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Section 5.1
Discrete Random Variables
With helpful added content
by D.R.S., University of Cordele
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Discrete Random Variables
Probability Distribution
A probability distribution is a table or formula that
gives the probabilities for every value of the random
variable X, where 0  P  X  x   1 and  P  X  xi   1.
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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.
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%
Total
Confirm – is it = 1 ?
Discrete Random Variables
Random variable
A random variable is a variable whose numeric value is
determined by the outcome of a probability
experiment.
The value is determined by chance,
Or it “could be” determined by “chance”.
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Example 5.1: Creating a Discrete Probability
Distribution
Create a discrete probability distribution for X, the sum
of two rolled dice.
Solution
To begin, let’s list all of the possible values for X.
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Example 5.1: Creating a Discrete Probability
Distribution (cont.)
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Example 5.1: Creating a Discrete Probability
Distribution (cont.)
When rolling two dice, there are 36 possible rolls, each
giving a sum between 2 and 12, inclusive. To find the
probability distribution, we need to calculate the
probability for each value.
1
P  X  2 
because there is only one way to get a
36 sum of 2:
2
because you may get the sum of 3 in
P  X  3 
36 two ways:
or
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Example 5.1: Creating a Discrete Probability
Distribution (cont.)
Continuing this process will give us the following
probability distribution.
Sum of Two Rolled Dice
x
2
3
4
5
6
7
8
9
10
11
12
P(X = x)
1
36
2 3
36 36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Check for yourself that the probabilities listed are the
true values for the probability distribution of X, the sum
of two rolled dice.
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Example 5.1: Creating a Discrete Probability
Distribution (cont.)
Note that all of the probabilities are numbers between
0 and 1, inclusive, and that the sum of the probabilities
is equal to 1. Check this for yourself.
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Another Example
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
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?)”
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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
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
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
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!
•
Excel calculations
The Mean of a Probability
Distribution is easy:
=SUMPRODUCT, just like for a
frequency distribution.
But computing the variance
and standard deviation in Excel
probably requires using the
primitive formulas.
Expected Value
Expected Value
The expected value for a discrete random variable X is
equal to the mean of the probability distribution of X
and is given by
E  X       xi  P  X  xi  
Where xi is the ith value of the random variable X.
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Example 5.2: Calculating Expected Values
Suppose that Randall and Blake decide to make a
friendly wager on the football game they are watching
one afternoon. For every kick the kicker makes, Blake
has to pay Randall $30.00. For every kick the kicker
misses, Randall has to pay Blake $40.00. Prior to this
game, the kicker has made 18 of his past 23 kicks this
season.
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Example 5.2: Calculating Expected Values (cont.)
a. What is the expected value of Randall’s bet for one
kick?
b. Suppose that the kicker attempts four kicks during
the game. How much should Randall expect to win
in total?
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Example 5.2: Calculating Expected Values (cont.)
Solution
a. There are two possible outcomes for this bet:
Randall wins $30.00 (x = 30.00) or Randall loses
$40.00 (x = -40.00). If the kicker has made 18 of his
past 23 kicks, then we assume that the probability
that he will make a kick—and that Randall will win
18
the bet—is
.
23
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Example 5.2: Calculating Expected Values (cont.)
By the Complement Rule, the probability that the kicker
18 5
will miss—and Randall will lose the bet—is 1 -  .
23 23
Randall’s Bet for One Kick
x
$30.00
-$40.00
P(X = x)
18
23
5
23
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Example 5.2: Calculating Expected Values (cont.)
Then we calculate the expected value as follows.
E  X     xi  P  X  xi  
But use TI-84
1-Var Stats
(giving it TWO
lists, one with
values and the
other with
frequencies.
 18 
 5
  30.00      -40.00   
 23 
 23 
540 200

23
23
340

23
 $14.78
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Example 5.2: Calculating Expected Values (cont.)
We see that the expected value of the wager is $14.78.
Randall should expect that if the same bet were made
many times, he would win an average of $14.78 per
bet.
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Example 5.2: Calculating Expected Values (cont.)
b. We know that over the long term Randall would win
an average of $14.78 per bet. So for four attempted
kicks, we multiply the expected value for one bet by
four: 14.78  4  59.12. If he and Blake place four
bets, then Randall can expect to win approximately
$59.12.
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Example 5.3: Calculating Expected Values
Peyton is trying to decide between two different
investment opportunities. The two plans are
summarized in the table below. The left column for
each plan gives the potential earnings, and the right
columns give their respective probabilities. Which plan
should he choose?
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Example 5.3: Calculating Expected Values (cont.)
Investment Plans
Plan A
Plan B
Earnings
Probability
Earnings
Probability
$1200
0.1
$1500
0.3
$950
0.2
$800
0.1
$130
0.4
-$100
0.2
-$575
0.1
-$250
0.2
-$1400
0.2
-$690
0.2
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Example 5.3: Calculating Expected Values (cont.)
Solution
It is difficult to determine which plan will yield the
higher return simply by looking at the probability
distributions. Let’s use the expected values to compare
the plans. Let the random variable X A be the earnings
for Plan A, and let the random variable X B be the
earnings for Plan B.
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Example 5.3: Calculating Expected Values (cont.)
For Investment Plan A:
E  X A     xi  P  X A  xi  
Again, use TI-84 1-Var Stats
with two lists is recommended,
instead of this by-hand
calculation.
 1200 0.1   950  0.2  130  0.4 
  -575 0.1   -1400  0.2
 120  190  52 - 57.5 - 280
 $24.50
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Example 5.3: Calculating Expected Values (cont.)
For Investment Plan B:
E  XB     xi  P  XB  xi  
Again, use TI-84 1-Var Stats
with two lists is recommended,
instead of this by-hand
calculation.
 1500  0.3   800  0.1
  -100  0.2   -250  0.2   -690  0.2
 450  80 - 20 - 50 - 138
 $322.00
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Example 5.3: Calculating Expected Values (cont.)
From these calculations, we see that the expected
value of Plan A is $24.50, and the expected value of
Plan B is $322.00. Therefore, Plan B appears to be the
wiser investment option for Peyton.
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Variance and Standard Deviation for a Discrete
Probability Distribution
Variance and Standard Deviation for a Discrete
Probability Distribution
The variance for a discrete probability distribution of a
random variable X is given by
s 2    xi 2  P  X  xi   - m2
2

   xi - m  P  X  xi  


Where x i is the i th value of the random variable X and
μ is the mean of the probability distribution.
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Formula
Variance and Standard Deviation for a Discrete
Probability Distribution (cont.)
The standard deviation for a discrete probability
distribution of a random variable X is the square root
of the variance, given by the following formulas.
s  s2 
2
2


x

P
X

x
m
  i  i 
2

   xi - m  P  X  xi  


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Example 5.4: Calculating the Variances and Standard
Deviations for Discrete Probability Distributions
Which of the investment plans in the previous example
carries more risk, Plan A or Plan B?
Solution
To decide which plan carries more risk, we need to look
at their standard deviations, which requires that we
first calculate their variances. Let’s calculate the
variance separately for each investment plan. To do
this, we will use a table to organize our calculations as
we compute the variance. We will use the expected
values that we calculated in the previous example as
the means.
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Example 5.4: Calculating the Variances and Standard
Deviations for Discrete Probability Distributions (cont.)
Again, use TI-84 1-Var Stats
with two lists is recommended,
instead of this by-hand
calculation.
For Investment Plan A,
m E  X A   $24.50.
Investment Plan A
x - 
2
 x -   P  X  x 
2
x
P(X = x)
x -
$1200
0.1
1175.50
1,381,800.25
138,180.025
$950
0.2
925.50
856,550.25
171,310.05
$130
0.4
105.60
11,130.25
4452.1
-$575
0.1
-599.50
359,400.25
35,940.025
-$1400
0.2
-1424.50
2,029,200.25
405,840.05
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Example 5.4: Calculating the Variances and Standard
Deviations for Discrete Probability Distributions (cont.)
2

    xi - m  P  X A  xi  


 138,180.025  171,310.05  4452.1
 35, 940.025  405,840.05
 755,722.25
2
  2
 755,722.25
 $869.32
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Again, use TI-84 1-Var Stats
with two lists is recommended,
instead of this by-hand
calculation.
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Example 5.4: Calculating the Variances and Standard
Deviations for Discrete Probability Distributions (cont.)
Again, use TI-84 1-Var Stats
with two lists is recommended,
instead of this by-hand
calculation.
For Investment Plan B,
  E  XB   $322.00.
Investment Plan B
 x - 2  P  X  x 
x
P(X = x)
x -
 x -  2
$1500
0.3
1178
1,387,684
416,305.2
$800
0.1
478
228,484
22,848.4
-$100
0.2
422
178,084
35,616.8
-$250
0.2
572
327,184
65,436.8
-$690
0.2
1012
1,024,144
204,828.8
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Example 5.4: Calculating the Variances and Standard
Deviations for Discrete Probability Distributions (cont.)
2

    xi - m  P  XB  xi  


 416,305.2  22,848.4  35,616.8
65,436.8  204,828.8
 745,036
Again, use TI-84 1-Var Stats
2
  2
with two lists is recommended,
instead of this by-hand
calculation.
 745,036
 $863.15
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Example 5.4: Calculating the Variances and Standard
Deviations for Discrete Probability Distributions (cont.)
What do these results tell us? Comparing the standard
deviations, we see that not only does Plan B have a
higher expected value, but its profits vary slightly less
than those of Plan A. We may conclude that Plan B
carries a slightly lower amount of risk than Plan A.
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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