Download 5.1

Discrete Probability Distributions
To accompany Hawkes lesson 5.1
Original content by D.R.S.
Examples of Probability Distributions
Rolling a single die
Total of rolling two dice
Value
Probability
Value
Prob.
Value
Prob.
1
1/6
2
1/36
8
5/36
2
1/6
3
2/36
9
4/36
3
1/6
4
3/36
10
3/36
4
1/6
5
4/36
11
2/36
5
1/6
6
5/36
12
1/36
6
1/6
7
6/36
Total
Total
1
(Note that it’s a two-column chart but
we had to typeset it this way to fit it
onto the slide.)
1
Example of a Probability Distribution
http://en.wikipedia.org/wiki/Poker_probability
Draw this 5-card poker hand
Probability
Royal Flush
0.000154%
Straight Flush (not including Royal Flush)
0.00139%
Four of a Kind
0.0240%
Full House
0.144%
Flush (not including Royal Flush or Straight Flush)
0.197%
Straight (not including Royal Flush or Straight Flush)
0.392%
Three of a Kind
2.11%
Two Pair
4.75%
One Pair
42.3%
Something that’s not special at all
50.1%
Total (inexact, due to rounding) 100%
Exact fractions avoid rounding errors
(but is it useful to readers?)
Draw this 5-card poker hand
Royal Flush
Probability
4 / 2,598,960
Straight Flush (not including Royal Flush)
Four of a Kind
36 / 2,598,960
624 / 2,598,960
Full House
3,744 / 2,598,960
Flush (not including Royal Flush or Straight Flush)
5,108 / 2,598,960
Straight (not including Royal Flush or Straight Flush)
10,200 / 2,598,960
Three of a Kind
54,912 / 2,598,960
Two Pair
123,552 / 2,598,960
One Pair
1,098,240 / 2,598,960
Something that’s not special at all
1,302,540 / 2,598,960
Total (exact, precise, beautiful fractions) 2,598,600 / 2,598,600
Example of a probability distribution
“How effective is Treatment X?”
Outcome
Probability
The patient is cured.
The patient’s condition improves.
There is no apparent effect.
The patient’s condition deteriorates.
85%
10%
4%
1%
A Random Variable
• The value of “x” is determined by chance
• Or “could be” determined by chance
• As far as we know, it’s “random”, “by chance”
• The important thing: it’s some value we get
in a single trial of a probability experiment
• It’s what we’re measuring
Discrete vs. Continuous
Discrete
• A countable number of
values
Continuous
• All real numbers in some
interval
• “Red”, “Yellow”, “Green”
• An age between 10 and 80
(10.000000 and 80.000000)
• 2 of diamonds, 2 of hearts,
… etc.
• A dollar amount
• 1, 2, 3, 4, 5, 6 rolled on a die
• A height or weight
Discrete is our focus for now
Discrete
• A countable number of
values (outcomes)
• “Red”, “Yellow”, “Green”
• “Improved”, “Worsened”
• 2 of diamonds, 2 of hearts,
… etc.
• What poker hand you draw.
• 1, 2, 3, 4, 5, 6 rolled on a die
• Total dots in rolling two dice
Continuous
• Will talk about continuous
probability distributions in
future chapters.
Start with a frequency distribution
General layout
• Outcome
Count of
occurrences
A specific made-up example
How many children
live here?
Number of
households
0
50
1
100
2
150
3
80
4
40
5
20
6 or more
10
Total responses
450
Include a Relative Frequency column
General layout
• Outcome Count
of
occurrences
A specific simple example
Relative
Frequency
=count ÷
total
# of
child
ren
Number of
households
Relative
Frequency
0
50
0.108
1
110
0.239
2
150
0.326
3
80
0.174
4
40
0.087
5
20
0.043
6+
10
0.022
Total
460
1.000
You can drop the count column
General layout
• Outcome
Relative
Frequency
=count ÷ total
A specific simple example
# of children
Relative
Frequency
0
0.108
1
0.239
2
0.326
3
0.174
4
0.087
5
0.043
6+
0.022
Total
1.000
Sum MUST BE EXACTLY 1 !!!
• In every Probability Distribution, the total of
the probabilities must always, every time,
without exception, be exactly 1.00000000000.
– In some cases, it might be off a hair because of
rounding, like 0.999 for example.
– If you can maintain exact fractions, this rounding
problem won’t happen.
Answer Probability Questions
What is the probability …
• …that a randomly selected
household has exactly 3
children?
• …that a randomly selected
household has children?
• … that a randomly selected
household has fewer than 3
children?
• … no more than 3 children?
A specific simple example
# of children
Relative
Frequency
0
0.108
1
0.239
2
0.326
3
0.174
4
0.087
5
0.043
6+
0.022
Total
1.000
Answer Probability Questions
Referring to the Poker probabilities table
• “What is the probability of drawing a Four of a
Kind hand?”
• “What is the probability of drawing a Three of a
Kind or better?”
• “What is the probability of drawing something
worse than Three of a Kind?”
• “What is the probability of a One Pair hand twice
in a row? (after replace & reshuffle?)”
Theoretical Probabilities
Rolling one die
Total of rolling two dice
Value
Probability
Value
Prob.
Value
Prob.
1
1/6
2
1/36
8
5/36
2
1/6
3
2/36
9
4/36
3
1/6
4
3/36
10
3/36
4
1/6
5
4/36
11
2/36
5
1/6
6
5/36
12
1/36
6
1/6
7
6/36
Total
1
Total
1
Tossing coin and counting Heads
One Coin
Four Coins
How many heads
Probability
How many heads
Probability
0
1/2
0
1/16
1
1/2
1
4/16
Total
1
2
6/16
3
4/16
4
1/16
Total
1
Tossing coin and counting Heads
How did we get this?
• Could try to list the entire
sample space: TTTT, TTTH,
TTHT, TTHH, THTT, etc.
• Could use a tree diagram to
get the sample space.
• Could use nCr
combinations.
• We will formally study The
Binomial Distribution soon.
Four Coins
How many heads
Probability
0
1/16
1
3/16
2
6/16
3
3/16
4
1/16
Total
1
Graphical Representation
Histogram, for example
Four Coins
How many heads
Probability
0
1/16
1
4/16
2
6/16
6/16
3
4/16
4/16
4
1/16
Total
1
Probability
1/16
0
1
2
3
4 heads
Shape of the distribution
Histogram, for example
Probability
Distribution shapes matter!
• This one is a bell-shaped
distribution
• Rolling a single die: its graph
is a uniform distribution
6/16
• Other distribution shapes
can happen, too
3/16
1/16
0
1
2
3
4 heads
Remember the Structure
Required features
• The left column lists the
sample space outcomes.
• The right column has the
probability of each of the
outcomes.
• The probabilities in the right
column must sum to exactly
1.0000000000000000000.
Example of a
Discrete Probability Distribution
# of children
Relative
Frequency
0
0.108
1
0.239
2
0.326
3
0.174
4
0.087
5
0.043
6+
0.022
Total
1.000
The Formulas
• MEAN: 𝜇 =
𝑋 ∙ 𝑃(𝑋)
• VARIANCE: 𝜎 2 =
𝑋 2 ∙ 𝑃(𝑋) − 𝜇2
• STANDARD DEVIATION: 𝜎 = 𝜎 2
TI-84 Calculations
• Put the outcomes into a
TI-84 List (we’ll use L1)
• Put the corresponding
probabilities into
another TI-84 List (we’ll
use L2)
• 1-Var Stats L1, L2
• You can type fractions
into the lists, too!
•
Practice Calculations
Rolling one die
Statistics
Value
Probability
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
Total
1
• The mean is 𝜇 = 3.5
• The variance is 𝜎 2 = 2.92
• The standard deviation is
𝜎 = 1.71
Practice Calculations
Statistics
Total of rolling two dice
• The mean is 𝜇 = 7
2
• The variance is 𝜎 = 5.83
• The standard deviation is
𝜎 = 2.42
Value
Prob.
Value
Prob.
2
1/36
8
5/36
3
2/36
9
4/36
4
3/36
10
3/36
5
4/36
11
2/36
6
5/36
12
1/36
7
6/36
Total
1
Practice Calculations
One Coin
Statistics
How many heads
Probability
0
1/2
1
1/2
Total
1
• The mean is 𝜇 = 0.5
• The variance is 𝜎 2 = 0.25
• The standard deviation is
𝜎 = 0.5
Practice Calculations
Statistics
Four Coins
• The mean is 𝜇 = 2
2
• The variance is 𝜎 = 1.00
• The standard deviation is
𝜎 = 1.00
How many heads
Probability
0
1/16
1
4/16
2
6/16
3
4/16
4
1/16
Total
1
Expected Value
• Probability Distribution with THREE columns
– Event
– Probability of the event
– Value of the event (sometimes same as the event)
• Examples:
– Games of chance
– Insurance payoffs
– Business decisions
Expected Value Problems
The Situation
• 1000 raffle tickets are sold
• You pay $5 to buy a ticket
• First prize is $2,000
• Second prize is $1,000
• Two third prizes, each $500
• Three more get $100 each
• The other ____ are losers.
What is the “expected value”
of your ticket?
The Discrete Probability Distr.
Outcome
Win first
prize
Net Value
Probability
$1,995
1/1000
Win second
prize
$995
1/1000
Win third
prize
$495
2/1000
Win fourth
prize
$95
3/1000
Loser
$ -5
993/1000
Total
1000/1000
Expected Value Problems
Statistics
• The mean of this probability
is $ - 0.70, a negative value.
• This is also called “Expected
Value”.
• Interpretation: “On the
average, I’m going to end up
losing 70 cents by investing
in this raffle ticket.”
The Discrete Probability Distr.
Outcome
Win first
prize
Net Value
Probability
$1,995
1/1000
Win second
prize
$995
1/1000
Win third
prize
$495
2/1000
Win fourth
prize
$95
3/1000
Loser
$ -5
993/1000
Total
1000/1000
Expected Value Problems
Another way to do it
• Use only the prize values.
• The expected value is the
mean of the probability
distribution which is $4.30
• Then at the end, subtract
the $5 cost of a ticket, once.
• Result is the same, an
expected value = $ -0.70
The Discrete Probability Distr.
Outcome
Net Value
Probability
Win first
prize
$2,000
1/1000
Win second
prize
$1,000
1/1000
Win third
prize
$500
2/1000
Win fourth
prize
$100
3/1000
Loser
Total
$0
993/1000
1000/1000
Expected Value Problems
The Situation
• We’re the insurance
company.
• We sell an auto policy for
$500 for 6 months coverage
on a $20,000 car.
• The deductible is $200
What is the “expected value” –
that is, profit – to us, the
insurance company?
The Discrete Probability Distr.
Outcome
Net Value
Probability
No claims
filed
_______
An $800
fender
bender
0.004
An $8,000
accident
0.002
A wreck,
it’s totaled
0.002
An Observation
• The mean of a probability distribution is really
the same as the weighted mean we have seen.
• Recall that GPA is a classic instance of
weighted mean
– Grades are the values
– Course credits are the weights
• Think about the raffle example
– Prizes are the values
– Probabilities of the prizes are the weights