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104 5 Discrete Distributions
Tip
Geometric distribution means that (k − 1) unsuccessful (failure) trials must
occur in order to achieve the first trial to be successful.
5.5 Geometric Distribution
Geometric distribution is used for discrete variables and is a process that meets all
the conditions of binomial distribution; only the number of trials is not fixed. This
means that we repeat the Bernoulli trial with probability of p on any one trial, until
the first success occurs on t he kth trial. Therefore, the random variable of x is the
number of Bernoulli trial and must be the result of the first (k − 1) trials be failure
and only the kth trial be a success. But in a binomial distribution, the number of
trials is predetermined and the random variable X is the number of successes. The
probability of geometric distribution is obtained from the following formula.
p
(
x = k
)
=
(
1 − p
)
k−1
× p.
The expected value (mean) and variance of the number of independent trials to
achieve the first success are equal to:
E
(
x
)
=
1
p
and Var
(
x
)
=
1 − p
p
2
.
The mean and variance of the number of failures before the first success are also
equal to:
E
(
X
)
=
1 − p
p
and σ
2
= V (X ) =
1 − p
p
2
.
The geometric distribution is also called the “waiting time” distribution. Because
if we assume each Bernoulli trial as a unit of time, the waiting time for the first
success is exactly equal to the random variable x, which has a geometric distribution.
Geometric distribution is used in time management to reach the success before a
certain time period. It is also useful for studying a rare phenomenon in population,
such as the presence of a rare blood disorder.
Example 10 A pharmaceutical company makes syrups such that 98% of them are
not defective. The produced syrups are selected at random and tested.
(a) What is the probability that the second selected syrup is the first to be non-
defective?
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5.5 Geometric Distribution 105
(b) What is the probability that the first non-defective syrup is randomly selected
after the 10th selection?
Solution: Let “a non-defective syrup” be a “success” with p = 98% = 0.98.
The second selected syrup is the first to be non-defective, which means that the
first syrup is defective and the second syrup is non-defective. Therefore:
p
(
x = 2
)(
1 − 0.98
)
2−1
×
(
0.98
)
= 0.0196.
The probability that the first non-defective syrup is randomly selected after the
10th selection means that the first to tenth selections must be defective. Therefore,
after the 10th selection means:
P(x > 10). Substitute n by 10 and p by 0.98 in the
formula
P
(
x > n
)
=
(
1 − p
)
n
.
P
(
x > 10
)
=
(
1 − 0.98
)
10
= 10
−20
≈ 0.
Tip
If we wait for predetermined number x of successes in geometric distribution
rather than waiting only one success, the Negative Binomial distribution, which
is generalized of geometric distribution, is used.
Example 11 A patient is waiting for a suitable matching kidney donor for a
transplant. The probability that a randomly selected donor is a suitable match is p
= 0.125. What is the expected number of donors who will be tested before a
matching donor is found?
Solution: With p = 0.125, the mean number of failures before the first success is
E
(
Y
)
=
1−p
p
=
1−0.125
0.125
= 7.0.
For the alternative formulation, where X is the number of trials up to and including
the first success, the expected value is
E
(
X
)
=
1
p
=
1
0.125
= 8.
Example 12 The probability of encountering a universal blood donor (with blood
type O and Rh negative) is 0.05. Calculate the probability that the fifth person chosen
will be the first to qualify as a universal blood donor.
Solution: In this example, p = 0.05. In order for the fifth blood donor to be the first
person with universal donor, it is equal to
p
(
x = 5
)
=
(
1 − 0.05
)
5−1
× 0.05 = 0.0407.
https://t.me/med1917
106 5 Discrete Distributions
5.6 Hypergeometric Distribution
The hypergeometric distribution is one of the most widely used distributions and is
very similar to binomial distribution. In the binomial distribution, the events were
independent and its necessity was the infinity of population or sampling with
replacement. However, in the hypergeometric distribution, the trials are not
independent of each other.
If the sample is selected from a small finite population without replacement,
the trials can no longer be considered independent of each other, the probability
of selecting subjects varies from time to time, and the binomial distribution will not
give exact probabilities. The smaller the population, the less accurate the calculations.
Therefore, if the outcomes occur in one of two types and the selection is without
replacement, the hypergeometric distribution should be used instead of the binomial
distribution. Obviously, in the case of without-replacement selection, the larger the
population size, the closer the accuracy of the binomial distribution will be to the
hypergeometric distribution. Consider a population that has A objects of one type
(such as female), while the remaining B objects are of the other type (such as male).
If n samples are selected without replacement that x objects are from type A and n −
x other objects are type B, Table
5.4 can be considered:
In this case, the probability of getting x objects of type A and n − x objects of
type B is obtained from the following formula.
p
(
x
)
=
A
x
B
n − x
A + B
n
=
A!
(
A−x
)
!×x!
×
B!
(
B−n+x
)
!×
(
n−x
)
!
(
A+B
)
!
(
A+B−n
)
!×n!
.
The easiest way to distinguish a hypergeometric distribution with a binomial
distribution is how to select samples which are selected without replacement in a
hypergeometric distribution. In the binomial distribution, trials must be independent
of each other. That is, the probability of each selection is equal to the other selections.
Example 13 A medical diagnostic laboratory performs 10 tests a day, which 3 cases
are suspected to Covid-19. If 4 samples are randomly selected from this laboratory,
what is the probability that 2 cases are Covid-19?
Table 5.4 Objects of
population and sample in
terms of two types
Type A Type B Total
Population A B A + B
Sample x n − x n
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5.6 Hypergeometric Distribution 107
Solution: Since the sample selection is without replacement, the hypergeometric
distribution is used. N = 10, A = 3, B = 7, n = 4, x = 2, and n − x = 2.
p
(
x = 2
)
=
A!
(
A−x
)
!×x!
×
B!
(
B−n+x
)
!×
(
n−x
)
!
(
A+B
)
!
(
A+B−n
)
!×n!
=
3!
(
3−2
)
!×2!
×
7!
(
7−4+2
)
!×
(
4−2
)
!
(
7+3
)
!
(
7+3−4
)
!×4!
=
3 × 21
210
= 0.3.
Example 14 In a medical diagnostic laboratory, 12 tests are performed daily, 5 of
which are not covered by any insurance. If we choose 3 of these tests randomly, what
is the probability that 2 of them are not insured?
Solution: In such cases where population is finite and we want to identify people
with a characteristic, the selection of people will be without replacement. Therefore,
a hypergeometric distribution can be used. In this example A = 5, B = 7, A + B =
12, n = 4, and x = 3.
p
(
x
)
=
A
x
.
B
n − x
A + B
n
=
A!
(
A − x
)
!.x!
×
B!
(
B − n + x
)
!
(
n − x
)
!
÷
(
A + B
)
!
(
A + B − n
)
!n!
=
5!
(
5 − 3
)
!.3!
×
7!
(
7 − 4 + 3
)
!
(
7 − 3
)
!
÷
12!
(
12 − 4
)
!4!
=
5!
2!.3!
×
7!
6!1!
÷
12!
8!4!
=
(
10 × 7
)
÷ 495 = 0.1414.
Example 15 The classroom consists of 7 female students and 3 male students. Find
the probability that the council will be consisted of two women and one man.
Solution: In the election of the council, individuals cannot be elected by replacement.
Therefore, a hypergeometric distribution can be used. In this example, A = 7, B =
3, A + B = 10, n = 3, and x = 2.
https://t.me/med1917
108 5 Discrete Distributions
p
(
x
)
=
A
x
.
B
n − x
A + B
n
=
A!
(
A − x
)
!.x!
×
B!
(
B − n + x
)
!
(
n − x
)
!
÷
(
A + B
)
!
(
A + B − n
)
!n!
=
7!
(
7 − 2
)
!.2!
×
3!
(
3 − 3 + 2
)
!
(
3 − 2
)
!
÷
10!
(
10 − 3
)
!3!
=
7!
5!.2!
×
3!
2!1!
÷
10!
7!3!
=
(
21 × 3
)
÷ 120 = 0.525.
5.7 Poisson Distribution
The Poisson distribution is a limiting form of binomial distribution.Inthe binomial
variable x counts the number of successes from n independent Bernoulli trials. If n
becomes extremely large and the probability of success becomes extremely small,
applying the binomial distribution is not accurate and it is recommended to employ
the Poisson distribution. Therefore, when rare events occur randomly in time or space,
the number of events observed in a continuous interval is defined as the variable of
interest. The interval may have the nature of time or refer to the length, or to an area
or to volume and space.
If n is large and p is small, so that np < 5 (and cautiously less than 10), the
probability of event occurrence is calculated by the Poisson distribution as follows:
p
(
x
)
= e
−λ
λ
x
x!
,
where e is a constant value of Napierian approximately equals 2.71828 and λ = n · p
and x is an integer figure of 0, 1, 2, …n.
Matsunawa provides accuracy thresholds for the Poisson–binomial
approximation. A general guideline indicates that the Poisson approximation to the
binomial is reliable when n is at least 20 and p is less than or equal to 0.05.
Moreover, it is highly accurate when n is at least 100 and p is less than or equal to
0.10.
Tip
A rule of thumb is that the Poisson approximation to the binomial is good, if
n ≥ 20 and p ≤ 0.05, and very good if n ≥ 100 and p ≤ 0.10.
https://t.me/med1917
5.7 Poisson Distribution 109
This distribution has only one non-negative parameter, because its two
components, n and p, are both non-negative. The shape of the distribution depends
on the quantity of
λ. The mean and variance of this distribution are also equal.
Because in Poisson distribution p is very small and therefore q = 1 − p is very
close to 1, then n.p.q approximately equals to n.p.
μ = σ
2
= λ = n · p.
The Poisson distribution is used in many areas of medical (particularly, in
laboratory, toxicology, bacteriology, and epidemiology) research, which we deal
with the expansion of a substance per unit of area, or particles per unit of volume
and space, or with the occurrence of accidents per unit of time and generally with a
rare event in a very large population.
Examples of rare events include the number of daily referrals from a fire to the
emergency department of a hospital, the number of white blood cells in a drop of
blood, the number of floating and suspended particles or the spread of bacteria in a
liquid, the mortality rate of pregnant mothers, etc.
In this distribution, the probability of an event occurring at a given interval must
be proportional to the length of that interval. This means that the distribution of
events at different interval must be the similar. That is, they are accidentally spread
everywhere. Therefore, if, for example, particles are precipitated in a liquid, the
Poisson distribution cannot be assumed for them.
Example 16 The University of Medical Sciences has 1095 students. If the
probability of birth is the same on different days of the year, find the probability of:
Four students were born today.
Less than 4 students were born today.
More than 4 students were born today.
Solution: Provided information is n = 1095 and p =
1
365
= 0.0027, λ = np =
1095 ×
1
365
= 3. λ = 3 means that we are expected to have 3 births a day in the
medical university. Using Poisson distribution:
For x = 4: p
(
x
)
= e
−λ
λ
x
x!
= 2.718
−3
4
4
4!
= 0.168.
X < 4 means that x can take values of 0, 1, 2, or 3.
p
(
x < 4
)
= p
(
x = 0
)
+ p
(
x = 1
)
+ p
(
x = 2
)
+ p
(
x = 3
)
= 2.718
−3
3
0
0!
+ 2.718
−3
3
1
1!
+ 2.718
−3
3
2
2!
+ 2.718
−3
3
3
3!
= 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472.
x > 4 means that x can take the values 5‚ 6… and 1095. Calculating the probability
for the above series is too long and time consuming, so using its complementary
probability we have:
https://t.me/med1917
110 5 Discrete Distributions
p
(
x > 4
)
= 1 − p
(
x ≤ 4
)
= 1 −
p
(
x < 4
)
+ p
(
x = 4
)
= 1 −
[
0.6472 + 0.1680
]
= 0.1848.
Example 17 Two percent of MRIs taken from patients are not of good quality and
the patient must be reimaged. If the center has 150 weekly visits, find the probability
of all MRIs which are taken well.
Solution: Provided information is n = 150 and p = 0.02 and λ = np = 150 ×
0.02 = 3
. λ = 3 means that we are expected to have 3 reimaging per week. When it
is said that all MRIs are done correctly, it means that none of them are damaged (x
= 0). Using the Poisson distribution, we have:
p
(
x
)
= e
−λ
λ
x
x!
= 2.718
−3
3
0
0!
= 0.0497.
Example 18
An average of 60 calls is made to an emergency center every hour.
What is the probability that the center will not be called at all in 4 min?
Solution: Provided information is n = 240 and p =
1
60
, λ = np = 240 ×
1
60
= 4.
λ = 4 means that we are expected to have 3 calls in 4 min (240 s). For (x = 0) using
the Poisson distribution, we have:
p
(
x
)
= e
−λ
λ
x
x!
= 2.718
−4
4
0
0!
= 0.018.
Example 19 There are two maternity hospitals in a city. The daily capacity of the
larger maternity hospital is 10 people, but at present it receives an average of 4 people
per day. The city’s health authorities plan to close the smaller maternity hospital,
which has in average of two daily visits. Obviously, its clients will inevitably go to
another maternity hospital. Rising demand for larger maternity hospital (more than
a daily 10 women) has worried officials. Assuming that the Poisson distribution can
be used for the number of daily maternity admissions with an average of 6, it is
desirable to estimate the number of days of the year when patient admissions exceed
the nominal capacity (10 people) of the maternity hospital (Table
5.5).
Solution: Officials should be worried if the daily maternity hospital exceeds 10
deliveries (i.e., x = 11‚ 12‚ 13‚ …). Calculating the probability for all these x’s is
difficult, long and even inaccurate, because the value of x is unlimited and infinite.
Therefore, we obtain the probabilities of 10 deliveries and less (x = 0‚ 2‚ 3‚ … 10)
and subtract the sum from 1 to obtain the probability of having more admissions.
Assuming λ = 6.1, these calculations are obtained, for example, for the
probability of no daily delivery and the probability of daily admission of 3 patients
as follows:
https://t.me/med1917
5.8 Exercises 111
Table 5.5 Probabilities
Daily delivery Probability
0 0.0022
1 0.0137
2 0.0417
3 0.0848
4 0.1294
5 0.1579
6 0.1605
7 0.1399
8 0.1066
9 0.0723
10 0.0440
Total 0.9530
p
(
x
)
= e
−λ
λ
x
x!
= 2.718
−6.1
6.1
0
0!
= 0.0022,
p
(
x
)
= e
−λ
λ
x
x!
= 2.718
−6.1
6.1
3
3!
= 0.0848.
The probabilities obtained for all x’s are summarized in Table 6.5.
p
(
x ≤ 10
)
= 0.953,
p
(
x > 10
)
= 1 − p
(
x ≤ 10
)
= 1 − 0.953 = 0.047.
The probability that the maternity capacity will exceed 10 deliveries per day is
equal to 0.047 or (4.7%) 17 days per year
(
0.047 × 365 = 17.16
)
in excess of the
capacity.
After ensuring that n is large and p is small, one way to investigate the use of
Poisson is to calculate the mean and variance of the sample distribution. If these two
measures are close together, the Poisson distribution can be used.
5.8 Exercises
1. From a radioactive object, an average of 15 electrons per minute (with Poisson
distribution) is thrown. Calculate the probability of being thrown in a given
second is 2 electrons?
2. Below a microscope field, the number of cells in 500 cells is as shown in
Table
5.6:
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112 5 Discrete Distributions
Table 5.6 Number of cells in a microscope field cells in 500 cells
No. of cells 0 1 2 3 4 5 6
Freq. 134 182 110 58 9 4 3
A. Can the distribution of this data be considered a Poisson distribution?
B. Assuming that the data distribution is Poisson, find the theoretical
distribution in the table above.
3. The probability of death after a certain surgery in a given period of time is
equal to 0.2. If we have 6 patients who have had surgery, find the following
probability:
A. None of these patients die.
B. A maximum of 2 patients die.
C. Maximum of 2 patients survive.
D. Exactly one patient dies.
4. Among angiography devices that need repair, 7 devices have cardiac monitoring
systems. We selected a sample of 20 of these devices. Find the probability that
5 devices have a cardiac monitoring system.
5. The number of white blood cells in healthy people is 6000 per cubic millimeter
of blood. We take a drop of blood with a volume of 0.001 cubic millimeter of
sample blood. How many white blood cells are expected to be present? Find
the probability that there are a maximum of 2 white blood cells in this drop of
blood?
6. Each series of ointment production in the pharmaceutical factory contains
25,000 tubes. According to previous inspections, for every 9000 tubes of
product, 2 tubes are filled with air and defective. Find the probability that:
A. In every production series of this factory, all the tubes are complete.
B. Only 2 tubes are defective.
7. A student in the exam session is not ready for the exam at all and has to mark the
answers by chance. If there are 10 four-answer questions in this exam, calculate
the probability that 6 questions will be correctly marked.
8. The probability that the syrup produced by a pharmaceutical factory is not
standardized and their bottle is broken or the lid is not completely closed and
dripping equal to 0.002. Calculate the probability that there is no defective bottle
in a 2000 shipment of this product.
9. Monthly consumption of pain relief pills in a city of 1000 households has a
Poisson distribution with an average of 5. How many painkillers should be
brought to the pharmacies of this city for sale each month (we know that no
family consumes more than 15 pills per month).
10.
There are n students in a classroom and n can take the values 5‚ 10‚ 15‚ 20‚ 25‚
30, and 35. Find the probability below:
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5.8 Exercises 113
A. Two students were born in the same day.
B. Then plot the above probabilities against n.
C. Two students were born today?
D. Explain the difference between parts A and C.
11. In a laboratory, 1095 unstable isotopes are stored. If it is assumed that the
probability of atomic disintegration is the same on all different days of the year,
consider the following probabilities:
A. Four isotopes today do disintegrate.
B. Less than 4 isotopes disintegrate today.
C. More than 4 isotopes disintegrate today.
D. Today or tomorrow 4 isotopes disintegrate.
12. The probability of a random accident between the electron and the target
material per second has a Poisson distribution with an average of 10. Find the
probability that no accident occurs in one second?
13. If the probability of a baby being born with white hair (zal) is
1
200,000
, among the
600,000 babies born annually in a population, find the following probabilities:
A. No infant with white-haired are born.
B. At least one infant is born with white-haired.
C. There are more than 5 white-haired infants.
14. Calculate the value of the expression A =
1 +
1
n
n
for n = 2, 4, 6, 8, and 10.
As n enlarges, does the expression A tend to the value of e = 2.71828? The e is
the basis of the natural logarithm used in the Poisson distribution.
15.
When n = 50 and p = 0.1, once, using a binomial distribution, obtain the
probability that x has values of 10, 15, and 20. Then obtain the same probability
using the Poisson distribution for
λ = n.p = 50 × 0.1 = 5 and compare the
results.
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