Transcript Chapter5.1to5.2

Probability Distributions and
Expected Value
Chapter 5.1 – Probability Distributions
and Predictions
Mathematics of Data Management
(Nelson)
MDM 4U
Probability Distributions of a Discrete
Random Variable
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a discrete random variable X is one that can
take on only a finite number 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 complete 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 Rolling A Die
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a 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:
outcome
1
=
2
3
4
5
6
probability <new>
1
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1/6
Rolling A Die
Histogram
1
Count
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0
1
2
3
4
outcome
5
6
7
Probability Distribution for 2 Dice
RollingDice
Two Dice
7
6
Count
5
4
3
2
1
2
sum
Histogram
4
6
8
sum
10
12
1
=
2
3
4
5
6
7
8
9
10
11
2
3
4
5
6
7
8
9
10
11
12
probability <new>
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
What would a probability distribution
graph for three dice look like?
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lets try it! Using three dice, figure out how many
possible cases there are
now find out how many possible ways there are to
create each of the possible cases
fill in a table like the one below
now you can make your graph
Outcome 3
# cases
1
4
5
6
7
8
9
…
So what does an experimental distribution
look like?
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A simulated dice
throw was done a
million times using a
computer program
and generated the
following data
What is the most
common outcome?
Does this make
sense?
Line Scatter Plot
Rolling 3 Dice
140000
120000
100000
Freq
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80000
60000
40000
20000
0
2
4
6
8
10
12
roll
14
16
18
20
Back to 2 Dice
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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
E (sum of 2 dice)
1
2
3
1
 2   3   4   ...  12 
36
36
36
36
252

7
36
n
 E ( X )   xi P( X  xi )
i 1
Example 1a: tossing 3 coins
X
P(X)
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0 heads 1 head 2 heads 3 heads
⅛
⅜
⅜
⅛
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
Example 1b: tossing 3 coins
X
P(X)
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0 heads 1 head 2 heads 3 heads
⅛
⅜
⅜
⅛
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
Example 2a: Selecting a Committee of
three people from a group of 4 men and 3
women
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What is the probability of having at least one
woman on the team?
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 have 3 women
Example 2a cont’d: selecting a committee
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X
0 women
1 woman
2 women
3 women
P(X)
4/35
18/35
12/35
1/35
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
Example 2b: selecting a committee
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X
0 women
1 woman
2 women
3 women
P(X)
4/35
18/35
12/35
1/35
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
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page 277 #1, 2, 3, 4, 5, 9, 12, 13
Thu begin 5.2 – Fri complete
independently
Pascal’s Triangle and the
Binomial Theorem
Chapter 5.2 – Probability Distributions
and Predictions
Mathematics of Data Management
(Nelson)
MDM 4U
How many routes are there to the top
right-hand corner?
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you need to move up 4
spaces and over 5
spaces
This is the same as
rearranging the letters
NNNNEEEEE
This can be calculated
by C(9,4) or C(9,5)
= 126 ways
Pascal’s Triangle
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1
1
1
the outer values
are always 1
 the inner values
are determined by
adding the values
of the two values
diagonally above
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2 1
1 3 3 1
1 4 6 4 1
 1 5 10 10 5 1
Pascal’s Triangle
sum of each row is a power of 2
 1
 1 = 20
 1 1
 2 = 21
 1 2 1
 4 = 22
 1 3 3 1
 8 = 23
 1 4 6 4 1
 16 = 24
 1 5 10 10 5 1
 32 = 25
 1 6 15 20 15 6 1
 64 = 26
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Pascal’s Triangle
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1
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1 1
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 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
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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
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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
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each diagonal is summed
up in the next value below
and to the left
called the “hockey stick”
property
there may even be music
hidden in it
http://www.geocities.com/Vi
enna/9349/pascal.mid
Pascal’s Triangle – Cool Stuff
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numbers
divisible by 5
similar
patterns exist
for other
numbers
http://www.shodor.
org/interactivate/ac
tivities/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
0
 
0
 1   1
   
 0   1
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 2
 
0
 2
 
1
 2
 
 2
 3   3  3   3
       
 0  1  2   3
 4  4  4  4  4
         
 0 1  2  3  4
5
 
0
6
 
0
5  5 5  5 5
         
1  2  3  4   5
6  6 6  6 6
6
         
 
5
1  2  3  4  
6
Pascal’s Triangle - Summary
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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
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sum of nth row is 2n
Begin count at 0
number inside a row is the sum of the two
numbers above it
The Binomial Theorem
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the term (a + b) can be expanded:
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(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
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remember that:
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(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
and the triangle’s 4th row is 1 4 6 4 1
so Pascal’s Triangle allows you to predict the
coefficients in the binomial expansion
notice also that the exponents on the variables
also form a predictable pattern with the
exponents of each term having a sum of n
The Binomial Theorem
( a  b) n
 n  n  n  n 1  n  n  2 2
 n  nr r
n n
  a   a b   a b  ...   a b  ...   b
0
1
 2
r
n
 n  nr r
so for the binomial (a  b) the t r 1 term is  a b
r
n
A Binomial Expansion
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lets expand (x + y)4
x  y 4
 4  4 0 0  4  41 1  4  4  2 2  4  43 3  4  4  4 4
   x y    x y    x y    x y    x y
 0
1
 2
 3
 4
 x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4
Another Binomial Expansion
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lets expand (a + 4)5
a  45
 5  50 0  5  51 1  5  5 2 2  5  53 3  5  5 4 4  5  55 5
  a 4   a 4   a 4   a 4   a 4   a 4
 0
1
 2
 3
 4
 5
 (1)a 5 (1)  (5)a 4 (4)  (10)a 3 (16)  (10)a 2 (64)  (5)a1 (256)  (1)a 0 (1024)
 a 5  20a 4  160a 3  640a 2  1280a  1024
Some Binomial Examples
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what is the 6th term in (a + b)9?
don’t forget that when you find the 6th term, r = 5
 9  9 5 5
4 5
 a b  126 a b
 5
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what is the 11th term of (2x + 4)12
12 
 (2 x)1210 410  66 (4 x 2 )1048576  276824064 x 2
10 
Look at the triangle in a different way
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n=0
n=1
n=2
n=3
n=4
n=5
n=6
r0
1
1
1
1
1
1
1
r1 r2 r3 r4 r5
1
2
3
4
5
6
1
3
6
10
15
1
4 1
10 5 1
20 15 6 1
for a binomial
expansion of
(a + b)5, the term for
r = 3 has a
coefficient of 10
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And one more thing…
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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:
 4  4  5
       
1  2  2
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which is an example of Pascal’s Identity
 n   n   n  1
   
  

 r   r  1  r  1 
For Example…
 6  6 7 
       
 3 4  4 
8   8   9 
       
5   6   6 
How can this help us solve our original
problem?
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so by overlaying
Pascal’s Triangle
over the grid we
can see that there
are 126 ways to
move from one
corner to another
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
How many routes pass through the green
square?
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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?
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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
Exercises / Homework
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Homework: read the examples on pages
281-287, in particular the example starting
on the bottom of page 287 is important
page 289 #1, 2aceg, 3, 4, 5, 6, 8, 9, 11, 13