need a sample space. A table is good here since we have 2 dice. Let’s create the table together!

Remember, P(AC_{) = 1 - P(A)}

P(doubles) = 6/36 = 1/6

P(not doubles) = 1 – P(doubles) = 1 - 1/6 = 5/6

1

Die

2

Die

1 2 3 4 5 6

1

2

3

4

5

6

1, 1 1, 2 1, 3 1, 4 1, 5 1, 6

2, 1 2, 2 2, 3 2, 4 2, 5 2, 6

3, 1 3, 2 3, 3 3, 4 3, 5 3, 6

4, 1 4, 2 4, 3 4, 4 4, 5 4, 6

5, 1 5, 2 5, 3 5, 4 5, 5 5, 6

6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

1, 1

1, 2 1, 3 1, 4 1, 5 1, 6

2, 1

2, 2

2, 3 2, 4 2, 5 2, 6

3, 1 3, 2

3, 3

3, 4 3, 5 3, 6

4, 1 4, 2 4, 3

4, 4

4, 5 4, 6

5, 1 5, 2 5, 3 5, 4

5, 5

5, 6

6, 1 6, 2 6, 3 6, 4 6, 5

6, 6

Use the Counting

Principle to check

your sample

space!

6

●

6 = 36 items

Checking for

understanding….

Why do we need

You Try!

Example 7:

A card is drawn at random from a standard

deck of cards. Find each of the following:

P(heart) ___________

P(black card) ____________

P(2 or jack) ____________

P(not a heart) ____________

1/4 (simplify 13/52)

1/2 (simplify 26/52)

2/13 (simplify 8/52)

If A and B are

independent

events, then

P(A and B) =

P(A) · P(B)

If A and B are

dependent

events, then

P(A, then B) =

P(A) · P(B after A)



P(A



B) =P(A) P(B l A)

**assume success on 1

draw**

If there are three events, then

Calculating Probabilities of Independent and Dependent Events

A game board in your closet has 7 purple game pieces, 4 red game pieces, and 3 green game pieces. You randomly

choose one game piece and then replace it. Then you choose a second game piece. Find each probability.

1)

P(red and green)

2)

P(green and purple)

3)

P(both red)

Calculating Probabilities of Independent and Dependent Events

You are folding the socks from the laundry basket, which contains 6 brown socks, 2 blue socks, and 5 black socks. You pick one sock at a time and don’t replace it. Find each probability.

4) P(blue, then black)

5) P(brown, then blue)

6) P(both black)

Rain or No Rain?

.60 .40

Win Lose .70 .30

Win Lose .95 .05

A COMPOUND EVENT is an event that is the result of more than one outcome.

To determine probabilities of compound events, we can use TREE DIAGRAMS.

Example 2: Create a Tree Diagram for the following scenario:

There is a 60% chance of rain on Wednesday.

If it rains, the track team has a 70% chance of winning.

To calculate probabilities using tree diagrams, we

use the

MULTIPLICATION RULE

for Compound

Events. To use this rule, all of the separate

outcomes that make up the compound event must

be

INDEPENDENT EVENTS

, which means that the

probabilities of the outcomes do not affect one

another.

To calculate probability of compound events:

1st)

MAKE A TREE DIAGRAM.

Venn Diagram Practice

There are 100 people in an office.

• *5 people have a dog, a cat, and a hamster.

• *8 people own ONLY a dog and a cat.

• *12 people own ONLY a dog and a hamster.

• *60 people own a dog.

• *0 people own ONLY a hamster and a cat.

• *20 people own hamsters

• *18 people own ONLY a cat

Fill in the above Venn-diagram.

How many people in this office own no cats, dogs, or hamsters?

Probability of

Mutually Exclusive Events

 _{To find the probability of one of two mutually exclusive events} occurring, use the following formula:

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

or

P(A  B) = P(A) + P(B)

*If A and B are mutually exclusive (no

overlap), then

Mutually Inclusive Events

 _{Suppose you are rolling a six-sided die. What is the probability that you} roll an odd number or a number less than 4?

 _{Can these both occur at the same time? If so, when?}

 _{Mutually Inclusive Events: Two events that can occur at the same time.}

Ex/ Probability of selecting someone with

brown hair or green eyes from the class.

Overlap or No overlap?

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

Another example of why we have to calculate

differently for OR probability with overlap!

Spinning an even number or a prime number on a single spin.

Are the events mutually exclusive?

Spinning an even number or a number less than 2 on a single spin.

They can both happen at once, 2 is even and also prime –

Not Mutually Exclusive

Example

What is the probability of choosing a card from a deck of

cards that is a club or a ten?

P(choosing a club or a ten)

= P(club) + P(ten) – P(10 of clubs)

= 13/52 + 4/52 – 1/52

= 16/52

= 4/13 or .308

The probability of choosing a club or a

ten is 4/13 or 30.8%

Are they mutually exclusive or mutually inclusive?

Conditional Probability

_{Contains a CONDITION that may LIMIT the sample space for an event}

_{Can be written with the notation P(B|A), which is read “the probability of}

event B, GIVEN event A”

_{How likely is one event to happen, given that another event HAS happened?}

_{Percentages/probability based on the ROW or COLUMN total of the given}

event -> for two-way table problems

_{More complex “given” problems}

may require use of this formula:

Probability of A given B =

(A and B)

(

)

( )

P

P A B

P B

Example

You are playing a game of cards where the winner is determined by drawing two cards of the same suit. What is the probability of drawing clubs on the second draw if the first card drawn is a

club?

P(clubclub)

= P(2nd _{club and 1}st _club)/P(1st _club)

= (13/52 x 12/51)/(13/52) = 12/51 or 4/17

The probability of drawing a club on the second draw given the first card is a club is 4/17 or 23.5%

(A and B)

(

)

( )

P

P A B

P B

In Mr. Jonas' homeroom, 70% of the students have

brown hair, 25% have brown eyes, and 5% have both

brown hair and brown eyes. A student is excused early

to go to a doctor's appointment. If the student has

brown hair, what is the probability that the student

also has brown eyes?

P(brown eyes



brown hair)

= P(brown eyes and brown hair)/P(brown hair)

= .05/.7

= .071

The probability of a student having brown eyes given he

or she has brown hair is 7.1%

(A and B)

(

)

( )

P

P A B

P B

Suppose we randomly select one student.

a. What is the probability that the student walked to school?

 _{88/500 = 22/125}

b. P(9th _{or 10}th _grader)

 _{210/500 = 21/50}

c. P(rode the bus OR 11th _{or 12}th _grader)

 _{147/500 + 290/500 – 41/500 = 396/500 = 99/125}

Bus

Walk

Car

Other

Total

9

or

10

106

30

70

4

210

11

or

12

41

58

184

7

290

d. What is the probability that a student is in 11th or 12th grade given that they rode in a car to school?

P(11th _{or 12}th  _car)

* We only want to look at the car column for this probability! = 11th or 12th graders in cars/total in cars

= 184/254 = 92/127 (or 72.4%)

The probability that a person is in 11th or 12th grade given that

they rode in a car is 72.4%

Bus Walk Car Other Total

9th or 10th 106 30 70 4 210

11th _or

12th

41 58 184 7 290

e. What is P(Walk|9th or 10th grade)?

= walkers who are 9th _{or 10}th _{/ all 9}th _{or 10}th

= 30/210

= 1/7 or 14.2%

The probability that a person walks to school given he or she is in 9th _{or 10}th

grade is 14.2%

Bus Walk Car Other Total

9th or 10th 106 30 70 4 210

11th _or

12th

41 58 184 7 290

Using Conditional Probability to

Determine if Events are

Independent

 _{If two events are statistically independent of each other, then:}

P(A



B) = P(A)

and

P(B



A) = P(B)

Ex: Suppose you manage a restaurant that serves chicken wings that are mild or hot, and boneless or regular. From your experience you know that of boneless wings bought, 75% of them are mild, and of the regular wings

bought, 70% are hot. Only 4 out of 10 costumers buy boneless wings.

1) Create a tree diagram for the scenario displaying all the possibilities and probabilities.

2) P(boneless and hot wings)

3) P(hot | boneless)

4) P(hot)

5) P(boneless |hot)

6) P( mild wings)

7) If a person orders regular wings, what is the probability they choose mild?

8) P(boneless |mild)

9) Of the boneless wings, what is the probability someone orders hot?

(.4)(.75) .1

10%

=

(.4)(.75) (.6)(.3) .48

48%



=

(.4)(.25)

.25 25%

(.4)

=

=

30%

(.4)(.25) (.6)(.7) .52 52%



=

=

(.4)(.75)

.7407 74.07%

(.4)(.75) (.6)(.3)



=

=