}{

Zero elements. One element. Two elements. Three elements.

{ a }

{ b }

{ c }

{ a, b }

{ b, c }

{ a, c }

{ a, b, c }

8 subsets.

Finding the number of subsets and the cardinal number of a set.

Cardinal number is a non-negative integer defining how many elements are in a set. It is denoted as n( ). Example, n( O ) = 0

Let A = { a, b }. How many subsets are there.

}{

Zero elements. One element. Two elements.

{ a } { b } { a, b }

4 subsets.

Let n be the cardinal number of set A. Set A has 2n subsets.

What would the formula for the number of Proper Subsets? 12 n

Define Union and Intersection of sets.

BAnBnAnBAn

Intersection of sets are the elements that are the same in both sets. BA

Union of sets are all the elements that are in each set. BA

Counting Principle (Addition).

In A and B.

In A or B.

In a survey of 100 college students, 35 were registered in College Algebra, 52 were registered in English, and 18 were registered in both courses.

How many were registered in College Algebra or English?

How many were registered in neither class?

35An 52En 18EAn

EAnEnAnEAn

union

69185235

3169100

Make a tree diagram for two appetizers, 4 entrees, and 2 desserts.

soup

salad

2

1

43

65

16

15

10987

14131211

Counting Principle (Multiplication).If a task consists of a sequence of individual events, then multiply the number of each event to find total possibilities.

The Student Union is having a lunch special value meal. You get to choose one of 4 sandwiches, one of 5 bags of chips, one of 7 drinks and one of 2 desserts. How many different lunch specials can you make? 2802754

____ ____ ____How many letters in the alphabet?26 26 26

576,17263

____ ____ ____26 25 24 600,15

!23

!23242526

!23

!26

)!326(

!26326

P 600,15

!3!2

!5

!3)!35(

!535

C!321

!345

102

!3!5

!8

!3)!38(

!838

C!5321

!5678

56

2625 CC men women !2!4

!6

!2!3

!5

12!4

!456

12!3

!345

1501510 2 3

This is the repeating letters.

There are 11 letters. n = 11 4 I’s, 4 S’s and 2 P’s

!2!4!4

!11

12!41234

!4567891011

2

650,345791011

1’s must be on the corners of the triangle.

06C 16C 26C 36C 46C 56C 66C

How to use the graphing calculator to find a row of Pascal’s triangle!

)5,0,,5( XnCrXseqFind the 5th row.

Decreasing power

Increasing power

Need the 4th row of Pascal’s Triangle. 1, 4, 6, 4, 1

1( )4( )0

( ) ( )

+ 4( )3( )1 + 6( )2( )2 + 4( )1( )3 + 1( )0( )42x – 3– 3 2x 2x 2x 2x– 3 – 3 – 3

0416 yx 1396 yx 22216 yx 31216 yx 4081 yx

812162169616 234 xxxx

Notice the powers on both variables add up to 4, the power on the binomial.

Remember from the last example that the powers on the x-term and y-term have to add up to be . . . n = 10

37310 )3()2( xC

Therefore, 10 – 7 = 3 = r. The y-term is always to the r power.

rrnrn yxC )()( )27(128120 7 x

7720,414 x

Remember that the Combination formula starts counting with zero for the r!r = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

rrnrn yxC )()(

n = 10

64610 )3()2( xC )729(16210 4 x

4440,449,2 x

r = 6

H

T

H

H

T

T

T

T

T

TH

H

H

H

Sample Space (Total = 8 = 23 )

H HHH THH HTH TTT HHT THT HTT TT

1st flip

2nd flip3rd flip

43

32

65

54

87

76

65

54

87

76

109

98

87

76

109

98

1211

1110

2

2

1

1

4

4

3

3

6

6

5

5

Diamonds

3

Hearts

2

Spades

5

Clubs

4 76 98 J10 KQ A

6 * 6 = 36 total 4 * 13 = 52 total

Sample Space

H HHH THH HTH TTT HHT THT HTT TT

Always find total number of outcomes 1st!

38

48 2

1

43

32

65

54

87

76

65

54

87

76

109

98

87

76

109

98

1211

1110

2

2

1

1

4

4

3

3

6

6

5

5

636 6

1

36

10

18

5

Diamonds

3

Hearts

2

Spades

5

Clubs

4 76 98 J10 KQ A

52

12

13

3

52

13

52

3

.).()( CFPHeartsP

4

1

13

3

52

13

52

12

.).(.).()( CFHPCFPHP

52

12

52

3

52

22

26

11

)5()4()3()2()1( TPTPTPTPTP 3225 Total NO WAY! What is the opposite of at least 1 tails?

No Tails or All Heads.

)5(1 HP

All Heads can only happen one way.

32

11

32

31

The complement is P( all different birthdays ), 1 – P( all different B-days ).There are 365 days in the year and every day is likely to occur. Find the size of the sample space…

365365365365365365365365365365

The 1st person has all 365 days available.

365

The 2nd person has 364 days available.

364

The 3rd person has 363 days available.

363362 361 360 359 358 357 356

10 probabilities would have a denominator of 365.

The numerators are an arrangement where order matters, therefore, we can use Permutation notation. n = 365, r = 10

10365P

P(different B-days) =

10365

1 – P( different B-days )

1010365

3651

P 1169481777.0

Binomial Probability. This probability has only two outcomes, success or failure.Let n = the number of trials.Let r = the number of successes.Let n – r = the number of failures.Let p = the probability of success in one trial.Let q = the probability of failure. 1 – p = q.

Binomial Probability Formula

P( r successes in n trials ) rnrrn qpC

Always need two categories. Other common categories are Good or Bad, Boys or Girls, True or False, etc. The sum of the two probabilities is 1.

The spirit club is 75% girls. What is the probability of randomly picking 5 people, where there are 3 boys and 2 girls?

P( 2 girls in 5 trials ) 3225 25.075.0 C 08789.0

A manufacture has determined that a bulb machine will produce 1 bad bulb for every 2000 bulbs it produces.

What is the probability that an order of 200 bulbs are all perfect?

There is at least one bad bulb?

There is one bad bulb?

There are two bad bulbs?

2000

1999)( GoodP

2000

1)( BadP

200

2000

1999)200(

GoodP 9048.0

)200(1 GoodP 0952.09048.01

We need the new Binomial Probability formula.rnr

rn BadPGoodPCrGoodP )()()(1199

199200 2000

1

2000

1999)199(

CGoodP 0905.0

2198

198200 2000

1

2000

1999)198(

CGoodP 0045.0

20Total

3.010

3

20

6

15.020

3

05.020

1

25.04

1

20

5

1.010

1

20

2

15.020

3

Total chips = 10

10

51st pick

10

322ndnd pick pick

10

23rd pick

100

3

10

51st pick

9

322ndnd pick pick

8

23rd pick

24

1

1 11

2 3 4

Notice that the denominator decreases by one on each pick because there is no replacement.

Same colors, but order doesn’t matter.

10

51st pick

10

322ndnd pick pick

10

23rd pick

100

3We have to consider all the different ways

the 3 colors can be picked. RWB, RBW, WRB, WBR, BRW, and BWR. 3! = 6

50

9

100

186

10

51st pick

9

322ndnd pick pick

8

23rd pick

24

1

1 11

2 3 4

Same colors, but order doesn’t matter.

We have to consider all the different ways the 3 colors can be picked. RWB, RBW, WRB, WBR, BRW, and BWR. 3! = 6

4

16

Total chips = 10

Another approach we can use with Probabilities that have NO REPLACEMENT, is to use Permutations or Combinations.

Since order matters, this is a Permutation. 310

121315

P

PPP

24

1

Since order doesn’t matter, this is a Combination. 310

121315

C

CCC

4

1