Jeff Bivin -- LZHS Permutation, Combinations, Probability, Oh My… Jeff Bivin Lake Zurich High...

Jeff Bivin -- LZHS

Permutation, Combinations,

Probability, Oh My…

Jeff BivinLake Zurich High School

[email protected]

ICTM ConferenceOctober 21, 2011

http://www.ictm.org/index.html

Jeff Bivin -- LZHS

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

)!(!rnnPrn

!26!30

)!430(!30

430

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

)!(!rnnPrn

!0!25

)!2525(!25

2525

P

!251!25

1.5511 x 1025

Jeff Bivin -- LZHS

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

245120

5!5

straight line

Divide by 5

Jeff Bivin -- LZHS

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

245120

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

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

403209

3628809!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 ?

4 3 2 1 4! 24

How many ways can you arrange the letters ABCDEF ?

6 5 4 3 2 16! 720

Jeff Bivin -- LZHS

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

4624

!324!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 in the non-word

A A B B C C C C D E F G G G G G G ?

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

Jeff Bivin -- LZHS

Combinations NO orderNO replacement

Use the

nCr formula

Typically

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(!30

430

C

Jeff Bivin -- LZHS

!3!9!12

!3)!312(!12

312

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

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?

5 35! 5!

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

Combinations

!)!(!rrn

nCrn

5 red6 white9 blue

3 red

have want

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?

!3!6!9

!3)!39(!9

39

C

Combinations

!)!(!rrn

nCrn

5 red6 white9 blue

3 blue

have want

Jeff Bivin -- LZHS

The OR factor.

10 84 94 5 red

6 white9 blue

3 redOR

3 blue

have want

5 3 9 35! 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 45! 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

5 1 9 25! 9!

(5 1)! 1! (9 2)! 2!C 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 1 red marble and 2 blue marbles?

The AND factor.

AND MULTIPLY

5 36 180 5 red

6 white9 blue

1 redand

2 blue

have want

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

And More

Jeff Bivin -- LZHS

(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2xy + 1y2

(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5

(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6

Let’s look at (x + y)p

(x + y)7 = _x7 + _x6y + _x5y2 + _x4y3 + _x3y4 + _x2y5 + _xy6 + _y7

Jeff Bivin -- LZHS

(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2xy + 1y2

(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5

(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6


Jeff Bivin -- LZHS

(x + y)7 = 1 7 21 35 35 21 7 1

(x + y)0 = 1

(x + y)1 = 1 1

(x + y)2 = 1 2 1

(x + y)3 = 1 3 3 1

(x + y)4 = 1 4 6 4 1

(x + y)5 = 1 5 10 10 5 1

(x + y)6 = 1 6 15 20 15 6 1

(x + y)8 = 1 8 28 56 70 56 28 8 1


Jeff Bivin -- LZHS

(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + 1y7

(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2xy + 1y2

(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5

(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6


Jeff Bivin -- LZHS

12! 479001600 19,958,4002! 3! 2! 24

In how many ways can you arrange the letters in the word M A T H E M A T I C A L ?

Jeff Bivin -- LZHS

52 yx

351445040

!3!4!7

212405040

!5!2!7

34 yx

In how many ways can you arrange the letters in the non-word xxxxyyy?

In how many ways can you arrange the letters in the non-word xxyyyyy?

(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

Jeff Bivin -- LZHS

Let’s look closer(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

071!0!7!7 yx 167

!1!6!7 yx 2521

!2!5!7 yx

3435!3!4!7 yx 4335

!4!3!7 yx 5221

!5!2!7 yx

617!6!1!7 yx 701

!7!0!7 yx

Jeff Bivin -- LZHS

An alternate look(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

77!0!7!7 C 67!1!6

!7 C 57!2!5!7 C

47!3!4!7 C 37!4!3

!7 C 27!5!2!7 C

17!6!1!7 C 07!7!0

!7 C

Jeff Bivin -- LZHS

(2x - y)4 = 16x4 - 32x3y + 24x2y2 - 8xy3 + y4

40444 161)()2(1

!0!4!4 xyxC

)(84)()2(4!1!3!4 313

34 yxyxC

)(46)()2(6!2!2!4 2222

24 yxyxC

)(24)()2(4!3!1!4 331

14 yxyxC

)(1)()2(1!4!0!4 440

04 yyxC

Jeff Bivin -- LZHS

(3x + 2y)5 = 243x5 + 810x4y + 1080x3y2 + 720x2y3 + 240xy4 + 32y5

50555 2431)2()3(1

!0!5!5 xyxC

yxyxC 2815)2()3(5!1!4!5 414

45

222335 42710)2()3(10

!2!3!5 yxyxC

323225 8910)2()3(10

!3!2!5 yxyxC

44115 1635)2()3(5

!4!1!5 yxyxC

55005 3211)2()3(1

!5!0!5 yyxC

Jeff Bivin -- LZHS

In how many ways can you arrange the

letters in the non-word

x x x x x y y y y y y y y y y ?

Given: (x + y)15

105515 3003

!10!5!15 yxC

1053003 yx

What is the coefficient of the term ____ x5y10 ?

Jeff Bivin -- LZHS

Given: (4x - 3y)10



x x x x x x x y y y ?37

710 )3()4(120!3!7!10 yxC

37 2716384120 yx

What is the coefficient of the term ____ x7y3 ?

37160,084,53 yx

Jeff Bivin -- LZHS

Expand: (x + y + z)2

(x + y + z) (x + y + z)

x2 + xy + xz + yx + y2 + yz + zx + zy + z2

x2 + 2xy + 2xz + y2 + 2yz + z2

Jeff Bivin -- LZHS

x3 + 3x2y + 3x2z + 3xy2 + 6xyz + 3xz2 + y3 + 3y2z + 3yz2 + z3

(x + y + z)2 (x + y + z)

x3 + x2y + x2z + 2x2y + 2xy2 + 2xyz + 2x2z + 2xzy + 2xz2 + y2x + y3 + y2z

+2yzx + 2y2z + 2yz2 + z2x + z2y + z3

Expand: (x + y + z)3

(x2 + 2xy + 2xz + y2 + 2yz + z2)(x + y + z)

We did this in the last example

Jeff Bivin -- LZHS

zyx 6!1!1!1!3

x3 + 3x2y + 3x2z + 3xy2 + 6xyz + 3xz2 + y3 + 3y2z + 3yz2 + z3

Given: (x + y + z)3

In how many ways can you arrange the letters in the non-word xyz ?

What is the coefficient of the term ____ xyz ?

In how many ways can you arrange the letters in the non-word xxz ?

zx 23!1!0!2

!3

What is the coefficient of the term ____ x2z ?

RE

ME

MB

ER

Jeff Bivin -- LZHS

Given: (x + y + z)15



xxyyyyyyyzzzzzz ?

672180180!6!7!2!15 zyx

672180,180 zyx

What is the coefficient of the term ____ x2y7z6 ?

Jeff Bivin -- LZHS

Given: (2x + 3y - z)9



x x x y y y y z z ?243 )()3()2(1260

!2!4!3!9 zyx

243480,816 zyx

What is the coefficient of the term ____ x3y4z2 ?

243 8181260 zyx

Jeff Bivin -- LZHS



a a a a a b b b b b b c c c c c c c d d ?

Given: (a + b + c + d)20

2765720,510,793,2!2!7!6!5

!20 dcba

2765720,510,793,2 dcba

What is the coefficient of the term ____ a5b6c7d2 ?

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Can we think of combinations as permutations with repetitions?

In a group of 7 people, how many different committees of 3 people can we select?

Two choices….

A member of the 3 person committee

A member of the 4 person non-committee

Jeff Bivin -- LZHS

In a group of 7 people, how many different committees of 3 people can we select?

Alex Betty Chuck Deb Ed Fiona Gabe

N C N C N N C

How many ways can you arrange 3 C’s and 4 N’s?

7! 353! 4!

Jeff Bivin -- LZHS

In a group of 7 people, a pair of 2 people are needed for one committee and 3 different people are need for a second committee. How many possibilities exist?

Alex Betty Chuck Deb Ed Fiona Gabe

C1 N C2 C1 N C2 N

How many ways can you arrange 2 C1’s and 2 C’s and 3 N’s?

7! 2102! 2! 3!

Jeff Bivin -- LZHS

In a group of 7 people, how may ways can a president, a vice-president and a secretary be selected? Alex Betty Chuck Deb Ed Fiona Gabe

N VP N P N N S

How many ways can you arrange 1 P and 1 VP and 1 S and 4 N’s?

7! 2101! 1! 1! 4!

Can we think of most permutations as permutations with repetitions?

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

PROBABILITY Defined

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 }

number of successtotal number of outcomesPr(H) = 1

2=

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= 1

3=

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

Jeff Bivin -- LZHS

number of failurestotal 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= 1

3=

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

1435

28610

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

14328

28656

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

14370

286140

286285

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?

# of successtotal # of outcomesPr(1R,1B,1W) = 5C1●8C1●7C1

20C3=

5 red8 blue7 white

have want

1 red

Total: 20 3

5714

1140280

1140785

1 blue1 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?

# of successtotal # of outcomesPr(2R,2W) = 7C2 ● 3C2

15C4=

7 red5 blue3 white

have want

2 red

Total: 15 4

653

136563

1365321

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?

# of successtotal # of outcomesPr(3H,2C) = 13C3 ● 13C2

52C5=

13 diamonds13 hearts13 clubs

13 spades

have want

3 hearts

Total: 52 5

6497405577

259896022308

259896078286

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?

# of successtotal # of outcomes

Probability – “or”

Pr(3R or 3B) = 5C3 + 8C3 13C3

=

5 red

8 blue

have want

3 red

Total: 13 3

133

28666

2865610

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

286101

286355610

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 132616325

22155

1326330

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

1310

286220

286285810

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

14345

28690

28610810

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

28610810285

2 blue

want

OR2 red

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

want

OR3 red

1 blue

143115

286230

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

14328

28656

3 blue

In the previous example we found

1431151Pr 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 )

143115

286230

2865611

313

38 CC

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

791CC

48624853

19448361

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)

77520303601

646393

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 Permutation, Combinations, Probability, Oh My… Jeff Bivin Lake Zurich High...

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