CHAPTER 2
DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
1.
2.
3.
4.
5.
Random Variable
Probability Distribution
Expected Value of a Random Variable
The Variance and Standard Deviation of a Random Variable
4.1. Arithmetic Properties Of Expected Value And Variance
4.1.1. Standardized Random Variable
The Binomial Distribution
5.1. Cumulative Probability Distribution
5.2. How to Use Excel to Find Binomial Probabilities
5.3. Expected Value and Variance of the Binomial Distribution
1. Random Variable
A random variable is a variable whose values are determined through a random experiment or process. In
other words, a random variable is a variable whose value cannot be predicted exactly. The value is not
known in advance; it is not known until after the random experiment is conducted.
Example 1
As a random experiment, toss a coin. This random experiment has two outcomes: H and T. Let’s assign the
value 0 to H, (H = 0), and 1 to T, (T = 1). Counting the number of 1’s in each outcome provides the values
assigned to x. If you toss two coins, then the number of tails is either 0, 1, or 2:
Outcomes of the random
experiment
(0,0)
(0,1), (1,0)
(1,1)
Values of random variable
x
Number of tails
0
1
2
Example 2
When you toss a pair of dice, let 𝑥 denote the sum of the number of dots appearing on top. These numbers, as
they appear in the top row below, are assigned to x through the outcomes of the random experiment. The
outcomes are shown below.
Chapter 2—Discrete Random Variables and Probability Distributions
Page 1 of 17
Outcomes of the random experiment
(1,1)
Values of random variable x
Sum of dots
2
(1,2), (2,1)
3
(1,3), (2, 2), (3,1)
4
(1,4), (2,3), (3,2), (4,1)
5
(1,5), (2,4), (3,3), (4,2), (5,1)
6
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
7
(2,6), (3,5), (4,4), (5,3), (6,2)
8
(3,6), (4,5), (5,4), (6,3)
9
(4,6), (5,5), (6,4)
10
(5,6), (6,5)
11
(6,6)
12
Example 3
When you are guessing the answers to a set of 5 multiple-choice questions, you are conducting a random
experiment. Assign 0 to the incorrect answer and 1 to the correct answer for each question, and let x denote
the number of possible correct answers: 0, 1, 2, 3, 4, 5. These numbers are assigned to x through the
outcomes of the random experiment as shown below. There are 32 possible outcomes in this random
experiment.1
x
Correct
guesses
Outcomes of the random experiment
(0,0,0,0,0)
0
(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1),
1
(1,1,0,0,0), (1,0,1,0,0), (1,0,0,1,0), (1,0,0,0,1), (0,1,1,0,0), (0,1,0,1,0), (0,1,0,0,1), (0,0,1,1,0), (0,0,1,0,1), (0,0,0,1,1)
2
(1,1,1,0,0), (1,1,0,1,0), (1,1,0,0,1), (1,0,1,1,0), (1,0,1,0,1), (1,0,0,1,1), (0,1,1,1,0), (0,1,1,0,1), (0,1,0,1,1), (0,0,1,1,1)
3
(1,1,1,1,0), (1,1,1,0,1), (1,1,0,1,1), (1,0,1,1,0), (0,1,1,1,1)
4
(1,1,1,1,1)
5
2. Probability Distribution
The probability distribution of a (discrete) random variable is the set of all possible values of the
random variable along with the probability corresponding to each value. Since a probability distribution
lists all possible values of the random variable, then the sum of the probabilities must equal one (1).
Example 4
Let 𝑥 be the random variable denoting the number of tails when tossing two coins as in Example 1. Write the
probability distribution of 𝑥.
As Example 1 showed, there are four outcomes when tossing two coins, each assigning a discrete value to 𝑥.
Since each outcome is equally likely then the probability distribution is:
𝑥
0
1
2
𝑓(𝑥)
0.25
0.50
0.25
1.00
The graph of the above probability distribution is:
Each trial has two outcomes (failure or success—0 or 1). The experiment has five trials. Therefore, the total number of
outcomes is 2⁵ = 32.
1
Chapter 2—Discrete Random Variables and Probability Distributions
Page 2 of 17
Probability Distribution of Number of Tails in
Tossing Two Coins
f(x)
0.6
0.50
0.5
0.4
0.3
0.25
0.25
0.2
0.1
0
0
1
2
x (number of tails)
Example 5
Let x be the random variable denoting the sum of dots appearing on top when tossing a pair of dice. Write the
probability distribution of x.
As Example 2 showed, there are 36 outcomes of (ordered) pairs of numbers generating the values of x. Since
each outcome is equally likely, then the probability distribution of x can be written as follows:
x
𝑓(𝑥)
2
3
4
5
6
7
8
9
10
11
12
1 ∕ 36 = 0.0278
2 ∕ 36 = 0.0556
3 ∕ 36 = 0.0833
4 ∕ 36 = 0.1111
5 ∕ 36 = 0.1389
6 ∕ 36 = 0.1667
5 ∕ 36 = 0.1389
4 ∕ 36 = 0.1111
3 ∕ 36 = 0.0833
2 ∕ 36 = 0.0556
1 ∕ 36 = 0.0278
Chapter 2—Discrete Random Variables and Probability Distributions
Page 3 of 17
Probability Distribution of Sum of Dots in Tossing Two Dice
f(x)
0.18
0.17
0.16
0.14
0.14
0.11
0.12
0.10
0.11
0.08
0.08
0.08
0.06
0.06
0.04
0.14
0.06
0.03
0.03
0.02
0.00
2
3
4
5
6
7
8
9
10
11
12
x (sum of dots)
3. Expected Value of a Random Variable
The expected value of the random variable x, denoted by 𝐄(𝒙), is simply the mean of all the values taken by
the random variable. Since E(𝑥) represents the mean of all possible values of 𝑥, we can use the symbol 𝜇
(population mean) also to represent the expected value. If probabilities assigned to each value were the
same, then the mean or the expected value would simply be the sum of all the values divided by the number
of values of 𝑥. But the assigned probabilities are rarely equal for all the values of a random variable.
Therefore, to find the expected value of 𝑥 the weighted average formula must be used. The weights assigned
to each value are the probabilities.
Example 6
The following table shows the probability distribution of the number of automobiles sold in any given
business day in a car dealership.
Number of
automobiles sold
𝑥
0
1
2
3
4
5
Probability
𝑓(𝑥)
0.18
0.39
0.24
0.14
0.04
0.01
Find the expected value of the number of automobiles sold in a day.
Chapter 2—Discrete Random Variables and Probability Distributions
Page 4 of 17
𝑥
0
1
2
3
4
5
𝑓(𝑥)
0.18
0.39
0.24
0.14
0.04
0.01
𝑥𝑓(𝑥) =
𝑥𝑓(𝑥)
0.00
0.39
0.48
0.42
0.16
0.05
1.50
The expected value of a random variable is the
sum of the products of the values of the random
variable and the corresponding probabilities.
E(𝑥) = 𝑥𝑓(𝑥)
Example 7
A state lottery issues 100,000 instant lottery tickets. The possible prizes and the number of tickets containing
each prize is shown below:
Prize
$0
1
2
5
10
25
500
5000
Number of Tickets
Issued
86,246
8,000
3,200
1,540
740
260
13
1
100,000
Let x denote the prize value. Using the number (frequency) of tickets associated with each prize develop the
probability distribution of x and compute the expected value of the prize amount.
The following worksheet shows the calculations. The first and second columns show the probability
distribution of x. The probability of winning each prize amount is simply the relative frequency of the tickets
containing that prize amount. The third column in the worksheet shows the calculation of E(x).
𝑥
0
1
2
5
10
25
500
5000
𝑓(𝑥)
0.86246
0.08000
0.03200
0.01540
0.00740
0.00260
0.00013
0.00001
1.00000
𝑥𝑓(𝑥)
0.000
0.080
0.064
0.077
0.074
0.065
0.065
0.050
E(𝑥) = 𝑥𝑓(𝑥) = 0.475
Chapter 2—Discrete Random Variables and Probability Distributions
Page 5 of 17
4. The Variance And Standard Deviation of a Random Variable
The variance of a random variable, denoted by 𝐯𝐚𝐫(𝒙), is a measure of the dispersion of the values of the
random variable. Generally, the variance, as explained in Chapter 1, is computed as the mean squared
deviation of a variable from the mean: σ2 = (𝑥 − µ)2 ⁄𝑁
If the variable is a random variable, then we must compute the weighted average of the squared deviations,
using the probabilities as the weights. The mean of the random variable is µ. Then the variance of the
random variable 𝑥 is defined as the expected value of the squared deviations:
var(𝑥) = E[(𝑥 − µ)2 ]
The above expression means that the variance of the random variable 𝑥 is the weighted mean of squared
deviations.
var(𝑥) = (𝑥 − 𝜇)2 𝑓(𝑥)
Variance of a random variable:
E[(𝑥 − 𝜇)2 ]
Is the expected value of the squared deviations
and is computed as
var(𝑥) = ∑(𝑥 − 𝜇)2 𝑓(𝑥)
the weighted mean of the squared deviations.
Example 8
Given E(𝑥) = 𝜇 = 1.5, find the variance of the number of cars sold in a day in Example 6.
𝑥
0
1
2
3
4
5
𝑓(𝑥)
0.18
0.39
0.24
0.14
0.04
0.01
(𝑥 − 𝜇)2
2.25
0.25
0.25
2.25
6.25
12.25
(𝑥 − 𝜇)2 𝑓(𝑥)
0.4050
0.0975
0.0600
0.3150
0.2500
0.1225
1.2500
The variance of 𝑥 is:
var(𝑥) = (𝑥 − 𝜇)2 𝑓(𝑥) = 1.25
The standard deviation of x is the square root of the variance:
sd(𝑥) = √(𝑥 − 𝜇)2 𝑓(𝑥) = √1.25 = 1.118
Chapter 2—Discrete Random Variables and Probability Distributions
Page 6 of 17
Example 9
𝐸(𝑥) = 𝜇 = 0.475, find the variance of prize values in the lottery example (Example 7)
𝑥
0
1
2
5
10
25
500
5000
The standard deviation is:
𝑓(𝑥)
0.86246
0.08
0.032
0.0154
0.0074
0.0026
0.00013
0.00001
(𝑥 − 𝜇)2
0.2256
0.2756
2.3256
20.4756
90.7256
601.4756
249525.2256
24995250.2256
(𝑥 − 𝜇)2 𝑓(𝑥)
0.1946
0.0221
0.0744
0.3153
0.6714
1.5638
32.4383
249.9525
285.2324
sd(𝑥) = √285.2324 = 16.89
Example 10
A company employs salespersons to market its product. Keeping track of sales for 100 weeks, the following
table shows the number of units of the product sold per week and the frequency of the weeks. For example,
in 15 of the weeks, a salesperson sold 11 units, and in 40 of the weeks, 12 units were sold.
Number of units sold
𝑥
10
11
12
13
14
Weeks
10
15
40
30
5
100
Using this table, we can generate the probability (relative frequency) distribution of the random variable 𝑥
(the number of units sold in a given week), as shown below.
Number of items sold
𝑥
10
11
12
13
14
𝑓(𝑥)
0.10
0.15
0.40
0.30
0.05
Find the expected value and the standard deviation of the number of items sold per salesperson per week.
The following worksheet shows the calculations.
Chapter 2—Discrete Random Variables and Probability Distributions
Page 7 of 17
𝑥
10
11
12
13
14
𝑓(𝑥)
0.10
0.15
0.40
0.30
0.05
(𝑥 − µ)2
4.2025
1.1025
0.0025
0.9025
3.8025
var(𝑥) =
sd(𝑥) =
𝑥𝑓(𝑥)
1.00
1.65
4.80
3.90
0.70
µ = 12.05
(𝑥 − µ)2 𝑓(𝑥)
0.4203
0.1654
0.0010
0.2708
0.1901
1.0475
1.0235
Each salesperson sells, on average, 12.05 units, and the number units sold deviate from the mean by 1.0235,
on average.
A simpler formula to compute the variance of a random variable
A simple algebraic manipulation of the variance formula provides an alternative, much simpler, way to
compute the variance of a discrete random variable.2
(𝑥 − 𝜇)2 𝑓(𝑥) = 𝑥2 𝑓(𝑥) − 𝜇2
Simpler formula to compute the variance:
E[(𝑥 − 𝜇)2 ] = E(𝑥 2 ) − 𝜇 2
(𝑥 − 𝜇)2 𝑓(𝑥) = 𝑥2 𝑓(𝑥) − 𝜇2
Use the computational formula to obtain the variance in Example 8 and Example 10.
Variance for Example 8
𝑥
0
1
2
3
4
5
𝑥𝑓(𝑥)
0.18
0.39
0.24
0.14
0.04
0.01
𝑥2
0
1
4
9
16
25
Variance for Example 10
𝑥 2 𝑓(𝑥)
0.00
0.39
0.96
1.26
0.64
0.25
3.50
𝑥2 𝑓(𝑥) − 𝜇2 = 3.5 − 1.52 = 1.25
𝑥
10
11
12
13
14
𝑥𝑓(𝑥)
0.10
0.15
0.40
0.30
0.05
𝑥2
100
121
144
169
196
𝑥 2 𝑓(𝑥)
10.00
18.15
57.60
50.70
9.80
146.25
𝑥2 𝑓(𝑥) − 𝜇2 = 146.3 − 12.052 = 1.0475
(𝑥 − 𝜇)2 𝑓(𝑥) = (𝑥 2 − 2𝜇𝑥 + 𝜇2 )𝑓(𝑥)
(𝑥 − 𝜇)2 𝑓(𝑥) = [𝑥 2 𝑓(𝑥) − 2𝜇𝑥𝑓(𝑥) + 𝜇2 𝑓(𝑥)]
(𝑥 − 𝜇)2 𝑓(𝑥) = 𝑥 2 𝑓(𝑥) − 2𝜇𝑥𝑓(𝑥) + 𝜇2 𝑓(𝑥)
(𝑥 − 𝜇)2 𝑓(𝑥) = 𝑥 2 𝑓(𝑥) − 2𝜇2 + 𝜇2
(𝑥 − 𝜇)2 𝑓(𝑥) = 𝑥 2 𝑓(𝑥) − 𝜇2
2
Chapter 2—Discrete Random Variables and Probability Distributions
Page 8 of 17
Example 11
In Example 10, suppose each salesperson receives a sales commission of $20 per unit sold plus a fixed weekly
wage of $500. What is the expected weekly income?
Here the random variable 𝑋 (the number of units sold) is linearly transformed to the random variable Y (the
weekly income), where
𝑦 = 500 + 20𝑥
By linearly transforming 𝑋 we have a new random variable 𝑌. However, the corresponding probabilities
remain intact. The following shows the probability distribution of 𝑦 and the expected value of 𝑦.
𝑦
𝑓(𝑦)
𝑦𝑓(𝑦)
500 + 20(10) = 700
500 + 20(11) = 720
500 + 20(12) = 740
500 + 20(13) = 760
500 + 20(14) = 780
0.10
0.15
0.40
0.30
0.05
E(𝑦) =
70
108
296
228
39
741
A salesperson expects to earn $741 per week. (He earns on average $741 per week.) Note that you can find
the expected value of 𝑦 using E(𝑥). Since 𝑦 is a linear transformation of 𝑥, then
E(𝑦) = E(500 + 20𝑥)
E(𝑦) = E(500) + E(20𝑥)
E(𝑦) = 500 + 20E(𝑥)
E(𝑦) = 500 + 20(12.05) = 741
4.1. Arithmetic Properties Of Expected Value And Variance
The arithmetic properties of E(𝑥) and var(𝑥) shows what happens to the expected value and variance of x if
the random variable is linearly transformed. The random variable 𝑦 is a linear transformation of 𝑥, if for any
constant 𝒂 and a nonzero constant 𝒃,
𝑦 = 𝑎 + 𝑏𝑥
Then for any linear transformation of 𝑥, the following holds: 3
E(𝑦) = E(𝑎 + 𝑏𝑥)
E(𝑦) = E(𝑎) + 𝑏E(𝑥)
E(𝑦) = a + 𝑏E(𝑥)
E(𝑎 + 𝑏𝑥) = ∑(𝑎 + 𝑏𝑥)𝑓(𝑥)
E(𝑎 + 𝑏𝑥) = ∑𝑎𝑓(𝑥) + ∑𝑏𝑥𝑓(𝑥)
E(𝑎 + 𝑏𝑥) = 𝑎∑𝑓(𝑥) + 𝑏∑𝑥𝑓(𝑥)
E(𝑎 + 𝑏𝑥) = 𝑎 + 𝑏E(𝑥)
3
Chapter 2—Discrete Random Variables and Probability Distributions
Page 9 of 17
and,
var(𝑦) = var(𝑎 + 𝑏𝑥)
var(𝑦) = var(𝑎) + var(𝑏𝑥)
var(𝑦) = 𝑏 2 var(𝑥)
(See the footnote4.) Note that the variance of a constant is always zero. Therefore, var(𝑎) = 0.
Also note that
sd(𝑦) = 𝑏sd(𝑥)
Example 12
Find the variance and standard deviation of the weekly income of the salespersons in Example 10.
First find the variance of y by using the worksheet method.
𝑦
𝑓(𝑦)
(𝑦 − µ)2 𝑓(𝑦)
700
720
740
760
780
0.10
0.15
0.40
0.30
0.05
168.10
66.15
0.40
108.30
76.05
var(𝑦) = (𝑦 − 𝜇)2 𝑓(𝑦) = 419.00
sd(𝑦) = 20.47
The standard deviation of income means that, on average, a salesperson should expect his income to vary by
$20.47 per week.
Next use the arithmetic property of variance to find var(𝑦).
var(𝑦) = var(500 + 20𝑥) = 202 var(𝑥) = 400(1.0475) = 419
and
sd(𝑦) = √419 = 20.47
or
sd(𝑦) = 𝑏sd(𝑥) = 20(1.0235) = 20.47
var(𝑎 + 𝑏𝑥) = ∑[𝑎 + 𝑏𝑥 − E(𝑎 + 𝑏𝑥)]2 𝑓(𝑥)
var(𝑎 + 𝑏𝑥) = ∑[𝑎 + 𝑏𝑥 − 𝑎 − 𝑏E(𝑥)]2 𝑓(𝑥)
var(𝑎 + 𝑏𝑥) = ∑(𝑏𝑥 − 𝑏µ)2 𝑓(𝑥)
var(𝑎 + 𝑏𝑥) = ∑𝑏 2 (𝑥 − µ)2 𝑓(𝑥) = 𝑏 2 ∑(𝑥 − µ)2 𝑓(𝑥)
var(𝑎 + 𝑏𝑥) = 𝑏 2 var(x)
4
Chapter 2—Discrete Random Variables and Probability Distributions
Page 10 of 17
Example 13
Given the following probability distribution of 𝑥, find the expected value and variance of x and the expected
value and variance of 𝑦 = 2 + 5𝑥.
First, find E(x) and E(y):
𝑥
1
2
3
4
5
𝑓(𝑥)
𝑥𝑓(𝑥)
0.05
0.05
0.15
0.30
0.35
1.05
0.30
1.20
0.15
0.75
E(𝑥) = 𝑥𝑓(𝑥) = 3.35
𝑦 = 2 + 5𝑥
7
12
17
22
27
𝑦𝑓(𝑥)
0.35
1.80
5.95
6.60
4.05
𝐸(𝑦) = 18.75
Using the linear transformation property E(𝑦) = 𝑎 + 𝑏E(𝑥).
E(𝑦) = 2 + 5E(𝑥) = 2 + 5(3.35) = 18.75
Once you determine E(𝑥), you can use the arithmetic properties of expected value and find E(𝑦). You do not
need to perform the worksheet computations.
Next, compute var(𝑥) and var(𝑦):
𝑥
1
2
3
4
5
𝑓(𝑥)
0.05
0.15
0.35
0.30
0.15
[𝑥 − E(𝑥)]2 𝑓(𝑥)
0.2761
0.2734
0.0429
0.1268
0.4084
var(𝑥) = 1.1275
𝑦 = 2 + 5𝑥
7
12
17
22
27
[𝑦 − E(𝑦)]2 𝑓(𝑥)
6.9031
6.8344
1.0719
3.1688
10.2094
var(𝑦) = 28.1875
Note, again, that:
var(𝑦) = var(2 + 5𝑥) = 52 var(𝑥) = 25(1.1275) = 28.1875
sd(𝑥) = √1.1275 = 1.0618
sd(𝑦) = 5sd(𝑥) = 5(1.0618) = 5.3092
5. The Binomial Distribution
Certain random experiments or processes, such as guessing the answers to a multiple choice test, tossing a
coin or a pair of dice, drawing a card from a deck of playing cards (with replacement) possess properties that
generate special random variables. The probability assigned to each value of such random variables are then
determined using a general formula. What are the properties of such random experiments?
1.
The experiment consists of identical and independent repeated trials.
2.
Each trial consists of two mutually exclusive outcomes: success or failure. The answer for each multiple
choice question is either correct (success) or incorrect (failure). Defining heads as success in tossing a
Chapter 2—Discrete Random Variables and Probability Distributions
Page 11 of 17
coin, or a double-six in tossing a pair of dice, then the other outcomes are “failures”. Or, you may
define drawing an ace a success. Then drawing any of the other cards is a failure.
3.
The probability of success, P(𝑆) = π, or failure, P(𝐹) = 1 − π, remains the same for all repetitive trials.
In the quiz example, 𝜋 = 0.25 and 1 − π = 0.75. In the coin toss 𝜋 = 0.5 and 1 − π = 0.5. In tossing a
pair of dice the probability of a double-six is π = 1⁄6. And in drawing an ace, π = 4⁄52.
In general, any random process having these properties is called a Bernoulli process. The probability
distribution of a random variable defining the number of "successes" in a Bernoulli process is called a
binomial distribution. Thus, the probability distribution of x, the number of correct answers in the multiple
choice quiz example, is a binomial distribution.
Each binomial distribution is defined by two parameters, or identifying characteristics: the number of trials,
n, and the probability of success, π. In symbols, the distribution is expressed as follows,
𝑋~B(𝑛, π)
which reads: “X is binomially distributed with parameters n and π”.
We can now explain the method to determine the probabilities associated with each value of a binomial
random variable using the following example.
Example 14
The table from Example 3 is reproduced below. When you are guessing the answers to a set of 5 multiplechoice questions, you are conducting a random experiment. Assign 0 to the incorrect answer and 1 to the
correct answer for each question, and let 𝑥 be the random variable representing the number of possible
correct answers out of 5 questions. The values of 𝑥 are shown at the top row of the table: 0, 1, 2, 3, 4, 5.
0
(0,0,0,0,0)
1
(1,0,0,0,0)
(0,1,0,0,0)
(0,0,1,0,0)
(0,0,0,1,0)
(0,0,0,0,1)
2
(1,1,0,0,0)
(1,0,1,0,0)
(1,0,0,1,0)
(1,0,0,0,1)
(0,1,1,0,0)
(0,1,0,1,0)
(0,1,0,0,1)
(0,0,1,1,0)
(0,0,1,0,1)
(0,0,0,1,1)
3
(1,1,1,0,0)
(1,1,0,1,0)
(1,1,0,0,1)
(1,0,1,1,0)
(1,0,1,0,1)
(1,0,0,1,1)
(0,1,1,1,0)
(0,1,1,0,1)
(0,1,0,1,1)
(0,0,1,1,1)
4
(1,1,1,1,0)
(1,1,1,0,1)
(1,1,0,1,1)
(1,0,1,1,1)
(0,1,1,1,1)
5
(1,1,1,1,1)
Develop the probability distribution of 𝑥 by determining the probability associated with each value of the
random variable.
Note that each value of 𝑥 has its own number of events. For example, there are five events that generate a
value of 𝑥 = 1; ten events generate 𝑥 = 2, etc. Each event has its own probability. If there are four choices per
question, the probability of answering each question, the probability of success, is P(1) = 1⁄4 = 0.25, and the
probability of failure is P(0) = 3⁄4 = 0.75. The probability that all questions are guessed incorrectly is then:
f(0) = P(0,0,0,0,0) = (0.75)(0.75)(0.75)(0.75)(0.75) = 0.2373
Chapter 2—Discrete Random Variables and Probability Distributions
Page 12 of 17
The probability that one question is guessed correctly is:
f(1) = P(1,0,0,0,0) = (0.25)(0.75)(0.75)(0.75)(0.75) = 0.25 × 0.754 = 0.0791 +
P(0,1,0,0,0) = (0.75)(0.25)(0.75)(0.75)(0.75) = 0.25 × 0.754 = 0.0791 +
P(0,0,1,0,0) = (0.75)(0.75)(0.25)(0.75)(0.75) = 0.25 × 0.754 = 0.0791 +
P(0,0,0,1,0) = (0.75)(0.75)(0.75)(0.25)(0.75) = 0.25 × 0.754 = 0.0791 +
P(0,0,0,0,1) = (0.75)(0.75)(0.75)(0.75)(0.25) = 0.25 × 0.754 = 0.0791
f(1) = 5 × 0.25 × 0.754 = 5 × 0.0791 = 0.3955
The probability of two correct guesses is:
f(2) = P(1,1,0,0,0) = (0.25)(0.25)(0.75)(0.75)(0.75) = 0.252 × 0.753 = 0.02637 +
P(1,0,1,0,0) = (0.25)(0.75)(0.25)(0.75)(0.75) = 0.252 × 0.753 = 0.02637 +

P(0,0,0,1,1) = (0.75)(0.75)(0.75)(0.25)(0.25) = 0.252 × 0.753 = 0.02637
f(2) = 10 × 0.252 × 0.753 = 10 × 0.02637 = 0.2637
f(3) = 10 × 0.253 × 0.752 = 10 × 0.00879 = 0.0879
f(4) = 5 × 0.254 × 0.75 = 5 × 0.0029 = 0.0146
f(5) = 1 × 0.255 × 0.750 = 0.0010
Putting these calculations compactly, we have,
𝑥
0
1
2
3
4
5
1 ×. 250 × 0.755
5 ×. 251 × 0.754
10 ×. 252 × 0.753
10 ×. 253 × 0.752
5 ×. 254 × 0.751
1 ×. 255 × 0.750
=
=
=
=
=
=
𝑓(𝑥)
0.2373
0.3955
0.2637
0.0879
0.0146
0.0010
The probability distribution is then:
𝑥
0
1
2
3
4
5
𝑓(𝑥)
0.2373
0.3955
0.2637
0.0879
0.0146
0.0010
Note that to find the number of ways you can arrange a given number of 𝑥 successes among 𝑛 trials is
obtained using the combination counting formula:
C(𝑛, 𝑥) =
𝑛!
(𝑛 − 𝑥)! 𝑥!
For example, the number of arrangements of 3 items selected without replacement from 7 items is:
Chapter 2—Discrete Random Variables and Probability Distributions
Page 13 of 17
C(7,3) =
7!
= 35
(7 − 3)! 3!
The general formula for the binomial distribution is:
The Binomial Distribution Formula
𝑓(𝑥) = C(𝑛, 𝑥)π𝑥 (1 − π)(𝑛−𝑥)
𝑛=
𝑥=
C(𝑛, 𝑥) =
𝜋=
Number of trial
Number of successes
Combinations (number of arrangements) of 𝑥 successes in 𝑛 trials
Probability of “success” in each trial
Example 15
Suppose you find a parking spot at IUPUI within ten minutes 65 percent of the time. You always arrive at the
campus ten minutes before your class starts. Let X be the random variable denoting the number of times you
will be late to your class in a seven-day period. Find the probability distribution of X.
Since you find a spot in ten minutes or under 65 percent of the time, the probability of being late to your class
each time is 1 – 0.65 = 0.35. X is binomially distributed with the following parameters:
𝑋~𝐵(𝑛 = 7, 𝜋 = 0.35)
𝑥
0
1
2
3
4
5
6
7
𝑓(𝑥)
C(7,0)(0.350)(0.657) =
C(7,1)(0.351)(0.656) =
C(7,2)(0.352)(0.655) =
C(7,3)(0.353)(0.654) =
C(7,4)(0.354)(0.653) =
C(7,5)(0.355)(0.652) =
C(7,6)(0.356)(0.651) =
C(7,7)(0.357)(0.650) =
0.0490
0.1848
0.2985
0.2679
0.1442
0.0466
0.0084
0.0006
1.0000
5.1. Cumulative Probability Distribution
The cumulative probability distribution is the probability that the random variable X takes on a value which is
at most (less than or equal to) a given value. In the previous example, what is the probability that you will be
late to class 3 or fewer times in a seven-day period? That is, find 𝑓(𝑥 ≤ 3). To answer this question, first
write the cumulative probability distribution of X.
Chapter 2—Discrete Random Variables and Probability Distributions
Page 14 of 17
𝑥
𝑓(𝑥)
𝑓(𝑋 ≤ 𝑥)
0
1
2
3
4
5
6
7
0.0490
0.1848
0.2985
0.2679
0.1442
0.0466
0.0084
0.0006
0.0490
0.2338
0.5323
0.8002
0.9444
0.9910
0.9994
1.0000
1.0000
Thus, f(x ≤ 3) = f(x = 0) + f(x = 1) + f(x = 2) + f(x = 3) = 0.8002.
Using the above schedule, find the following probabilities:
1.
f(2 ≤ x ≤ 5)
f(2 ≤ x ≤ 5) = f(x = 2) + f(x = 3) + f(x = 4) + f(x = 5) = 0.7572
Alternatively, using the cumulative probability column,
f(2 ≤ x ≤ 5) = f(x ≤ 5) − f(x ≤ 1) = 0.9910 − 0.2338 = 0.7572
2.
f(3 ≤ x < 5)
f(3 ≤ x < 5) = f(x = 3) + f(x = 4) = 0.4121
Alternatively, using the cumulative probability column,
f(3 ≤ x ≤ 4) = f(x ≤ 4) − f(x ≤ 2) = 0.9444 − 0.5323 = 0.4121
3.
f(x ≥ 5)
f(x ≥ 5) = f(x = 5) + f (x = 6) + f(x = 7) = 0.0556
Alternatively, using the cumulative probability column,
f(x ≥ 5) = 1 − f(x ≤ 4) = 1 − 0.9444 = 0.0556
5.2. How to Use Excel to Find Binomial Probabilities
In Excel, you can find the above binomial probability using the following function:
=BINOM. DIST(number_s, trials, probability_s, cumulative)
For example, given X ~ B(n = 7, π = 0.35):
Find f(x = 5)
=BINOM.DIST(5,7,0.35,0) = 0.0466
Note that “0” is entered for “cumulative” in the formula. This means that we want 𝑥 = 5, not 𝑥 ≤ 5
Chapter 2—Discrete Random Variables and Probability Distributions
Page 15 of 17
For cumulative probability, “1” replaces “0” in the formula.
Find 𝑓(𝑥 ≤ 5)
=BINOM.DIST(5,7,0.35,1) = 0.9910
Example
Toss a coin 10 times. Let X define the number of tails. Find the following probabilities using Excel.
1. f(x = 4)
=BINOM.DIST(4,10,0.5,0) = 0.2051
2. f(x ≤ 4)
=BINOM.DIST(4,10,0.5,1) = 0.3770
3. 𝑓(𝑥 ≥ 4) = 1 − 𝑓(𝑥 ≤ 3)
=1 – BINOM.DIST(3,10,0.5,1) = 0.8281
4. 𝑓(4 ≤ 𝑥 ≤ 7) = 𝑓(𝑥 ≤ 7) − 𝑓(𝑥 ≤ 3) =BINOM.DIST(7,10,0.5,1)-BINOM.DIST(3,10,0.5,1) = 0.7734
5. 𝑓(4 < 𝑥 < 7) = 𝑓(𝑥 ≤ 6) − 𝑓(𝑥 ≤ 3) =BINOM.DIST(6,10,0.5,1)-BINOM.DIST(4,10,0.5,1) = 0.4512
5.3. Expected Value and Variance of the Binomial Distribution
To show how to find the expected value and variance of a binomial distribution use the following familiar
example.
Example
Suppose you find a parking spot at IUPUI within ten minutes 65 percent of the time. You always arrive at the
campus ten minutes before your class starts. Let X define the number of times you will be late to your class in
a seven-day period. What is the average number of times you will be late to class in a seven day period?
What is the variance and standard deviation of x? Here you are looking for μ = E(x) and var(x). Use the
probability distribution determined above.
𝑥
𝑓(𝑥)
0
1
2
3
4
5
6
7
0.0490
0.1848
0.2985
0.2679
0.1442
0.0466
0.0084
0.0006
𝑥𝑓(𝑥)
(𝑥 − 𝜇)²𝑓(𝑥)
0.0000
0.1848
0.5970
0.8036
0.5770
0.2330
0.0502
0.0045
µ = 2.4500
0.2943
0.3885
0.0604
0.0810
0.3465
0.3030
0.1054
0.0133
var(𝑥) = 1.5925
Since this is a binomial distribution, there are specific formulas that allow you to find the expected value and
the variance of the distribution without using a worksheet. These formulas are:
The mean or expect value is:
The variance is:
The standard deviation is
µ = E(𝑥) = 𝑛π
var(𝑥) = 𝑛π(1 − π)
sd(𝑥) = √𝑛π(1 − π)
Chapter 2—Discrete Random Variables and Probability Distributions
Page 16 of 17
Thus E(𝑥) = 7(0.35) = 2.45 and var(𝑥) = 7(0.35)(0.65) = 1.5925
Chapter 2—Discrete Random Variables and Probability Distributions
Page 17 of 17