1. Discrete Random Variables
Section 5.1
Discrete Random Variables
With helpful added content
by D.R.S., University of Cordele

2. 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

3. Discrete is our focus for now
Continuous
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
Will talk about continuous probability distributions in future chapters.

4. Examples of Probability Distributions
Rolling a single die
Total of rolling two dice
Value
Probability 1
1/6 2 3 4 5 6
Total
Value
Prob. 2
1/36 8
5/36 3
2/36 9
4/36 4
3/36 10 5 11 6 12 7
6/36
Total 1
(Note that it’s a two-column chart but we had to typeset it this way to fit it onto the slide.)

5. Example of a Probability Distribution http://en. wikipedia
Draw this 5-card poker hand
Probability
Royal Flush %
Straight Flush (not including Royal Flush) %
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%

6. Exact fractions avoid rounding errors (but is it useful to readers?)
Draw this 5-card poker hand
Probability
Royal Flush
4 / 2,598,960
Straight Flush (not including Royal Flush)
36 / 2,598,960
Four of a Kind
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

7. Example of a probability distribution “How effective is Treatment X?”
Outcome
Probability
The patient is cured.
85%
The patient’s condition improves.
10%
There is no apparent effect. 4%
The patient’s condition deteriorates. 1%
Total
Confirm – is it = 1 ?

8. 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”.

9. 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.

10. Example 5.1: Creating a Discrete Probability Distribution (cont.)

11. 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. because there is only one way to get a sum of 2: because you may get the sum of 3 in two ways: or

12. Example 5.1: Creating a Discrete Probability Distribution (cont.)
Continuing this process will give us the following probability distribution. Check for yourself that the probabilities listed are the true values for the probability distribution of X, the sum of two rolled dice.
Sum of Two Rolled Dice x 2 3 4 5 6 7 8 9 10 11 12
P(X = x)

13. 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.

14. Another Example Start with a frequency distribution
General layout
A specific made-up example
Outcome
Count of occurrences
How many children live here?
Number of households 50 1
100 2
150 3 80 4 40 5 20
6 or more 10
Total responses
450

15. Include a Relative Frequency column
General layout
A specific simple example
Outcome
Count of occur-rences
Relative
Frequency
=count ÷ total
# of children
Number of households
Relative Frequency 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

16. You can drop the count column
General layout
A specific simple example
Outcome
Relative
Frequency
=count ÷ total
# of children
Relative Frequency
0.108 1
0.239 2
0.326 3
0.174 4
0.087 5
0.043 6+
0.022
Total
1.000

17. Answer Probability Questions
What is the probability …
A specific simple example
…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?
# of children
Relative Frequency
0.108 1
0.239 2
0.326 3
0.174 4
0.087 5
0.043 6+
0.022
Total
1.000

18. 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?)”

19. Tossing coin and counting Heads
One Coin
Four Coins
How many heads
Probability
1 / 2 1
Total
How many heads
Probability
1/16 1
4/16 2
6/16 3 4
Total

20. Graphical Representation
Histogram, for example
Four Coins
How many heads
Probability
1/16 1
4/16 2
6/16 3 4
Total
Probability
6/16
4/16
1/16
heads

21. Shape of the distribution
Histogram, for example
Distribution shapes matter!
This one is a bell-shaped distribution
Rolling a single die: its graph is a uniform distribution
Other distribution shapes can happen, too
Probability
6/16
3/16
1/16
heads

22. Remember the Structure
Required features
Example of a
Discrete Probability Distribution
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
# of children
Relative Frequency
0.108 1
0.239 2
0.326 3
0.174 4
0.087 5
0.043 6+
0.022
Total
1.000

23. The Formulas
MEAN: 𝜇= 𝑋∙𝑃(𝑋) VARIANCE: 𝜎 2 = 𝑋 2 ∙𝑃(𝑋) − 𝜇 2 STANDARD DEVIATION: 𝜎= 𝜎 2

24. Practice Calculations
Rolling one die
Statistics
Value
Probability 1
1/6 2 3 4 5 6
Total
The mean is 𝜇=3.5
The variance is 𝜎 2 =2.92
The standard deviation is 𝜎=1.71

25. Practice Calculations
Statistics
Total of rolling two dice
The mean is 𝜇=7
The variance is 𝜎 2 =5.83
The standard deviation is 𝜎=2.42
Value
Prob. 2
1/36 8
5/36 3
2/36 9
4/36 4
3/36 10 5 11 6 12 7
6/36
Total 1

26. Practice Calculations
One Coin
Statistics
How many heads
Probability
1 / 2 1
Total
The mean is 𝜇=0.5
The variance is 𝜎 2 =0.25
The standard deviation is 𝜎=0.5

27. Practice Calculations
Statistics
Four Coins
The mean is 𝜇=2
The variance is 𝜎 2 =1.00
The standard deviation is 𝜎=1.00
How many heads
Probability
1/16 1
4/16 2
6/16 3 4
Total

28. 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!

29. 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.

30. 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 Where xi is the ith value of the random variable X.

31. 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 $ For every kick the kicker misses, Randall has to pay Blake $ Prior to this game, the kicker has made 18 of his past 23 kicks this season.

32. 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?

33. 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 = ). 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 the bet—is

34. Example 5.2: Calculating Expected Values (cont.)
By the Complement Rule, the probability that the kicker will miss—and Randall will lose the bet—is
Randall’s Bet for One Kick x
$30.00
-$40.00
P(X = x)

35. Example 5.2: Calculating Expected Values (cont.)
Then we calculate the expected value as follows.
But use TI-84
1-Var Stats
(giving it TWO lists, one with values and the other with frequencies.

36. Example 5.2: Calculating Expected Values (cont.)
We see that the expected value of the wager is $ Randall should expect that if the same bet were made many times, he would win an average of $14.78 per bet.

37. 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: If he and Blake place four bets, then Randall can expect to win approximately $59.12.

38. 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?

39. Example 5.3: Calculating Expected Values (cont.)
Investment Plans
Plan A
Plan B
Earnings
Probability
$1200
0.1
$1500
0.3
$950
0.2
$800
$130
0.4
-$100
-$575
-$250
-$1400
-$690

40. 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 be the earnings for Plan A, and let the random variable be the earnings for Plan B.

41. Example 5.3: Calculating Expected Values (cont.)
Again, use TI-84 1-Var Stats
with two lists is recommended, instead of this by-hand calculation.
For Investment Plan A:

42. Example 5.3: Calculating Expected Values (cont.)
Again, use TI-84 1-Var Stats
with two lists is recommended, instead of this by-hand calculation.
For Investment Plan B:

43. 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 $ Therefore, Plan B appears to be the wiser investment option for Peyton.

44. 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 Where is the value of the random variable X and μ is the mean of the probability distribution.

45. 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.

46. 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.

47. 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,
Investment Plan A x
P(X = x)
$1200
0.1
1,381,800.25
138,
$950
0.2
925.50
856,550.25
171,310.05
$130
0.4
105.60
11,130.25
4452.1
-$575
359,400.25
35,
-$1400
2,029,200.25
405,840.05

48. 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.

49. 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,
Investment Plan B x
P(X = x)
$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
572
327,184
65,436.8
-$690
1012
1,024,144
204,828.8

50. 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.

51. 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.

52. Expected Value Problems
The Situation
The Discrete Probability Distr.
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?
Outcome
Net Value
Probability
Win first prize
$1,995
1/1000
Win second prize
$995
Win third prize
$495
2/1000
Win fourth prize
$95
3/1000
Loser
$ -5
993/1000
Total
1000/1000

53. Expected Value Problems
Statistics
The Discrete Probability Distr.
The mean of this probability is $ , 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.”
Outcome
Net Value
Probability
Win first prize
$1,995
1/1000
Win second prize
$995
Win third prize
$495
2/1000
Win fourth prize
$95
3/1000
Loser
$ -5
993/1000
Total
1000/1000

54. Expected Value Problems
Another way to do it
The Discrete Probability Distr.
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
Outcome
Net Value
Probability
Win first prize
$2,000
1/1000
Win second prize
$1,000
Win third prize
$500
2/1000
Win fourth prize
$100
3/1000
Loser
$ 0
993/1000
Total
1000/1000

55. Expected Value Problems
The Situation
The Discrete Probability Distr.
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?
Outcome
Net Value
Probability
No claims filed
_______
An $800 fender bender
0.004
An $8,000 accident
0.002
A wreck, it’s totaled

56. 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