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4.7 Odds and Probability 83
Tips
The PPV and NPV describe the performance of a diagnostic test or other
statistical measure.
They depend also on the prevalence of a disease.
A high result can be interpreted as indicating the accuracy of such a statistic.
4.7 Odds and Probability
Although the words odds and probability are closely related, they are not equivalent.
These two terms are widely used in biostatistics and epidemiology and this section
gives a brief description of the two terms. The term of odds is defined as the probability
that the event will occur divided by the probability that the event will not occur. In
other words, odds is the ratio of chance of “winning” a bet by the chance of “losing”
that bet. We know that the probability of something not happening (or not winning)
is equal to the “one minus the probability of happening or winning” of that thing.
Therefore:
Odds =
p(A)
1 − p(A)
.
By simple algebraic operations in this equation, the probability of occurrence of
the event is equal to:
p(A) =
Odds
1 + Odds
.
The relationship between odds and probability is summarized in the table below.
Probability Odds Interpretation
1.00 ∞ The event will definitely happen
0.90 9.0 An event happens in 90% of cases (that is, it happens nine times versus
not happening once)
0.75 3.0 The event occurs in 75% of cases (i.e., three times against one time)
0.667 2.0 An event occurs in two-thirds of cases (i.e., twice versus once)
0.50 1.0 The event happens in half of the cases (that is, it happens once versus not
happening once)
0.333 0.5 The event happens in a third of cases (that is, it happens once versus not
happening twice, it is half)
0.25 0.333 The event happens in 25% of cases (i.e., it happens once versus not
happening three times, one third)
0.10 0.111 The event happens in 10% of cases (that is, it happens once versus nine
times not happening, one ninth)
0.00 0 An event never happens
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84 4 Probability
As can be seen from the above table, the probability varies between zero and one,
while “odds” can take any non-negative value. For very low probabilities, the value
of odds is very close to the probability value. But as the probability increases, so
does the odds. So that when the probability reaches 0.5, the odds is equal to one.
When the probability reaches 1, the value of odds tends to infinity. This definition of
odds is also consistent with the dialog language. For example, if the hope of losing a
bet is three to one (3:1), it means that your hope of “losing” in the bet is three times
your hope of “winning”. So you lose the bet with a probability of 0.75.
4.8 Counting Principle
The counting principle is a rule and very helpful to figure out the total number of
possible outcomes (sample space) in a situation. For example, if there are m possible
ways for an event to occur and n possible ways for another event to occur, there are
m x n possible ways for both events to occur.
In other words, the counting principle is used to determine how many different
ways one can choose/do certain events and is very useful in calculating the probability
of an event. It is especially necessary to calculate the number of possible events and
favorable events. These methods include factorial, permutation, and combination and
are briefly described below.
Factorial
A factorial, noted by n! with integer number of n (between 1 and infinite), is a product
of consecutive counting numbers starting at 1 and ending at n.
n!= 1 × 2 × 3 × ··· × (n − 2) × (n − 1) × n.
The number 0 is an exception. 0!= 1 and 1!= 1.
It can be easily said:
n!= n × (n − 1)!.
It is very useful for when we’re trying to count how many different orders there
are for things or how many different ways we can combine things.
Example 12
How many different ways can we arrange 5 books in the shelf?
Solution: We have 5 choices for the first book. For second book, we are left with 5
− 1 = 4, so that we have 5 × 4 choices for the first two books, in order. Now, for the
third book, we have 5–2 = 3 choices giving us 5 × 4 × 3 in order. And so on, until
we get down to just one book remains, and then just one thing remaining. Altogether,
we have 5 × 4 × 3 × 2 × 1 = 120.
https://t.me/med1917
4.8 Counting Principle 85
Permutation
The number of ways to select and arrange objects from among distinct objects is
permutation and denoted by
P
r
n
. In permutation, order of objects matters. This means
that ABC, ACB, BAC, BCA, CBA, and CAB are different.
P
r
n
=
n!
(
n − r
)
!
.
Example 13 In how many ways, we can pick 3 letters from 5 letters (A, B, C, D and
E)?
Solution: We have 5 choices for the first letter. For the second letter, we are left with
5 − 1 = 4, so that we have 5 × 4 choices for the first two letters, in order. For the
third letter, we have 5 − 2 = 3 choices giving us 5 × 4 × 3 in order. Then, we have
5 × 4 × 3 = 60.
P
r
n
=
n!
(
n − r
)
!
=
5!
(
5 − 3
)
!
= 60.
Example 14 A coded lock is made from digits 0–9 (without repetition). We choose
a three-digit code by chance. What is the probability that the lock will be open?
Solution: The number of possible outcomes (sample space) is equal to n
(
S
)
=
P
3
10
= 10 × 9 × 8 = 720, and the number of favorable outcomes is equal to 1. So
the probability of unlocking is
1
720
in the first time.
Combination
Combination is the total number of ways of selecting r distinct combinations of n
objects, irrespective of order (i.e., order NOT important). That is, AB is the same
as BA and is not different. The number of combinations of r object and n object is
obtained from the following formula:
C
r
n
=
n
r
=
n!
r!
(
n − r
)
!
.
It can be easily shown that the following relations are always established.
C
0
n
= C
n
n
= 1 C
1
n
= n
C
r
n
= C
n−r
n
C
r
n−1
+ C
r−1
n−1
= C
r+1
n
.
Example 15 A health center has 5 employees. With these 5 staffs, in how many
ways we can create:
A.
A queue of 5 including all 5 staffs?
B. A queue of 3 staffs?
C. A committee of 3 staffs.
https://t.me/med1917
86 4 Probability
Solution:
A. Forming a queue means that when people swap places, a new queue is formed.
Therefore, factorial should be used.
5!= 5 × 4 × 3 × 2 × 1 = 120.
B. To form a queue of 3 out of 5 people, you must use the permutation of 5 members
3 to 3. Since in the queue, the order of people is matter:
P
3
5
=
5!
(
5 − 3
)
!
=
5 × 4 × 3 × 2 × 1
2 × 1
= 60.
C. In order to form a three-member committee, the arrangement of individuals is
not important, i.e., the ABC is no different from the BCA and CAB. So we use
combination and have:
5
3
=
5!
3!
(
5 − 3
)
!
=
5!
2!3!
= 10.
Example 16 We have 4 books with different cover designs. With these 4 books, in
how many ways we can:
Arrange them on the library shelf?
Gave two of them as a gift to someone?
Solution:
A factorial should be used to count the number of cases in which all four books can
be arranged on the shelf. 4!= 4 × 3 × 2 × 1 = 24.
To donate two of the four available books, we need to use the permutations of
4 books 2 by 2, because the order of their placement creates a different donation.
P
2
4
=
4!
(
4−2
)
!
=
4×3×2×1
2×1
= .
Example 17 There are two populations, the first one with 7 patients and 18 healthy
individuals and the second with 11 patients and 28 healthy individuals. One person
randomly migrates from the first population to the second population and settles in
that population. After that, we select another person from the second population.
What is the probability that the selected person is ill?
Solution: We denote event A = immigrant person and event B = the illness of the
selected person. The probability of being ill the immigrant person is
p
(
A
)
=
7
25
and
being healthy is
A
=
18
25
.
If the immigrant person is ill, the second population will have 12 patients and
28 healthy ones. As a result, the probability that the second person be ill is equal
p
(
B|A
)
=
12
40
.
https://t.me/med1917
4.9 Expected Value 87
If the immigrant person is healthy, the second population will have 11 patients
and 29 healthy. As a result, the probability that the second person be ill is equal
p
B|A
=
11
40
.
Therefore, the probability that the second person be ill is equal to:
p
(
B
)
= p
(
B|A
)
× p(A) + p
B|A
× p(p
B|A
)
=
12
40
×
7
25
+
11
40
×
18
25
= 0.282.
Example 18 From a city where 60% of the population are literate, we randomly
select 10 people. Find the following probabilities:
i. Every 10 people be literate.
ii. The first 3 selected people only be literate.
iii. Only the first two and the last two be literate.
Solution: If we denote event A = be literate, P(A) = 0.6 and:
i. In this case, x = 10 and p = 0.6, then p
(
x = 10
)
=
(
p
)
x
= 0.6
10
= 0.006.
Since we want only the first three selected persons to be literate, which means
that the other seven are illiterate.
ii. In this case, p = 0.6 and x = 3, then p
(
x
1
, x
2
, x
3
)
=
(
p
)
x
·
(
1 − p
)
n−x
= 0.6
3
×
(1 − 0.6)
7
= 0.00035, because only the first three people are important to us,
and the other 7 people are not important to be literacy or illiteracy.
iii. Because we want the first two and the last two to be literate, which means that
having literacy and not having the rest of the six don’t matter, and the probability
of that six is equal to one.
p
(
x
1
, x
2
, x
9
, x
10
)
= 0.6
2
×1
6
×0.6
2
= 0.6
4
= 0.1296.
4.9 Expected Value
Expected value or mathematical expectation of a discrete random variable of X
characterized by E(x) is a generalization of the weighted average. When the
variable X can take the values of
x
1
· x
2
· ··· · x
n
, respectively, with the
corresponding probabilities of
p
1
· p
2
· ··· · p
n
(sum of probabilities is 1), means
that the probability of having
x
1
is p
1
, the probability of having x
2
is p
2
and …. the
probability of having
x
n
is p
n
. The following table can be drawn.
X-values x
1
x
2
x
3
··· x
n
Probability p
1
p
2
p
3
··· p
n
Then, the expected value of X is:
E
(
X
)
= p
1
x
1
+ p
2
x
2
+ p
3
x
3
+ ··· + p
n
x
n
=
n
1
p
i
x
i
.
https://t.me/med1917
88 4 Probability
There are two important properties for expected value:
The expected value is always between the smallest
(
x
min
)
and the largest values
(
x
max
)
, i.e., X
min
≤ E
(
X
)
≤ X
max
.
For any constant values a and b,wehave: E
(
a + bX
)
= a + bE(X ).
Insurance companies operate with expected value to estimate the cost of life
insurance policy.
An insurance firm offers a 1-year term life insurance plan to a 50-year-old male
customer. The customer pays a premium of $300. In the event of his death within the
year, the company will provide a payout of $60,000 to his beneficiary. According to
the U.S. Centers for Disease Control and Prevention, the likelihood of a 70-year-old
man being alive 1 year later is 0.9715. Let’s denote X as the profit earned by the
insurance company. Determine the probability distribution and the expected value
of the profit.
Example 19 An insurance company wants to figure out the break-even point without
any expenses (premium money) for 50 years old for the next year. If he dies within one
year, the company will pay $60,000 to his beneficiary. Assuming that the probability
of dying 50 years old is 0.005 and the probability of living for the next year is 0.995,
figure out the break-even point.
Solution: A break-even point means that the total sales equal the total expenses.
This means that the insurance company is bringing in the same amount of money that
they need to cover all of their expenses. To figure out it, the mathematical expectation
must be zero.
E
(
X
)
= 0,
p
(
living
)
× paid amount + p
(
death
)
× premium money = 0,
0.005 × 60000 + 0.995 × premium money = 0,
300 + 0.995 × premium money = 0,
premium money =
300
0.995
= 301.507.
Then, the minimum expected premium without any extra expenses is 301.507.
p
(
living
)
× paid amount + p
(
death
)
× premium money = 0,
0.005 × 60000 + 0.995 × premium money = 0,
300 + 0.995 × premium money = 0,
premium money =
300
0.995
= 301.507.
Example 20
Variable X is the cost of treatment for each patient in a hospital. In this
hospital, six types of treatments A, B, C, D, E, and F are performed which cost 1.5,
4, 2.5, 7, 5.2, and 3.5 currencies, respectively. If patients refer to treatments A, B, C,
D, E, and F, respectively, obtain the mathematical expectation of this variable (the
cost per patient in this hospital).
https://t.me/med1917
4.10 Exercises 89
Solution: We create the table below and write down in another row the multiplication
of cost of any type of treatment by the probability of it. The sum of this row is equal
to the mathematical expectation of the above variable E(x) = 3.495.
Treatment A B C D E F Sum
X 1.5 4 2.5 7 2.5 3.5 –
P 0.25 0.18 0.20 0.10 0.15 0.12 1
x.p 0.375 0.72 0.50 0.70 0.78 0.42 3.495
4.10 Exercises
1. According to Table 4.1 of this chapter, find the probability that a 55 years old
will die by the age of 60.
2. If a family has five children, find the probability that have:
A. Two consecutive sons and 3 consecutive daughters.
B. Four consecutive sons.
3. The class has 18 female students and 12 male students; 6 girls and 4 boys
wear glasses. We randomly select 4 students from the attendance list. Find the
probability that:
A. All 4 students are wearing glasses.
B. None of them wears glasses.
4. In the above class, 2 students are selected from the list, and find the conditional
probability of:
A. Both students are wearing glasses, if both of them are boys.
B. Neither of them wears glasses, if both of them are girls.
Table 4.1 Distribution of blood type by gender and blood type
Blood type Female Male Tot a l
Known A 15 2 17
B 16 1 17
O 1 2 3
AB 27 0 27
Total 59 5 64
Don’t know 96 40 136
Grand total 155 45 200
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90 4 Probability
5. With digits 0–9 how many figures can be formed:
A. With 3 digits.
B. With 3 digits without zero.
C. With 3 digits and multiplicative of 5.
D. With 3 digits but even.
6. In a survey of several GPs, 64 out of 200 people identified their blood type. If
the number of visitors to a GP on a particular day is 7, find the probability that:
A. Patients know their blood type one every other.
B. Only the first three patients know their blood type.
C. Only three patients know their blood type.
7. If in the above study the distribution of blood type is as shown in Table 4.1,
calculate the probabilities below:
A. A visitor knows his/her blood type.
B. A female visitor knows her blood type.
C. A male visitor has blood type of O.
8. Half of the population is over 50 years old. If the probability of developing
hypertension in people over 50 is 35% and in people under 50 is 15% and a
person from this population is chosen randomly, calculate the following
probabilities:
A. That person is over 50 year’s old and developed hypertension.
B. The person is less than 50 years old and does not have blood pressure.
C. That person developed blood pressure.
9. The probability of recovery after a surgery operation is 60%. We randomly
select 8 patients who have undergone surgery under the same condition. Find
out the probability that:
A. The first 3 patients have recovered and this method is ineffective for the
other 5 patients.
B. Only 3 patients recover.
10.
There are 10 yellow apple seedlings and 5 red apple seedlings in a laboratory
field. An unknowing person accidentally picked and planted three saplings. Find
the probability that:
A. All three seedlings are red apples.
B. Maximum of two seedlings is yellow apples.
C. At least one apple is yellow.
https://t.me/med1917
4.10 Exercises 91
11. In a laboratory greenhouse, 640 of the 800 seeds sown germinated. We choose
six seeds by accident. We select six seeds randomly. Find out the probability
that:
A. None of them has sprouted.
B. All have sprouted.
12. A livestock vaccine with a probability of 0.80 has a positive effect. If we select
6 livestock that have been reacted in the same conditions, find out:
A. The probability that this vaccine has had a positive effect on the first four
livestock and is unaffected for the last two l ivestocks.
B. Only affect 4 livestocks.
13. Assuming that the two events A and B are independent of each other, which of
the following is independent of each other?
A. A and B.
B. A and B.
C. Both.
D. A prosthetic factory produces 170 hand prostheses and 330 foot prostheses
daily. We choose 30 prostheses randomly. Find out the probability that both
prostheses are of the same type?
14. Two sports teams A and B play in a three-set match. The final winner of the
match is the team that has won two sets. The history of the two teams shows
that the two teams have faced each other 10 times, and team A has won 6 times
and team B 4 times. Find out the following possibilities:
A. The match will be finished in three sets.
B. The match ends in two sets.
To solve this exercise, a tree diagram with probability of 0.6 and 0.4 can be used.
15. Sixty % of patients with dental lesions prefer TENS to anesthetic. Among the
8 patients being treated, consider the f ollowing probabilities:
A. The first three patients preferred the use of TENS and the rest preferred
anesthetic.
B. Three of these patients use TENS and the rest use anesthetic.
Notice: The rest of this chapter’s exercises are suitable for medical equipment
engineering students.
16. The probability of performing pneumatic, mercury, and gear rotations in a cobalt
CO
60
device is 0.40, 0.10, and 0.50, respectively. In the studies of this device,
it has been found that 3, 4, and 2 percent of these rotations are faced with
problems, respectively. Find out the probability that:
A. A rotation that encounters difficulty, is it mercury type?
B. Pneumatic rotation but without problems?
https://t.me/med1917
92 4 Probability
17. In a sample of 30 angiography devices, 9 devices have film changer. Four
angiographic devices are selected randomly, find out the probability that:
A. All 4 devices have film changer.
B. None of them has a film changer.
C. A maximum of 2 devices have a film changer.
D. Have at least one film changer.
18. In one study, 70 signals out of 100 real and even signals in one book and 50
signals out of 120 signals in another book were detected without Fourier series
coefficients. We now randomly select 15 signals from both books. What is the
probability that these 15 signals have Fourier coefficients (if we know that the
signals were selected from the second book)?
19. The probability of using Step Down model and Step Up model
autotransformers by an engineer in a certain circuit is equal to 0.05 and 0.03.
What is the probability that the engineer will use at least one of these models
in his circuit structure?
20. Half of the diagnostic radiology devices available in a medical center have been
used for more than 20 years. The probability of failure of devices with a lifespan
of more than 20 years is equal to 0.35 and the probability of failure of devices
with a lifespan of less than 20 years is equal to 0.15. If a device is selected at
random, find the following possibilities:
A. The device has been working for more than 20 years and is broken.
B. Has worked for less than 20 years and is worked properly.
C. Is broken.
21. Eighty % of medical devices produced in a factory have a certificate. We
select 7 devices randomly. Find out the probability that the first 5 devices have
certificates and the last 2 devices do not have certificates.
22. In an electronics lab, there are 2 analog multimeters and 3 digital multimeters.
We choose a multimeter at random. Then add a multimeter opposite to what
came out of the container. Then take another multimeter out of the container.
What is the probability that the second multimeter is analog? If the second
multimeter is analog, what is the probability that the first multimeter is digital?
23.
In a specialized hospital, 640 out of every 800 patients suspected of cancer
are imaging with PET system. Six patients will be selected to evaluate the test
results. Find the probability that:
A. No PET system has been used for any patient.
B. All 6 patients used PET system.
C. Only two patients used pet system.
24. The probability of a sustained response in orbit is equal to
1
7
. What is the
probability that the number of sustained responses in the study of 5 of these
circuits is more than 3?
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