Download Chapter5.1

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 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 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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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
# ways
1
4
5
6
7
8
9
…
Probability Distribution for 3 Dice
Outcome 3
4
5
6
7
8
9
# cases
3
6
10
15
21
28
1
10
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/are the
most common
outcome(s)?
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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p. 277 #1, 2, 3, 4, 5, 9, 12, 13