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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 - PowerPoint PPT Presentation
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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
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