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4.10 Exercises 93
25. Examining a number of continuous time intermittent signals, it was found that
about 20% of these signals have dirikle conditions. Ten signals are selected
randomly, what is the probability that exactly two signals have dirikle
conditions?
26. There are 4 inductors and 11 capacitors in the circuit that we want to place these
elements in parallel in the circuit. What is the probability that all four inductors
will fall behind each other?
27. In the circuit, five inductors with values of 1, 2, 3, 4, and 5 H are connected in
series. We choose two inductors at random and obtain their equivalents. What
is the probability that the equivalent inductor size is greater than 5?
https://t.me/med1917
Chapter 5
Discrete Distributions
5.1 Introductions
The primary objective of probability models in any random experiment involves
two fundamental components: the sample space and the assignment of probabilities
to each elementary outcome. The method presented in the previous chapter for
calculating the probability of a random outcome is not universally applicable to all
“sample space” scenarios. Wider sample spaces necessitate more comprehensive
approaches, and probability distributions serve as effective tools for such cases.
These distributions are typically represented through tables, graphs, mathematical
formulas, etc., encapsulating all possible outcomes of the random variable along
with their assigned probabilities.
A random variable is a variable whose value is contingent upon chance,
randomness, and the outcome of a random experiment. When a real number is
assigned to each situation in a “sample space”, the outcome is referred to as a
random variable. Random variables are denoted by uppercase letters, such as X and
Y , while observed values are indicated by lowercase letters, such as x and y.
Events unfold in specific patterns, and probability rules model the occurrence of
these events, enabling the determination of probabilities across the entire s ample
space. These models, including binomial, geometric, hypergeometric, multinomial,
Poisson distributions, as well as normal, t-student, Chi-squared, and others, provide
a framework for understanding and predicting outcomes. This chapter focuses on
the most essential discrete probability distributions, such as binomial, geometric,
hypergeometric, multinomial, and Poisson distributions, presenting their key
properties and applications.
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2024
S. H. Saneii and H. Doosti, Practical Biostatistics for Medical and Health Sciences,
https://doi.org/10.1007/978-981-97-3083-4_5
95
https://t.me/med1917
96 5 Discrete Distributions
5.2 Binomial Distribution
To facilitate a better understanding of the binomial distribution, we begin with the
concept of a Bernoulli trial, followed by an exploration of binomials raised to different
powers.
Bernoulli trial
A Bernoulli trial, also known as a binomial trial, is a random experiment featuring
precisely two possible outcomes—often framed as “success-failure” or “yes–no”. In
this trial, the probability of success remains constant with each iteration. Named after
the seventeenth-century Swiss mathematician Jacob Bernoulli, these trials should be
conducted independently. Let p represents the probability of success in a Bernoulli
trial, and q denotes the probability of failure. Notably, the sum of the probability of
success (p) and the probability of failure (q) is always equal to one, as success and
failure are complementary and mutually exclusive. This relationship is expressed by
the equation
p + q = 1.
Now, let’s extend the concept to multiple trials. If we independently repeat the
Bernoulli trial n times and denote the number of successes as X, where X is a discrete
random variable ranging from 0 to n, we can calculate the probabilities associated
with different outcomes. For instance, X can take values such as 0 (no success), 1
(one success), 2 (two successes), and so forth, up to n (success in all trials). While
computing the probabilities for a small number of trials is straightforward, it becomes
challenging and time consuming for a large number of experiments. In the following
sections, we will delve into a general method for calculating these probabilities
efficiently (Table
5.1).
In this table, the first column shows the number of trials from 1 to n and the first
row shows the number of successes from 0 to the number of trials (n). For example,
if the Bernoulli trial is repeated twice (
n = 2), the variable X can have no success
Table 5.1 Corresponding probability for Bernoulli trial
No. of trials No. of success General form
X = 0 X = 1 X = 2 X = 3 X = 4 ··· X = n
n = 1 q p – – – ··· – (p + q)
1
n = 2 q
2
p · q p
2
– – ··· – (p + q)
2
n = 3 q
3
3pq
2
3p
2
q p
3
– ··· – (p + q)
3
n = 4 q
4
4pq
3
6p
2
q
2
4p
3
q p
4
··· – (p + q)
4
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
···
.
.
.
n = n q
n
npq
n−1
··· ··· ··· ··· p
n
(p + q)
n
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5.2 Binomial Distribution 97
Table 5.2 Pascal triangle and binomial expansion (p + q)
n
n The coefficient of C
1 1
2 1 1
3 1 2 1
4 1 3 3 1
5 1 4 6 4 1
6 1 5 10 10 5 1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(X = 0) or one success (X = 1) or two successes (X = 2). Because Bernoulli’s trials
are independent, the probability of no success or two failures equals
qq = q
2
,the
probability of one success and one failure equals
qp + pq = 2pq, and the probability
of two consecutive success equals
pp = p
2
.
Algebraically, the probability of the number of successes in n trials can be obtained
by expanding the binomial coefficients of
(p + q)
n
.
(p + q)
1
= p + q,
(p + q)
2
= p
2
+ 2pq + q
2
,
(p + q)
3
= p
3
+ 3p
2
q + 3pq
2
+ q
3
,
(p + q)
4
= p
4
+ 4p
3
q + 6p
2
q
2
+ 4pq
3
+ q
4
.
The binomial expansion (p + q)
n
can be shown by the general form Cp
x
·q
n−x
that
the coefficient C can be found from the “Pascal triangle” in Table
5.2. The initial
combination of these coefficients was discovered by the Iranian mathematician Omar
Khayyam.
1
The general form of this expansion is as below:
(p + q)
n
=
n
x=0
n
x
p
x
q
(
n−x
)
.
That is the probability of x successes in n independent trials, denoted by P
n
(x)
and called the binomial distribution, is equal to:
P
n
(
x
)
= C
x
n
× p
x
q
(n−x)
=
n!
x!
(
n − x
)
!
× p
x
· q
(n−x)
The mean and variance of Binomial distribution are μ = np and σ
2
= npq
(Table 5.3).
1
Kennedy, Evelyn (1966). “Omar Khayyam”. The Mathematics Teacher. LIX(3):140–142. https://
doi.org/10.5951/MT.59.2.0140. JSTOR27957296.
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98 5 Discrete Distributions
Table 5.3 Corresponding probability for Bernoulli experiment
The probability
of
x
successes
in
n
trials
The No. of events
with exactly
x
successes in
n
trials
Example 1 It is asserted that a new pain relief treatment effectively cures muscular
pain in 70% of cases. When administered to 10 patients, determine the probability
that: (a) No more than 3 patients will experience relief. (b) The number of patients
cured falls between 6 and 8, inclusive. (c) At least 5 patients will be cured.
Solution: Suppose the patients are independent, the binomial distribution with n =
10 and p = 0.7 can applied for X = number of patients who are cured. To compute
the required probabilities, we consult the binomial table for n = 10 and p = 0.70.
(a)
p
(
X ≤ 3
)
= p
(
X = 0
)
+ p
(
X = 1
)
+ p
(
X = 2
)
+ p
(
X = 3
)
= C
0
10
p
0
q
(
10−0
)
+ C
1
10
p
1
q
(
10−1
)
+ C
2
10
p
2
q
(
10−2
)
+ C
3
10
p
3
q
(
10−3
)
=
10!
0!10!
0.7
0
× 0.3
10
+
10!
1!9!
0.7
1
× 0.3
9
+
10!
2!8!
0.7
2
× 0.3
8
+
10!
3!7!
0.7
3
× 0.3
7
= 0.0000059
+ 0.000138 + 0.00116 + 0.0090 = 0.103,
(b)
p
(
6 ≤ X ≤ 8
)
= p
(
X = 6
)
+ p
(
X = 7
)
+ p
(
X = 8
)
=
10!
6!4!
0.7
6
× 0.3
4
+
10!
7!3!
0.7
7
× 0.3
3
+
10!
8!2!
0.7
8
× 0.3
2
= 0.2001 + 0.2668
+ 0.2335 = 0.7004
,
(c) p
(
X ≥ 4
)
= 1 − p
(
X ≤ 3
)
= 1 − 0.103 = 0.897.
Example 2 Thirty percent of university students have poor vision. We select 4
students randomly. Find the probability that:
(a) All 4 students have vision problem.
(b)
No one has visual impairment.
(c) At least one person has visual impairment.
Solution: In this example n = 4, p = 0.3, and q = 1 − p = 1–0.3 = 0.7. So:
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5.2 Binomial Distribution 99
(a) p
(
x = 3
)
= C
4
4
p
4
q
(
4−4
)
=
4!
4!0!
0.3
4
× 0.7
0
= 0.0081.
(b) p
(
x = 0
)
= C
0
4
p
0
q
(
4−0
)
=
4!
0!4!
0.3
0
× 0.7
4
= 0.2401.
(c) p
(
x ≥ 1
)
= 1 − p
(
x = 0
)
= 1 − 0.2401 = 0.7599.
Example 3 20% of digital blood pressure monitors do not have the required accuracy
and are returned to the factory within 1-year warranty. If 6 units have been sold in a
given week, what is the probability that in the 1-year warranty period?
(a) None of those devices is corrupted.
(b) Fail at least one of the devices.
(c) Four digital blood pressures’ monitor breaks down and does not work properly.
Solution: Assuming the known values (n = 0.6, p = 0.20, and q = 1 − p = 1–
0.2 = 0.8) in this example, we can apply the binomial distribution to calculate the
probabilities.
(a) p
(
x = 0
)
= p
6
(
0
)
=
6
0
(0.2)
0
(0.8)
6
= 0.262.
(b)
p
(
x ≥ 1
)
= p
(
x = 1
)
+ p
(
x = 2
)
+ p
(
x = 3
)
+ p
(
x = 4
)
+ p
(
x = 5
)
+ p
(
x = 6
)
= 1 − p
(
x = 0
)
= 1 −
6
0
(
0.2
)
0
(
0.8
)
6
= 1 − 0.262 = 0.738.
.
(c) p
(
x = 4
)
= p
6
(
4
)
=
6
4
(0.2)
4
(0.8)
2
= 0.001024.
The binomial distribution is used for discrete and countable variables and has two
parameters n and p. The shape of the binomial distribution depends on the values of
n and p. The closer the values of q and p are to each other, the more symmetrical
the distribution of the binomials becomes. So that if q = p, this distribution will be
symmetric for all values of n. The farther the values q and p are from each other,
the more skewed the distribution. As the value of n increases, the skewness of the
distribution decreases, so much so that if n is large enough, the binomial distribution
tends to return to the normal distribution.
It should be noted that the use of binomial distribution in infinite populations is
desirable when the probability of an event can be assumed to be constant. But in finite
populations, the replacement sampling method should be used. That is, each sample
is returned to its place immediately and then the next sample is selected. But it often
happens that this distribution is used in finite population without replacement which
is problematic and disrupts the independence assumption of repeated experiments,
unless the ratio of population size to sample size (
N
n
) is at least 10 (Fig. 5.1).
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100 5 Discrete Distributions
Fig. 5.1 Different binomial distributions by p and n values
5.3 Multinomial Distribution
As mentioned, the binomial distribution deals with variables that their outcome is
only in two categories (yes and no or male and female), which is called binomial
variable (Bernoulli trial). Now if the outcome of variable X involves 3 or more
mutually exclusive types or categories or events (
E
1
· E
2
···· E
k
) with corresponding
https://t.me/med1917
5.3 Multinomial Distribution 101
probability of occurring (p
1
· p
2
· ··· p
k
), that p
1
+ p
2
+ p
3
+ ··· + p
k
= 1, it named
multinomial variable and we employ multinomial distribution instead of binomial
distribution. For example, blood types can occur in types of A, B, AB,or O.The
multinomial distribution is a generalization of the binomial distribution. If we repeat
an experiment n times independently and each time one of the k disjoint events
occurs, the probability that X will occur (
x
1
· x
2
· ···x
k
) times is obtained from the
following formula:
p
(
x
1
· x
2
· ··· x
k
; n · p
1
· p
2
· ··· p
k
)
=
n!
(x
1
)!(x
2
)! ....(x
k
)!
p
x
1
1
× p
x
2
2
× ··· × p
x
k
k
,
where x
1
+ x
2
+ ··· + x
k
= n and p
1
+ p
2
+ ··· + p
k
= 1.
Example 4 Consider drawing one card randomly from a standard deck of playing
cards, replacing it in the deck, and repeating this process 5 times. Determine the
probability of drawing 2 face cards, 1 ace card, and 2 numeric cards.
Solution: We apply the multinomial distribution formula to solve the example. The
known information is below:
The experiment consists of 5 trials, so n = 5.
The 5 trials produce 2 face cards, 1 ace card, and 2 numeric cards; so x
1
= 2, x
2
=
2, and x
3
= 1.
On any particular trial, the probability of drawing a face, ace, and numeric cards
is
p
1
=
12
52
= 0.231 · p
2
=
4
52
= 0.077 and p
3
=
36
52
= 0.692, respectively.
We substitute this information into the multinomial distribution:
p
(
x
1
· x
2
· x
3
; n · p
1
· p
2
· p
3
)
=
n!
x
1
!× x
2
!× x
3
!
p
x
1
1
× p
x
2
2
× p
x
3
3
=
5!
2!× 1!× 2!
× 0.231
2
× 0.077
1
× 0.692
2
= 0.059.
Example 5 In an urn, there are 4 white balls, 3 red balls, and 3 green balls. A ball is
randomly selected, and its color is noted before being replaced each time. Determine
the probability that, after selecting 5 balls, 2 are white, 2 are red, and 1 is green.
is:
The experiment consists of 5 trials, so n = 5.
The 5 trials produce 2 white balls, 2 red balls, and 1 green ball, so x
1
= 2 x
2
=
2 and x
3
= 1.
On any particular trial, the probability of drawing a white, red, and green is
p
1
=
4
10
= 0.3, p
2
=
3
10
= 0.3 and p
3
=
3
10
= 0.3, respectively.
We replace this information into the multinomial distribution:
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102 5 Discrete Distributions
p
(
x
1
· x
2
· x
3
; n · p
1
· p
2
· p
3
)
=
n!
x
1
!× x
2
!× x
3
!
p
x
1
1
× p
x
2
2
× p
x
3
3
=
5!
2!× 2!× 1!
× 0.4
2
× 0.3
2
× 0.3
1
= 0.1296.
Example 6 From a standard deck of 52 playing cards, we draw 9 cards. Find the
probability that there are 2 ace, 3 face, and 4 numeric cards.
Solution: When the selection of cards from a playing deck card is with replacement,
the probability of occurrence is equal each time. In this case: the probability of
coming an ace equal to
p
(
A
)
=
4
52
=
1
13
, the probability of coming a face card
equal to
p
(
F
)
=
12
52
=
4
13
, and the probability of coming a numeric card equal to
p
(
N
)
=
36
52
=
9
13
.Also x
1
+ x
2
+ x
3
= 2 + 3 + 4 = 9 and p
(
A
)
+ p
(
F
)
+ p
(
N
)
=
1
13
+
4
13
+
9
13
= 1.
p
(
2, 3, 4
)
=
n!
(x
1
)!(x
2
)!(x
3
)!
× p
x
1
1
· p
x
2
2
· p
x
3
3
=
9!
2!3!4!
×
1
13
2
×
4
13
3
×
9
13
4
= 0.499.
Example 7 The distribution of blood types in population is assumed as the table
below.
Blood type Percentage +RH −RH
O 44 37.4 6.6
A 42 35.7 6.3
B 10 8.5 1.5
AB 4 3.4 0.6
At a blood donation center, 12 people donate blood daily. What is the probability
that an equal number (3 individuals) of all four blood types will be donated on a
given day?
Solution: In this example,
p
(
O
)
= 0.44, p
(
A
)
= 0.42, p
(
B
)
= 0.10 and p
(
AB
)
=
0.04
, x
1
= x
2
= x
3
= x
4
= 3,
(
O
)
+ p
(
A
)
+ p
(
B
)
+ p
(
AB
)
= 0.44 + 0.42 + 0.10 +
0.04 = 1
, x
1
+ x
2
+ x
3
+ x
4
= 3 + 3 + 3 + 3 = 12.
p
(
3, 3, 3, 3
)
=
n!
(x
1
)!(x
2
)!(x
3
)!(x
4
)!
× p
x
1
1
· p
x
2
2
· p
x
3
3
· p
x
4
4
=
12!
3!3!3!3!
×
(
0.44
)
3
×
(
0.42
)
3
×
(
0.10
)
3
×
(
0.04
)
3
= 0.00015.
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5.4 The Limit of the Binomial Distribution 103
5.4 The Limit of the Binomial Distribution
As mentioned, the binomial distribution has two parameters n and p. By changing
(larger or smaller) these two parameters, it is sometimes difficult to calculate the
probability. For example, by increasing n or decreasing the value of p too much,
this calculation is difficult, impossible, time consuming, and approximate. To better
understand this difficulty, pay attention to the following two examples.
Example 8
One of three patients with headache is treated with acupuncture. Find
out the probability of recovery x out of 200 patients with acupuncture method.
Solution: In this example, n = 200, p =
1
3
and q =
2
3
and x = x. Using the binomial
distribution, calculations
p
200
(x) =
200
x
1
3
x
2
3
200−x
must be performed.
It is difficult to calculate this expression for any x-value. If x = 0, calculating
the expression
2
3
200
is difficult and impossible. If x = 100, it is also difficult to
calculate the expression
2
3
100
1
3
100
. Finally, if x = 200 is obtained, it is practically
impossible to compute the expression
1
3
200
.
Example 9
According to a study in Kermanshah province, the prevalence of
pulmonary tuberculosis in urban women is 0.002. We select a sample of 150 urban
women in t his province. Find the probability that x people have TB.
Solution: In this example, n = 150, p = 0.002, q = 0.998, and x = x.
p
150
(
x
)
=
150
x
(
0.002
)
x
(
0.998
)
150−x
.
The calculation of this probability is also difficult and very approximate.
It is mathematically proven that in some cases of binomial distribution, other
distributions such as the normal distribution and the Poisson distribution can be used
instead.
The calculation of these expressions (Examples 6 and 7) is very difficult due to
the small p or large n and the answers are very approximate and other methods must
be employed.
It is mathematically proved that in some cases, other distributions such as the
normal distribution and the Poisson distribution can be used instead of the binomial
distribution.
Also, when experiments are performed in a binomial distribution without
replacement, i.e., the probability of success changes in each test, the
Hypergeometric distribution is used.
In addition, if the number of experiments in the binomial distribution is not fixed
and the experiments continue until the first success is achieved, the geometric
distribution should be used.
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