05/04/2025 1
Probability andprobability distribution
At the end of this chapter, students are expected to
understand the following points
Probability (definition of terms, probability rules)
The difference between probability and probability distribution
Conditional probability
Distribution for categorical variable
Distribution for continuous variable
Different distribution tables
2.
05/04/2025 2
Probability definition
Chanceof observing a particular outcome or
Likelihood of an event
Assumes a “stochastic” or “random” process:
i.e.. the outcome is not predetermined there
‐
is an element of chance
An outcome is a specific result of a single trial
of a probability experiment
3.
05/04/2025
Chance
• When ameteorologist states that the chance of rain is 50%,
the meteorologist is saying that it is equally likely to rain or
not to rain. If the chance of rain rises to 80%, it is more likely
to rain. If the chance drops to 20%, then it may rain, but it
probably will not rain.
• These examples suggest the chance of an
occurrence of some event of a random variable.
3
05/04/2025
Why Probability inMedicine?
• Because medicine is an inexact science,
physicians seldom predict an outcome with
absolute certainty.
• E.g., to formulate a diagnosis, a physician must
rely on available diagnostic information about a
patient
– History and physical examination
– Laboratory investigation, X-ray findings, ECG, etc
5
6.
05/04/2025 6
Cont…
• Anunderstanding of probability is fundamental
for quantifying the uncertainty that is inherent in
the decision-making process.
• Probability theory also allows us to draw
conclusions about a population based on
known information about a sample which drown
from that population.
05/04/2025 8
Terminology
Random experiment/random variable: is one
in which the out comes occur at random or
cannot be predicted with certainty.
e.g. A single coin tossing experiment is a random
as the occurrence of Head(H) and Tail(T)
Trial: A physical action , the result of which
cannot be predetermined
9.
05/04/2025 9
Terminology…
Sample Space:The set of all possible outcomes of an
experiment .
In die throwing, S={1,2,3,4,5,6}
Events: Collections of basic outcomes from the sample space.
We say that an event occurs if any one of the basic outcomes in
the event occurs.
Any subset of sample space.
- Event of getting even number A={2,4,6}
Success/ favorable case: Outcome that entail the happening of
a desired event.
10.
05/04/2025 10
Equally likelyevents:
If in a random experiment all out comes have
equal chance of occurrence.
- In tossing coin both H and T have equal chance to occur
Mutually Exclusive Events (Disjoint Events)
If the occurrence of one event prevent the
occurrence of the other.
- In tossing coin the occurrence of Head prevent the
occurrence of Tail.
11.
05/04/2025 11
Cont…
Independent events(mutualindependence)
The occurrence or non-occurrence of one event
doesn’t affect the occurrence or non-occurrence
of the other event in repeated trials, conduction
of a random experiment.
While tossing of two coin simultaneously, the occurrence of
head in one coin does not affect the occurrence of tail on the
other.
12.
05/04/2025
Two Categories ofProbability
• Objective and Subjective Probabilities.
• Objective probability
1) Classical probability and
2) Relative frequency probability.
12
13.
05/04/2025 13
Types ofprobability
Classical Method
Is based on gambling ideas
If there are n equally likely possibilities, of
which one must occur and m are regarded as
favorable, or as a “success,” then the probability
of a “success” is m/n.
P(A) = m/n
What is the probability of rolling a 6 with a well-balanced
die? Ans.
In this case, m=1 and n=6, so that the probability is 1/6
= 0.167
14.
05/04/2025
Relative Frequency Probability
•In the long run process …..
• The proportion of times the event A occurs —
in a large number of trials repeated under
essentially identical conditions
• Definition: If a process is repeated a large
number of times (n), and if an event with the
characteristic E occurs m times, the relative
frequency of E,
Probability of E = P(E) = m/n.
14
15.
05/04/2025 15
Relative freq…
Examples
•If you toss a coin 100 times and head comes up 40
times,
P(H) = 40/100 = 0.4.
• If we toss a coin 10,000 times and the head
comes up 5562,
P(H) = 0.5562.
• Therefore, the longer the series and the longer
sample size, the closer the estimate to the true
value (0.5)
16.
05/04/2025 16
Subjective Probability
Personalistic (An opinion or judgment by a decision
maker about the likelihood of an event)
Personal assessment of which is more effective to
provide cure traditional/modern
‐
Personal assessment of which sports team will win a
match
Also uses classical and relative frequency methods to
assess the likelihood of an event, but does not rely
on repeatability of any process.
17.
05/04/2025
Properties of Probability
1.The numerical value of a probability always
lies between 0 and 1, inclusive.
0 P(E) 1
A value 0 means the event can not occur
A value 1 means the event definitely will occur
A value of 0.5 means that the probability that
the event will occur is the same as the
probability that it will not occur.
17
18.
05/04/2025 18
2. Thesum of the probabilities of all mutually
exclusive outcomes is equal to 1.
P(E1
) + P(E2
) + .... + P(En
) = 1.
3. For two mutually exclusive events A and B,
P(A or B ) = P(AUB)= P(A) + P(B).
If not mutually exclusive:
P(A or B) = P(A) + P(B) - P(A and B)
19.
05/04/2025 19
4. Thecomplement of an event A, denoted by
Ā or Ac
, is the event that A does not occur
– Consists of all the outcomes in which event A
does NOT occur
P(Ā) = P(not A) = 1 – P(A)
– Ā occurs only when A does not occur.
– These are complementary events.
20.
05/04/2025
Unions of TwoEvents
“If A and B are events, then the union of A and B, denoted
by AUB, represents the event composed of all basic
outcomes in A or B.”
Intersections of Two Events
“If A and B are events, then the intersection of A and B, denoted by A n
B, represents the event composed of all basic outcomes in A and
B.”
Unions and Intersections of Two Events
20
B =With lung
cancer
A=Cigarette
smoking
A n B=Smokers with lung cancer
21.
05/04/2025 21
Additive Lawof Probability
Let A and B be two events in a sample space S. The
probability of the union of A and B is
( ) ( ) ( ) ( ).
P A B P A P B P A B
B
A A n B
22.
05/04/2025
22
Mutually Exclusive Events
MutuallyExclusive Events: Events that have no basic
outcomes in common, or equivalently, their intersection is
empty set.
S
B
A
Let A and B be two events in a sample space S. The probability of the
union of two mutually exclusive events A and B is:
( ) ( ) ( ).
P A B P A P B
23.
05/04/2025 23
Two eventsare independent if the occurrence of one of the
events does not affect the probability of the other event.
That is, A and B are independent if :
P (B |A) = P (B) or if P (A |B) = P (A).
Independent Events
Example:
Let event A stands for “the sex of the first child from a mother is female”;
and event B stands for “the sex of the second child from the same
mother is female”
Are A and B independent?
Solution
P(B/A) = P(B) = 0.5 The occurrence of A does not affect the probability of B,
so the events are independent.
24.
05/04/2025 24
Multiplication rule
–If A and B are independent events, then
P(A ∩ B) = P(A) × P(B)
– More generally,
P(A ∩ B) = P(A) P(B|A) = P(B) P(A|B)
P(A and B) denotes the probability that A and
B both occur at the same time.
25.
05/04/2025 25
Conditional probabilitiesand the multiplicative law
Sometimes the chance a particular event happens depends on
the outcome of some other event. This applies obviously with
many events that are spread out in time.
Example: The chance a patient with some disease survives the next
year depends on his having survived to the present time. Such
probabilities are called conditional.
The notation is Pr(B/A), which is read as “the probability event B
occurs given that event A has already occurred .”
Let A and B be two events of a sample space S. The conditional
probability of an event A, given B, denoted by
Pr ( A/B )= P(A n B) / P(B) , P(B) not = 0.
Similarly, P(B/A) = P(A n B) / P(A) , P(A)not =0. This can be taken as an
alternative form of the multiplicative law.
26.
05/04/2025
Conditional Probability
The conditionalprobability of the event A given that
event B has occurred is denoted by P(A|B).
Then, P(A|B) =P(A ∩ B)/P(B) , P(B) > 0.
Similarly,
P(B|A) = P(A ∩ B)/P(A), P(A) > 0
when do you use conditional probability ???
Sensitivity and specificity
26
27.
05/04/2025 27
Example 1
Calculatingprobability of an event
Table 1: Shows the frequency of cocaine use by sex
among adult cocaine users
_______________________________________________________________________________________________
Life time frequency Male Female Total
of cocaine use
_______________________________________________________________________________________________
1-19 times 32 7 39
20-99 times 18 20 38
more than 100 times 25 9 34
--------------------------------------------------------------------------------------------
Total 75 36 111
---------------------------------------------------------------------------------------------
28.
05/04/2025 28
Questions…
1. Whatis the probability of a person randomly picked is a
male?
2. What is the probability of a person randomly picked uses
cocaine more than 100 times?
3. Given that the selected person is male, what is the
probability of a person randomly picked uses cocaine
more than 100 times?
4. Given that the person has used cocaine less than 100
times, what is the probability of being female?
5. What is the probability of a person randomly picked is a
male and uses cocaine more than 100 times?
29.
05/04/2025 29
Solution
1. Pr(m)=Totaladult males/Total adult cocaine users
=75/111 =0.68 .
2. Pr(c>100)=All adult cocaine users more than 100 times
Total adult cocaine users
=34/111=0.31.
3. Pr (c>100m)=25/75=0.33.
4. Pr(fc<100)=(7+20)/77 =0.35
5. Pr(m ∩ c>100)= Pr(m) × Pr (c>100)=75/111 ×25/75 or
25/34 x 34/111=25/111=0.23.
05/04/2025 31
Application ofprobability of categorical
variables
• Calculating the probability of an event in epidemiological
studies, we can estimate prevalence of certain diseases
in a given population.
– Prevalence of a disease (e.g. Tuberculosis, diabetes,
heart disease),
– Prevalence of certain characteristics (e.g. high blood
pressure, low birth weight) or
– prevalence of certain behavior (e.g. smoking, drug use,
condom use).
32.
05/04/2025
Probability Distributions
• Aprobability distribution is a device used to describe the
behavior that a random variable may have by applying the
theory of probability.
• It is a list of the probabilities associated with the
values of the random variable obtained in an
experiment
• It is the way data are distributed, in order to draw
conclusions about a set of data
• Random Variable = Any quantity or characteristic that is
able to assume a number of different values such that any
particular outcome is determined by chance
32
33.
05/04/2025 33
Probability distribution…
♣A probability distribution of a random variable
can be displayed by a table or a graph or a
mathematical formula.
♣ With categorical variables, we obtain the
frequency distribution of each variable.
♣ With numeric variables, the aim is to determine
whether or not normality may be assumed.
34.
05/04/2025 34
Therefore, theprobability distribution of a
random variable is a table, graph, or
mathematical formula that gives the
probabilities with which the random variable
takes different values or ranges of values.
35.
05/04/2025
A. Discrete ProbabilityDistributions
• For a discrete random variable, the probability
distribution specifies each of the possible
outcomes of the random variable along with the
probability that each will occur
• Examples can be:
– Frequency distribution
– Relative frequency distribution
– Cumulative frequency
35
05/04/2025 37
• Whatis the probability that a patient receives
exactly 3 diagnostic services?
P(X=3) = 0.031
• What is the probability that a patient receives
at most one diagnostic service?
P (X≤1) = P(X = 0) + P(X = 1)
= 0.671 + 0.229
= 0.900
38.
05/04/2025 38
• Whatis the probability that a patient
receives at least four diagnostic services?
P (X≥4) = P(X = 4) + P(X = 5)
= 0.010 + 0.006
= 0.016
39.
05/04/2025
Probability distributions canalso
be displayed using a graph
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 1 2 3 4 5
No. of diagnostic services, x
Probability,
X=x
39
40.
05/04/2025
Binomial Distribution
• Itis one of the most widely encountered discrete
probability distributions.
• Consider dichotomous (binary) random variable
• Is based on Bernoulli trial
– When a single trial of an experiment can result in only
one of two mutually exclusive outcomes (success or
failure; dead or alive; sick or well, male or female)
40
41.
05/04/2025
A binomial probabilitydistribution occurs when
the following requirements are met.
1. The procedure has a fixed number of trials.
2. The trials must be independent.
3. Each trial must have all outcomes that fall into
two categories.
4. The probabilities must remain constant for each
trial [P(success) = p].
41
42.
05/04/2025
Binomial Distribution
A processthat has only two possible outcomes is called a
binomial process. In statistics, the two outcomes are
frequently denoted as success and failure. The
probabilities of a success or a failure are denoted by p and
q, respectively. Note that p + q = 1. The binomial
distribution gives the probability of exactly k successes in
n trials
P(k)
n
k
pk 1 p
n k
42
43.
05/04/2025
Binomial distribution, generally
43
X
n
X
n
X
p
p
)
1
(
1-p = probability of
failure
p = probability of
success
X = #
successes out
of n trials
n = number of trials
Note the general pattern emerging if you have only two possible
outcomes (call them 1/0 or yes/no or success/failure) in n
independent trials, then the probability of exactly X “successes”=
44.
05/04/2025 44
• ndenotes the number of fixed trials
• x denotes the number of successes in
the n trials
• p denotes the probability of success
• q denotes the probability of failure (1- p)
=
• Represents the number of ways of selecting x objects out of n
where the order of selection does not matter.
• where n!=n(n-1)(n-2)…(1) , and 0!=1
45.
05/04/2025
Example 2
• Supposewe know that 40% of a certain population
are cigarette smokers. If we take a random sample
of 10 people from this population, what is the
probability that we will have exactly 4 smokers in
our sample?
45
46.
05/04/2025 46
• Ifthe probability that any individual in the
population is a smoker to be P=.40, then the
probability that x=4 smokers out of n=10
subjects selected is:
P(X=4) =10C4(0.4)4
(1-0.4)10-4
= 10C4(0.4)4
(0.6)6
= 210(.0256)(.04666)
= 0.25
• The probability of obtaining exactly 4 smokers in
the sample is about 0.25.
47.
05/04/2025 47
• Wecan compute the probability of observing zero
smokers out of 10 subjects selected at random,
exactly 1 smoker, and so on, and display the
results in a table, as given, below.
• The third column, P(X ≤ x), gives the cumulative
probability. E.g. the probability of selecting 3 or
fewer smokers into the sample of 10 subjects is
P(X ≤ 3) =.3823, or about 38%.
05/04/2025
The probability inthe above table can
be converted into the following graph
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
No. of Smokers
Probability
49
50.
05/04/2025
II. Probability distributionof continuous
variables
• Under different circumstances, the outcome of a random
variable may not be limited to categories or counts.
– E.g. Suppose, X represents the continuous variable
‘Height’; rarely is an individual exactly equal to 170cm tall.
– X can assume an infinite number of intermediate values
170.1, 170.2, 170.3 etc.
• Because a continuous random variable X can take on an
uncountable, infinite number of values, the probability
associated with any particular one value is almost equal
to zero.
50
51.
05/04/2025 51
Continuous ProbabilityDistributions
There are infinite number of continuous random variables
We try to pick a model that
Fits the data well
Allows us to make the best possible inferences
using the data.
f (x)
x
Uniform Normal Skewed
52.
05/04/2025 52
Properties ofNormal Distributions
The most important probability distribution in statistics is the
normal distribution.
A normal distribution is a continuous probability distribution
for a random variable, x.
The graph of a normal distribution is called the normal
curve.
Normal curve
x
53.
05/04/2025 53
The NormalDistribution
The formula that generates the normal probability distribution is:
Where, s = Population variance
µ = population mean
e =2.718…, π= 3.14…
2
)
(
2
1
2
1
)
(
x
e
x
f
This is a bell shaped curve
with different centers and
spreads depending on
and
54.
05/04/2025
Normal Curve Characteristics
1.It is a probability distribution of a continuous variable.
It extends from minus infinity to plus infinity.
2. It is unimodal, bell-shaped and symmetric.
3. The mean, the median and mode are all equal
4. The curve approaches, but never meets, the abscissa
at both high and low ends.
5. The total area under the curve is 1. (This is a
requirement of any probability density function.)
6. It is determined by two quantities: its mean and SD .
Changing mean alone shifts the entire normal curve to
the left or right. Changing SD alone changes the
degree to which the distribution is spread out.
54
05/04/2025 56
7. Theheight of the frequency curve, which is
called the probability density, cannot be taken as
the probability of a particular value.
• An observation from a normal distribution can
be related to a standard normal distribution
(SND) which has a published table.
57.
05/04/2025
The standard normaldistribution
Since a normal distribution could be an infinite number
of possible values for its mean and SD, it is impossible
to tabulate the area associated for each and every
normal curve.
Instead only a single curve for which μ = 0 and σ = 1 is
tabulated.
The curve is called the standard normal distribution
(SND).
57
58.
05/04/2025
The Standard NormalDistribution
To find P(a < x < b), we need to find the area under the
appropriate normal curve.
To simplify the tabulation of these areas, we standardize
each value of x by expressing it as a z-score, the number
of standard deviations s it lies from the mean m.
58
x
z
59.
05/04/2025
The Standard Normal(z)
Distribution
Mean = 0; Standard deviation = 1
When x = m, z = 0
Symmetric about z = 0
Values of z to the left of center are negative
Values of z to the right of center are positive
Total area under the curve is 1.
59
05/04/2025
Using normal table
61
Thefour digit probability in a particular row and column of Table
1 gives the area under the z curve to the left that particular value
of z.
Area for z = 1.36
62.
05/04/2025
P(z 1.36) =.9131
P(z >1.36)
= 1 - .9131 = .0869
P(-1.20 z 1.36)
= .9131 - .1151 = .7980
Example-4
62
Use Table 1 to calculate these probabilities:
63.
05/04/2025
Example:5 Probability Zis between
–2.59 and 1.31
63
P(-2.59 Z 1.31)
= P(0 < Z 1.31) + P(-2.59 < Z 0 )
= 0.4049 +0.4952 = 0.9001
64.
05/04/2025
Exercises 2
Find theprobability of the following under the SND
– Above 1.96?
– Below –1.96 , 1.96 ?
– Between –1.28 and 1.28?
– Between –1.65 and 1.08? 0.8502
– What level cuts the upper 25%?
– What level cuts the middle 99%?
64
65.
05/04/2025 65
Example: Theaverage weight of pregnant women attending a
prenatal care in a clinic was 78kg with a standard deviation of
8kg. If the weights are normally distributed:
a) Find the probability that a randomly selected pregnant woman
weights less than 90kg.
Probability and Normal Distributions
P(x < 90) = P(z < 1.5) = 0.9332
-
90-78
=
8
x μ
z
σ
= 1.5
The probability that a
randomly selected pregnant
woman weights less than
90kg. is 0.9332.
μ =0
z
?
1.5
90
μ =78
P(x < 90)
μ = 78
σ = 8
x
66.
05/04/2025 66
Example:
b) Basedon the above example, find the probability that a
pregnant woman weights greater than 85kg.
Probability and Normal Distributions
P(x > 85) = P(z > 0.88) = 1 P(z < 0.88) = 1 0.8106 = 0.1894
85-78
= =
8
x - μ
z
σ
= 0.875 0.88
The probability that a
randomly selected pregnant
woman weights greater than
85kg. is 0.1894.
μ =0
z
?
0.88
85
μ =78
P(x > 85)
μ = 78
σ = 8
x
67.
05/04/2025 67
Example:
From theabove example, find the probability that a randomly
selected pregnant woman weights between 60 and 80.
Probability and Normal Distributions
P(60 < x < 80) = P(2.25 < z < 0.25) = P(z < 0.25) P(z < 2.25)
- -
1
60 78
= =
8
x μ
z
σ
-
= 2.25
The probability that a
randomly selected pregnant
women weights between 60
and 80 is 0.5865.
2
- -
80 78
=
8
x μ
z
σ
= 0.25
μ =0
z
?
? 0.25
2.25
= 0.5987 0.0122 = 0.5865
60 80
μ =78
P(60 < x < 80)
μ = 78
σ = 8
x
68.
05/04/2025 68
Exercise 3
Calculatethe following
probabilities when X is taken
different value with mean 35 and
SD 2?
A. P(x<37),
B. P(x>40),
C. P(38<x<40)
69.
05/04/2025 69
Exercise
1. Whatproportion of newborns will weight above 2700 grams?
2. What is the probability that a randomly selected newborns
will weight between 2800 and 2700 grams?
3. What is the 75 percentile in gram for the distribution of
weight of newborns?
4. What is the probability that a randomly selected newborns
will weight exactly 2900 grams?
A population of newborn infant have a
mean weight of 2800 grams with
standard deviation of 400 grams.
Based on this information give a short
answer to the following questions.
#35 The probability distribution of a discrete random variable is a table, graph, formula, or other device used to specify all possible values of a discrete random variable along with their respective probabilities. The relationship between the values and their associated probabilities is called a probability mass function.
#37 Many random variables are displayed in tables or figures in terms of a cumulative distribution function rather than a distribution of probabilities of individual values. The basic concept is to assign to each individual value the sum of probabilities of all values that are no larger than the value being considered. Thus, the cumulative distribution function of a random variable X is denoted by F(X) and, for a specific value x of X, is defined by P(X≤ x) and denoted by F(x).
#40 In a sample of n independent trials, each of which can have only two possible outcomes. denoted as “success” and “failure”.
#49 Note that, although p is a fraction, its sampling distribution is discrete and not continuous, since it may take only a limited number of values for any given sample size.
As the sample size n increases the binomial distribution becomes very close to a normal distribution, and this can be used to calculate confidence intervals and carry out hypothesis tests.
In fact the normal distribution can be used as a reasonable approximation to the binomial distribution if both np and n-np are 10 or more. This approximating normal distribution has the same mean and standard error as the binomial distribution.