Expected Value Reprise

CP Canoe Club Duck Derby Maximum of 2 000 tickets sold $5/ticket

Prizes 12 VIP tickets to Cirque du Soleil ($2,000) 32gb BlackBerry Playbook ($600) $250

Should Mr. Lieff buy a ticket?

Expected Value Calculation

2 000 tickets E(X) =1/2000 (2000) + 1/2000(600)+1/2000(250) = 1.425

Mr. Lieff receives a hot tip that only 1 500 tickets will be sold. E(X) =1/1500 (2000) + 1/1500(600)+1/1500(250) = 1.9

Probability Distributions and Expected Value

Chapter 5.1 – Probability Distributions and PredictionsLearning goals: Calculate probabilities and expected value

Probability Distributions of a Discrete Random Variable a discrete random variable X is a variable

that can take on only a finite set of values for example, rolling a die can only produce

numbers in the set {1,2,3,4,5,6} rolling 2 dice can produce only numbers in

the set {2,3,4,5,6,7,8,9,10,11,12} choosing a card from a standard deck

(ignoring suit) can produce only the cards in the set {A,2,3,4,5,6,7,8,9,10,J,Q,K}

Probability Distribution the probability

distribution of a random variable X, is a function which provides the probability of each possible value of X

this function may be represented as a table of values, a graph or a mathematical expression

for example, rolling a die:C

ount

1

outcome0 1 2 3 4 5 6 7

Rolling A Die Histogram

Rolling A Die

=outcome probability <new>

123456

1 1/62 1/63 1/64 1/65 1/66 1/6

Probability Distribution for 2 Dice

Cou

nt

1234567

sum2 4 6 8 10 12

Two Dice HistogramRollingDice

=sum probability <new>

1234567891011

2 1/363 2/364 3/365 4/366 5/367 6/368 5/369 4/3610 3/3611 2/3612 1/36

What would a probability distribution graph for three dice look like? We will try it! Using three dice, figure out how many

outcomes there are Then find out how many possible ways there are to

create each of the possible outcomes Fill in a table like the one below Now you can make the graph

Outcome 3 4 5 6 7 8 9 …

# ways 1

Probability Distribution for 3 DiceOutcome 3 4 5 6 7 8 9 10

# cases 1 3 6 10 15 21 28

So what does an experimental distribution look like? A simulated dice

throw was done a million times using a computer program and generated the following data

What is/are the most common outcome(s)?

Does this make sense?

Fre

q

0

20000

40000

60000

80000

100000

120000

140000

roll2 4 6 8 10 12 14 16 18 20

Rolling 3 Dice Line Scatter Plot

Back to 2 Dice

What is the expected value of throwing 2 dice?

How could this be calculated?

So the expected value of a discrete variable X is the sum of the values of X multiplied by their probabilities

1

(sum of 2 dice)1 2 3 12 3 4 ... 12

36 36 36 36252 736

( ) ( )n

i ii

E

E X x P X x

Example 1a: tossing 3 coins

What is the likelihood of at least 2 heads? It must be the total probability of tossing 2 heads

and tossing 3 heads P(X = 2) + P(X = 3) = ⅜ + ⅛ = ½ so the probability is 0.5

X 0 heads 1 head 2 heads 3 heads

P(X) ⅛ ⅜ ⅜ ⅛

Example 1b: tossing 3 coins

What is the expected number of heads? It must be the sums of the values of x multiplied

by the probabilities of x 0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3) = 0(⅛) + 1(⅜) + 2(⅜) + 3(⅛) = 1½ So the expected number of heads is 1.5

X 0 heads 1 head 2 heads 3 heads

P(X) ⅛ ⅜ ⅜ ⅛

Example 2a: Selecting a Committee of three people from a group of 4 men and 3 women What is the probability of having at least one

woman on the committee? There are C(7,3) or 35 possible teams C(4,3) = 4 have no women C(4,2) x C(3,1) = 6 x 3 = 18 have one woman C(4,1) x C(3,2) = 4 x 3 = 12 have 2 women C(3,3) = 1 has 3 women

Example 2a cont’d: selecting a committee

What is the likelihood of at least one woman? It must be the total probability of all the cases

with at least one woman P(X = 1) + P(X = 2) + P(X = 3) = 18/35 + 12/35 + 1/35 = 31/35

X 0 women 1 woman 2 women 3 women

P(X) 4/35 18/35 12/35 1/35

Example 2b: selecting a committee

What is the expected number of women? 0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3) = 0(4/35) + 1(18/35) + 2(12/35) + 3(1/35) = 1.3 (approximately)

MSIP / Homework: p. 277 #1-5, 9, 12, 13

X 0 women 1 woman 2 women 3 women

P(X) 4/35 18/35 12/35 1/35

MSIP / Homework

p. 277 #1-5, 9, 12, 13

Pascal’s Triangle and the Binomial Theorem

Chapter 5.2 – Probability Distributions and PredictionsLearning goal: Generate Pascal’s Triangle and use it to solve counting problemspascals_triangle.mp3

How many routes are there to the top right-hand corner? Move up 4 spaces and

over 5 spaces Same as rearranging the

letters NNNNEEEEE Calculated by

9! 4!5!

C(9,4) or C(9,5) = 126 ways

Permutation AND Combination? As a permutation:

There are 9 moves 9! However, 4 are identical N moves and 5 are

identical E moves Divide by 4! due to identical arrangements of Ns Divide by 5! due to identical arrangements of Es 9! ÷ 4!5! = 126

As a combination: There are 9 moves required _ _ _ _ _ _ _ _ _ Choose 4 to be Ns, the rest are Es C(9, 4) = 126 Choose 5 to be Es, the rest are Ns C(9, 5) = 126

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

Pascal’s Triangle outer values: always 1

inner values: add the two values diagonally above

Pascal’s Triangle

1 1 1

1 2 1 1 3 3 1

1 4 6 4 1 1 5 10 10 5 1

1 6 15 20 15 6 1

sum of each row is a power of 2 1 = 20

2 = 21

4 = 22

8 = 23

16 = 24

32 = 25

64 = 26

Pascal’s Triangle 1

1 1 1 2 1

1 3 3 1 1 4 6 4 1

1 5 10 10 5 1 1 6 15 20 15 6 1

Uses? binomial theorem combinations!

e.g. choose 2 items from 5 go to the 5th row, the 2nd

number = 10 (always start counting at 0)

modeling the electrons in each shell of an atom (google ‘Pascal’s Triangle electron’)

Pascal’s Triangle – Cool Stuff 1

1 1 1 2 1

1 3 3 1 1 4 6 4 1

1 5 10 10 5 1 1 6 15 20 15 6 1

each diagonal is summed below and to the left

called the “hockey stick” property

may be music hidden in it pascals_triangle.mp3

Music: Three Parts in D Alternating Octatonic The piece of music has three parts: 1) A treble part where the center numbers 1, 2 and 6 are the tonic D

of the D alternating octatonic scale. The other vertical lines of numbers represent one scale step; a descending scale step to the left and an ascending scale step to the right. All ways of reaching all the numbers in the triangle are mapped to consecutive sixteenths, starting from the top and going down south west first to every new row, inwards. This part is repeated.

2) The main treble part. It has the structure A1 B A2 B. It starts with the left side of the symmetric triangle, including the center (A1), the numbers are the number of scale steps to ascend (south east arrows) or descend (south west arrows). After an intermission (B) where this part doubles the first part, the right side is initiated (A2) and it is the mirror image of A1.

3) A walking bass part (quarter notes), same as the first treble part but consecutive equal tones are tied. This part is repeated.

Playing time: 4' 47".

Pascal’s Triangle – Cool Stuff e.g., numbers

divisible by 5 similar

patterns for other numbers

http://www.shodor.org/interactivate/activities/pascal1/

Pascal’s Triangle can also be seen in terms of combinations n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

00

01

24

34

44

05

15

25

35

11

02

12

22

03

13

23

33

04

14

45

55

06

16

26

36

46

56

66

Pascal’s Triangle - Summary symmetrical down the middle outside number is always 1 second diagonal values match the row

numbers sum of each row is a power of 2

sum of nth row is 2n

Begin count at 0 number inside a row is the sum of the two

numbers above it

How many paths are there from X to Y?

X

Ya)As a checkerboardb)As a grid

Warm upHow many ways are there to spell the word SNOWMAN?

The Binomial Theorem

the term (a + b) can be expanded: (a + b)0 = 1 (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Blaise Pascal (for whom the Pascal computer language is named) noted that there are patterns of expansion, and from this he developed what we now know as Pascal’s Triangle. He also invented the second mechanical calculator.

So what does this have to do with the Binomial Theorem?

remember that the binomial expansion: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

and the triangle’s 4th row is 1 4 6 4 1 Pascal’s Triangle allows you to determine the

coefficients The exponents on the variables form a

predictable pattern The exponents of each term sum to n

The Binomial Theorem

nrrnnnn

n

bnn

barn

ban

ban

anba

......210

)(

221

rrnn barn

ba

is term t the)( binomial for the so 1r

A Binomial Expansion

Expand (x + y)4

432234

444334224114004

4

464

44

34

24

14

04

yxyyxyxx

yxyxyxyxyx

yx

Another Binomial Expansion

Expand (a – 4)5

1024128064016020

)1024()1()256()5()64()10()16()10()4()5()1()1(

)4(55

)4(45

)4(35

)4(25

)4(15

)4(054

2345

012345

555445335225115005

5

aaaaa

aaaaaa

aaaaaa

a

Some Binomial Examples

what is the 6th term in (a + b)9? don’t forget that when you find the 6th term, r = 5

what is the 11th term of (2x + 4)12

54559 12659

baba

22101012 2768240641048576)4(664)2(1012

xxx

Look at the triangle in a different way r0 r1 r2 r3 r4 r5 n = 0 1 n = 1 1 1 n = 2 1 2 1 n = 3 1 3 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 n = 6 1 6 15 20 15 6 1

for a binomial expansion of

(a + b)5, the term for r = 3 has a coefficient of 10

And one more thing… remember that for the inner numbers in the

triangle, any number is the sum of the two numbers above it

for example 4 + 6 = 10 this suggests the following:

which is an example of Pascal’s Identity

25

24

14

11

1 rn

rn

rn

rn

1rn

11

rn

For Example…

69

68

58

47

46

36

How can this help us solve our original problem?

1 5 15 35 70 126

1 4 10 20 35 56

1 3 6 10 15 21

1 2 3 4 5 6

1 1 1 1 1

so by overlaying Pascal’s Triangle over the grid we can see that there are 126 ways to move from one corner to another

How many routes pass through the green square? to get to the green

square, there are C(4,2) ways (6 ways)

to get to the end from the green square there are C(5,3) ways (10 ways)

in total there are 60 ways

How many routes do not pass through the green square? there are 60 ways

that pass through the green square

there are C(9,5) or 126 ways in total

then there must be 126 – 60 = 66 paths that do not pass through the green square

MSIP/ Homework

Read the examples on pages 281-287, in particular the example starting on the bottom of page 287 is important

Complete p. 289 #1, 2aceg, 3, 4, 5, 6, 8, 9, 11, 13