jeff bivin -- lzhs counting and probability by: jeffrey bivin lake zurich high school...

Jeff Bivin -- LZHS

Counting and Probability

By: Jeffrey BivinLake Zurich High School

[email protected]

Last Updated: April 16, 2008

Jeff Bivin -- LZHS

Fundamental Counting PrincipalHow many different meals can be made if 2 main courses, 3 vegetables, and 2 desserts are available?

M1 M2

V1 V2 V3 V1 V2 V3

D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2

1 2 3 4 5 6 7 8 9 10 11 12

Let’s choose a

main course

Now choose a

vegetable

Finally choose

A dessert

Jeff Bivin -- LZHS

Linear Permutations

A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible?

president vice-president secretary treasurer

Jeff Bivin -- LZHS

A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible?

president vice-president secretary treasurer

30P4

Linear Permutations

Jeff Bivin -- LZHS

Permutation Formula

)!(

!

rn

nPrn

!26

!30

)!430(

!30430

P

27282930!26

!2627282930

Jeff Bivin -- LZHS

Linear Permutations

There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible?

seat 1 seat 2 seat 3 seat 4 seat 5

1.5511 x 1025

Jeff Bivin -- LZHS

Linear Permutations

There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible?

seat 1 seat 2 seat 3 seat 4 seat 5

25P25

1.5511 x 1025

Jeff Bivin -- LZHS

Permutation Formula

)!(

!

rn

nPrn

!0

!25

)!2525(

!252525

P

!251

!25 1.5511 x 1025

Jeff Bivin -- LZHS

More PermutationsThere are 5 people sitting at a round table, how many different seating arrangements are possible?

245

120

5

!5

straight line

Divide by 5

Jeff Bivin -- LZHS

More PermutationsThere are 5 people sitting at a round table, how many different seating arrangements are possible?

245

120

5

!5

straight line

Treat all permutations as if linear

Now consider the circular issue

When circular, divide by the number of items in the circle

Jeff Bivin -- LZHS

More PermutationsThere are 9 people sitting around a campfire, how many different seating arrangements are possible?

403209

362880

9

!9

straight line

Treat all permutations as if linear

Is it circular?

Yes, divide by 9

Jeff Bivin -- LZHS

There are 5 people sitting at a round table with a captain chair, how many different seating arrangements are possible?

More Permutations

120!5

straight line

NOTE:

Jeff Bivin -- LZHS

More PermutationsHow many ways can you arrange 3 keys on a key ring?

12

236

3!3

straight line

Treat all permutations as if linearIs it circular?

Now, try it. . .PROBLEM:Turning it over results in the same outcome.

Yes, divide by 3

So, we must divide by 2.

Jeff Bivin -- LZHS

More PermutationsHow many ways can you arrange the letters MATH ?

24!4

How many ways can you arrange the letters ABCDEF ?

720!6

Jeff Bivin -- LZHS

Permutations with RepetitionHow many ways can you arrange the letters AAAB?

46

24

!3

24!4

Let’s look at the possibilities:

AAABAABAABAABAAA

Are there any others?What is the problem?

If a permutation has repeated items, we divide by the number of ways of arranging the repeated items (as if they were different).

Divide by 3!

Jeff Bivin -- LZHS

How many ways can you arrange 5 red, 7 blue and 8 white flags on the tack strip across the front of the classroom?

240,768,99!8!7!5

!20

If all were different, how may ways could we arrange

20 items?

There are 5 repeated red flags Divide by 5!

There are 7 repeated blue flags Divide by 7!

There are 8 repeated white flags Divide by 8!

Jeff Bivin -- LZHS

How many ways can you arrange the letters AABBCCCCDEFGGGGGG ?

800,940,145,5!6!4!2!2

!17

If all were different, how may ways could we arrange

17 items?

There are 2 repeated A’s Divide by 2!

There are 2 repeated B’s Divide by 2!

There are 4 repeated C’s Divide by 4!

There are 6 repeated G’s Divide by 6!

Jeff Bivin -- LZHS

Permutations ORDER

Multiply the possibilities

Divide by the numberof items in the circle

Divide by 2

Divide by the factorial of thenumber of each duplicated item

Assume the itemsare in a straight line! Use the nPr formula

(if no replacement)

or

Are the items in a circle??

Can the itembe turned over??

Are there duplicateitems in your

arrangement??

Jeff Bivin -- LZHS

How many ways can you put 5 red and 7 brown beads on a necklace?

!7!5212

!12

How may ways could we arrange 12 items in a straight line?

Is it circular? Yes divide by 12

Can it be turned over? Yes divide by 2

Are there repeated items? Yes divide by 5! and 7!

33

Jeff Bivin -- LZHS

How many ways can you arrange 5 red and 7 brown beads on a necklace that has a clasp?

!7!52

!12

How may ways could we arrange 12 items in a straight line?

Is it circular? N0 the clasp makes it linear

Can it be turned over? Yes divide by 2

Are there repeated items? Yes divide by 5! and 7!

396

Jeff Bivin -- LZHS

How different license plates can have 2 letters followed by 3 digits (no repeats)?

A straight line?

Is it circular? No

Can it be turned over? No

Are there repeated items? No

468,000

26 ∙ 25 ∙ 10 ∙ 9 ∙ 8lette

rletter number number number

Jeff Bivin -- LZHS

How different license plates can have 2 letters followed by 3 digits with repeats?

A straight line?

Is it circular? No

Can it be turned over? No

Are there repeated items? Yes, but because we are using multiplication andnot factorials, we do not need to divide by anything.

676,000

26 ∙ 26 ∙ 10 ∙ 10 ∙ 10lette

rletter number number number

Jeff Bivin -- LZHS

Combinations NO orderNO replacement

Use the

nCr formula

Jeff Bivin -- LZHS

Combinations An organization has 30 members and must select a committee of 4 people to plan an upcoming function. How many different committees are possible?

!)!(

!

rrn

nCrn

!4!26

!30

!4)!430(

!30430

C

Jeff Bivin -- LZHS

!3!9

!12

!3)!312(

!12312

C

Combinations

!)!(

!

rrn

nCrn

A plane contains 12 points, no three of which are co-linear. How many different triangles can be formed?

Jeff Bivin -- LZHS

!3!6

!9

!3)!39(

!939

C

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 blue marbles?

Combinations

!)!(

!

rrn

nCrn

Jeff Bivin -- LZHS

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 red marbles?

!3!2

!5

!3)!35(

!535

C

Combinations

!)!(

!

rrn

nCrn

Jeff Bivin -- LZHS

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 blue marbles or 3 red marbles?

The OR factor.

OR ADD

!3)!35(

!5

!3)!39(

!93539

CC

941084

Jeff Bivin -- LZHS

The OR factor.

10 84 94 5 red

6 white9 blue

3 redOR

3 blue

have want

5 3 9 3

5! 9!

(5 3)! 3! (9 3)! 3!C C

want

OR ADD

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 red marbles or 3 blue marbles?

Jeff Bivin -- LZHS

The OR factor.

5 70 75 5 red

8 blue

4 redOR

4 blue

have

5 4 8 4

5! 8!

(5 4)! 4! (8 4)! 4!C C

OR ADD

wantwant

An jar contains 13 marbles – 5 red and 8 blue. If four are selected at random, how many ways can you select 4 red marbles or 4 blue marbles?

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red or all three are blue?

# of successtotal # of outcomes

Probability – “or”

Pr(3R or 3B) = 5C3 + 8C3

13C3=

5 red

8 blue

have want

3 red

Total: 13 3

13

3

286

66

286

5610

3 blue

want

OR

Jeff Bivin -- LZHS

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 2 blue marbles and 1 red marble?

The AND factor.

AND MULTIPLY

!1)!15(

!5

!2)!29(

!91529

CC

180536

Jeff Bivin -- LZHS

3B2NB or 4B1NB or 5B

At least

3 or 4 or 5 blue

591114921139 CCCCC 126111265584

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If five marbles are selected at random, how many ways can you select at least 3 blue marbles?

6132

Jeff Bivin -- LZHS

0R5Nr or 1R4NR

At most

0 or 1 red

4151551505 CCCC 1365530031

An jar contains 20 marbles – 5 red, 6 white and 9 blue. If five marbles are selected at random, how many ways can you select at most 1 red marbles?

9828

Jeff Bivin -- LZHS

PROBABILITY

Definition:

number of successtotal number of outcomes

The ratio

Jeff Bivin -- LZHS

Probability

A coin is tossed, what is the probability that you will obtain a heads?

Look at the sample space/possible outcomes:

{ H , T }


Pr(H) = 12=

Jeff Bivin -- LZHS

number of success

Probability

A die is tossed, what is the probability that you will obtain a number greater than 4?

Look at the sample space/possible outcomes:

total number of outcomesPr(>4) =

26=

13=

{ 1 , 2 , 3 , 4 , 5 , 6 }

Jeff Bivin -- LZHS

number of failures

total number of outcomes

total number of outcomesnumber of success

Probability – Success & Failure

A die is tossed, what is the probability that you will obtain a number greater than 4?

Pr(>4) = 26=

13=

What is the probability that you fail to obtain a number greater than 4?

Pr(>4) = 46

23= =

TOTAL = Pr(success) + Pr(failure) = 1

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red?


Probability

Pr(3R) = 5C3

13C3=

5 red

8 blue

have want

3 red

Total: 13 3

143

5

286

10

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are blue?


Probability

Pr(3B) = 8C3

13C3=

5 red

8 blue

have want

3 blue

Total: 13 3

143

28

286

56

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one is red and two are blue?


Probability – “and”

Pr(1R2B) = 5C1 ● 8C2

13C3=

5 red

8 blue

have want

1 red

Total: 13 3

143

70

286

140

286

285

2 blue

multiply

Jeff Bivin -- LZHS

A jar contains 5 red, 8 blue and 7 white marbles. If 3 marbles are selected at random, what is the probability that one of each color is selected?


Pr(1R,1B,1W) = 5C1●8C1●7C1

20C3=

5 red8 blue7 white

have want

1 red

Total: 20 3

57

14

1140

280

1140

785

1 blue

1 white

1 red, 1 blue, & 1 white

Jeff Bivin -- LZHS

A jar contains 7 red, 5 blue and 3 white marbles. If 4 marbles are selected at random, what is the probability that 2 red and 2 white marbles are selected?


Pr(2R,2W) = 7C2 ● 3C2

15C4=

7 red5 blue3 white

have want

2 red

Total: 15 4

65

3

1365

63

1365

321

2 white

Jeff Bivin -- LZHS

Five cards are dealt from a standard deck of cards. What is the probability that 3 hearts and 2 clubs are obtained?


Pr(3H,2C) = 13C3 ● 13C2

52C5=

13 diamonds13 hearts13 clubs

13 spades

have want

3 hearts

Total: 52 5

649740

5577

2598960

22308

2598960

78286

2 clubs

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red or all three are blue?


Probability – “or”

Pr(3R or 3B) = 5C3 + 8C3

13C3=

5 red

8 blue

have want

3 red

Total: 13 3

13

3

286

66

286

5610

3 blue

want

OR

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles and 7 yellow marbles. If 3 marbles are selected at random, what is the probability that all three are the same color?

5C3 + 8C3 + 7C3 # of success

total # of outcomesPr(3R or 3B or 3w) = 20C3

=

5 red8 blue

7 yellow

have want

3 red

Total: 20 3

286

101

286

355610

3 blue

want

OR

3 red or 3 blue or 3 yellow ?

want

3 yellowOR

Jeff Bivin -- LZHS

26C2 + 4C2 – 2C2 # of successtotal # of outcomes

Probability – “or” with overlap

Pr(2R or 2B) = Pr(2R) + Pr(2K) – Pr(2RK)

52C2=

26 red26 black

have want

2 red

Total: 52 2

2 kings

want

OR

2 red kings

overlap 1326

16325

221

55

1326

330

If two cards are selected from a standard deck of cards, what is the probability that both are red or both are kings?

4 kings48 other

Jeff Bivin -- LZHS

5C2● 8C1 + 5C1 ● 8C2# of success

total # of outcomes

Probability – “and” with “or”

Pr(2R1B or 1R2B) = 13C3

=

5 red

8 blue

have want

2 red

Total: 13 3

13

10

286

220

286

285810

1 blue

want

OR1 red

2 blue

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that two are red and one is blue or that one is red and two are blue?

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least two red marbles are selected?

5C2● 8C1 + 5C3# of success

total # of outcomes

Probability – “at least”

Pr(at least 2Red) = 13C3

=

5 red

8 blue

have want

2 red

Total: 13 3

143

45

286

90

286

10810

1 blue

want

OR3 red

2 red or 3 red2 red and 1 blue or 3 red

Pr(2R1B or 3R) =

Jeff Bivin -- LZHS

5C1● 8C2 + 5C2 ● 8C1 + 5C3

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least one red marble is selected?


Pr(at least 1Red) = 13C3

5 red

8 blue

have want

1 red

Total:13 3

286

10810285

2 blue

want

OR2 red

Pr(1R2B or 2R1B or 3R) =

want

OR3 red

1 blue

143

115

286

230

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that NO red marbles are selected?

8C3


Pr(0R3B) = 13C3

5 red

8 blue

have want

Total:13 3

143

28

286

56

3 blue

In the previous example we found

143

1151Pr red

Pr(success) + Pr(failure) = 1

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least one red marble is selected?


Pr(>1 red) = 1 – Pr( 0 red )

143

115

286

230

286

5611

313

38 C

C

Pr(success) + Pr(failure) = 1

Pr(success) = 1 - Pr(failure)

Pr(3 blue)

Jeff Bivin -- LZHS

A jar contains 8 red and 9 blue marbles. If 7 marbles are selected at random, what is the probability that at least one red marbles is selected?


Pr(at least 1Red)

Pr(1R6B or 2R5B or 3R4B or 4R3B or 5R2B or 6R1B or 7R)

Pr(0Red) Pr( 0R7B )

success

failure

FASTEST

Pr(at least 1Red) = 1 - Pr(0R7B) = 717

791C

C

4862

4853

19448

361

Jeff Bivin -- LZHS

720

51228612187121C

CCCCC

A jar contains 8 red, 9 blue and 3 white marbles. If 7 marbles are selected at random, what is the probability that at least three red marbles are selected?


Pr(> 3Red) Pr(3-7 red)

Pr(< 3Red) Pr(0-2 red)

success

failure

FASTEST

1 - Pr(0R7NR or 1R6NR or 2R5NR)

77520

303601

646

393

Jeff Bivin -- LZHS

Probability – “with replacement”

2197320

138

138

135

138

138

135

Must use fractions! R B B

Note: In this example

an order is specified

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one red followed by two blue marbles are selected if each marble is replaced after each selection?

Jeff Bivin -- LZHS

A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one red and two blue marbles are selected if each marble is replaced after each selection?

Probability – “with replacement”

2197960

138

138

135

138

138

135

23 3 C

Must use fractions!

Must account of any order!

Problem:

Fractions imply order! R B B

jeff bivin -- lzhs counting and probability by: jeffrey bivin lake zurich high school...

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