probability distributions and expected value
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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. - PowerPoint PPT PresentationTRANSCRIPT
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Probability Distributions and Expected Value
Chapter 5.1 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U
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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}
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Probability Distribution 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:
Rolling A Die
=outcome probability <new>
123456
1 1/62 1/63 1/64 1/65 1/66 1/6
Co
un
t
1
outcome0 1 2 3 4 5 6 7
Rolling A Die Histogram
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Probability Distribution for 2 Dice
Co
un
t
1
2
3
4
5
6
7
sum2 4 6 8 10 12
Two Dice Histogram
RollingDice
=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
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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
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Probability Distribution for 3 DiceOutcome 3 4 5 6 7 8 9 10
# cases 1 3 6 10 15 21 28
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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
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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
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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
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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
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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 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
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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
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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)
X 0 women 1 woman 2 women 3 women
P(X) 4/35 18/35 12/35 1/35
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MSIP / Homework
p. 277 #1, 2, 3, 4, 5, 9, 12, 13
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Pascal’s Triangle and the Binomial Theorem
Chapter 5.2 – Probability Distributions and PredictionsMathematics of Data Management (Nelson)MDM 4U
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How many routes are there to the top right-hand corner? 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
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11 1
1 2 11 3 3 1
1 4 6 4 11 5 10 10 5 1
Pascal’s Triangle 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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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
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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’)
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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 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/Vienna/9349/pascal.mid
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Pascal’s Triangle – Cool Stuff numbers
divisible by 5 similar
patterns exist for other numbers
http://www.shodor.org/interactivate/activities/pascal1/
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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
0
1
2
4
3
4
4
4
0
5
1
5
2
5
3
5
1
1
0
2
1
2
2
2
0
3
1
3
2
3
3
3
0
4
1
4
4
5
5
5
0
6
1
6
2
6
3
6
4
6
5
6
6
6
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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
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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.
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So what does this have to do with the Binomial Theorem?
remember that: (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
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The Binomial Theorem
nrrnnnn
n
bn
nba
r
nba
nba
na
n
ba
......210
)(
221
rrnn bar
nba
is term t the)( binomial for the so 1r
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A Binomial Expansion
lets expand (x + y)4
432234
444334224114004
4
464
4
4
3
4
2
4
1
4
0
4
yxyyxyxx
yxyxyxyxyx
yx
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Another Binomial Expansion
lets expand (a + 4)5
1024128064016020
)1024()1()256()5()64()10()16()10()4()5()1()1(
45
54
4
54
3
54
2
54
1
54
0
5
4
2345
012345
555445335225115005
5
aaaaa
aaaaaa
aaaaaa
a
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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 1265
9baba
22101012 2768240641048576)4(664)2(10
12xxx
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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
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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
2
5
2
4
1
4
1
1
1 r
n
r
n
r
n
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For Example…
6
9
6
8
5
8
4
7
4
6
3
6
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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
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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
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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
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Exercises / Homework
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