Lecture-3 Probability and probability distribution.ppt

Probability and probability distribution
University of Gondar
College of Medicine and Health Science
Institute of Public Health
Department of Epidemiology and
Biostatistics
a b

Larson & Farber, Elementary Statistics: Picturing the World, 3e 2
Objective of the chapter
At the end of this chapter, students are expected to
understand the following
Probability
The difference between probability and probability distribution
Conditional probability
Distribution for categorical variable
Distribution for continuous variable
Different distribution tables
Normal distribution
Student t-distribution
Chi-square distribution
Lemma Derseh, department Epidemiology and Biostatistics, University of Gondar

Probability
 Because medicine is not an exact science, physicians
seldom can 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 studies, X-ray findings, ECG, etc
 Although no test result is absolutely accurate, it does
affect the probability of the presence (or absence) of a
disease.

Probability cont…
 An understanding 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 of patients based on known information
about a sample of patients drawn from that population.

Probability cont…
5
Conclusions/Inferences in science are using probability

Probability experiment is an action through which specific
results/outcomes (counts, measurements or responses) are
obtained.
Basic terms
Example:
Tossing a coin and observing the face showing up is a
probability experiment.
Outcome: It is the result of a single trial in a probability
experiment. It is also called simple event.
Example: the outcome of the sex of a newborn from a
mother in delivery room is either Male or female

Basic terms cont…
 Sample space: The set of all possible outcomes for an
experiment
 Example: The sample space for the sex of newborns when
two mothers are in the gynecology ward to give birth is:
{MM, MF, FM, FF}
 An event consists of one or more outcomes and is a subset of
the sample space
 Example: From the above experiment, an event consisting of
at least one female is E = {MF, FM, FF}

Basic terms cont…
 Random variable: is a function that associates a unique
numerical value with every outcome of an experiment.
 Probability function: A function that for each possible
value of a discreet random variable takes on the
probability of that value occurring
 Probability density function: A curve that specifies by
means of the area under the curve over an interval, the
probability that a continuous random variable falls
within the interval.

Unions of Two Events
“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
AnB, represents the event composed of all basic outcomes in A and B.”
Unions and Intersections of
events
9
A=Cigarette
smoking
B =With lung
cancer
AnB=Smokers with lung cancer

Additive Law of 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
    
10
B
A A n B

Mutually Exclusive Events
Mutually Exclusive 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
  
11

Two events are 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.

Multiplicative rule of probability
If A and B are two events in a sample space S, the
probability of the joint occurrence of both A and B is given
by:
P(A n B) = P(A)P(B/A)
or
P(A n B) = P(B n A) = P(B)P(A/B)
However, if A and B are independent events, then
P(A n B) = P(A)P(B)

Conditional probability and its
application
Disease status Total
D+ D-
Exposure
Status
E+ a b a+b
E- c d c+d
Total a+c b+d a+b+c+d

Application of conditional prob.
cont…
OR = Odds of diseased among exposed
Odds of diseased among non-exposed

Calculating probability of an
event
Table bellow shows the frequency of cocaine use by gender
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
---------------------------------------------------------------------------------------------

Questions
a. What is the probability of a person randomly picked is a male?
b. What is the probability of a person randomly picked uses cocaine
more than 100 times?
c. Given that the selected person is male, what is the probability of a
person randomly picked uses cocaine more than 100 times?
d. Given that the person has used cocaine less than 100 times, what is
the probability of being female?
e. What is the probability of a person randomly picked is a male and
uses cocaine more than 100 times?
17

Counting Rules
We have three different counting rules.
 Basic multiplication rule
 Permutations
 Combinations

Counting Rules cont…
Basic multiplication rule
If we have an experiment with k parts (such as
k tosses), and
Each part has n possible outcomes (such as
heads & tails), then
The total number of possible outcomes for the
experiment is nk
This is the simplest multiplication rule.

Basic multiplication rule cont…
E.g. Assume we have a coin & a die. If we toss a
coin first and then the die, how many possible
outcomes does the experiment have?
We have: n1xn2 = 2 x 6 = 12 possibilities

Permutations
The number of possible permutations is the number of
different orders in which particular events occur. The number
of possible permutations are
where r is the number of events in the series, n is the number
of possible events, and n! denotes the factorial of
n = the product of all the positive integers from 1 to n.
)!
(
!
r
n
n
r
p
n


21

Permutations cont…
 Example: Five different new drugs are given
simultaneously to each of the five patients. The drugs are
compared by the length of time taken to cure the patients.
(assume that the five patients are same in all other
characteristics like: disease type, severity status, sex, age
etc. )
a) How many possible drugs we have for the 1st place (the
fastest to cure).
b) How many possible arrangements we have for the first
three drugs?
c) How many possible arrangements of all the drugs we
have?

Combinations
When the order in which the events occurred is of no interest,
we are dealing with combinations. The number of possible
combinations is
Where r is the number of events in the series, n is the number
of possible events, and n! denotes the factorial of n = the
product of all the positive integers from 1 to n.
23

Combinations cont…
 To carry out an experimental study on the vaccine trial,
twelve healthy individuals are to be randomized to the new
drug and placebo groups with 1 : 1 ratio.
 How many different types of samples can be assigned to
the new drug?
Ans. 12c6

Sampling with and without replacement
cont…
 Example: From a population of 10 TB patients who are
numbered 1 up to 10, two are randomly selected
sequentially and assigned to a new drug treatment .
 How many possible samples of size 2 can we form if
sampling is:
a. With replacement?
a. Without replacement?

Sampling with and without … (answer)
 If sampling is with replacement:
 If sampling is without replacement:
Here it is the same as 10p2
Question:
 If sampling was not sequential (taking the two patients
simultaneously), find all the possible samples
 What is the probability of getting patients numbered 1
and 2 in the above three situations?
Number of ways = 10 x 10 = 100
Number of ways= 10 X 9 = 90

Probability distribution of
categorical variables

Probability distribution
 Every random variable has a corresponding probability
distribution.
 and probability distribution or just distribution refers to
the way data are distributed, in order to draw conclusions
about a set of data.
 It also refers to the underlying, usually unknown, distribution
of the population or random variable.
 The probability distribution of a categorical variable tells
us with what probability the variable will take on the different
possible values (outcomes).
 With numeric variables, the aim is to determine whether or not
normality may be assumed. 28

Probability distribution of a categorical
variables
Example on categorical random variable
 Consider the value on the face showing up from tossing a die. The
probability distribution of this variable is
Value on face 1 2 3 4 5 6
Probability 1/6 1/6 1/6 1/6 1/6 1/6
 In any prob. distribution, each probability must be between
0 and 1 and that their sum must be 1.
29

Probability distribution of categorical
cont…
30
 On the other hand, frequency distribution refers to the
observed distribution of the sample.
 When the number of observations is large, the observed
relative frequency distribution (empirical probability)will
tend to look the true probability distribution (theoretical).
 E.g. If the number of trials in throwing a die increases from 100 to
10,000, the frequency distribution approaches the theoretical
probability distribution of 1/6 for each of the six outcomes
 A probability distribution of a random variable can be displayed
by a table or a graph or a mathematical formula.
Lemma Derseh, Department of Epidemiology and Biostatistics, University of Gondar

Binomial distribution
31
For X (the number of successes in n trials) to be Bi(n, p), we
must have n independent Bernoulli trials
In general the binomial distribution involves three assumptions
I. There are fixed n number of Bernoulli trials each of
which results in one of two mutually exclusive outcomes.
II. The outcomes of n trials are independent.
III. The probability of “success” is constant for each trial
Pr (X=success) = Pr (X=1) = p
 Pr (X=failure) = Pr (X=0) = 1-p

Binomial distribution, generally
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
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” is:
32

Binomial distribution….
33
Example 1: Suppose that in a certain population, 52% of all recorded
births are males. If we select randomly 10 birth records, what is the
probability that exactly
5 will be males?
Given n=10, x=5 and p = 0.52
Pr (X= x) = n! p x (1- p) n- x
x ! (n -x )!
Therefore, Pr (X=5) = 10! X 0.52 5 x (1- 0.52)10-5 =0.24
5!(10-5)!
3 or more will be females?
Pr(X≥3) = 1- Pr (X<3) = 1-[Pr(X=0)+Pr(X=1)+Pr(X=2)]
=1-[0.001+0.013+0.055]= 1-0.069=0.931

Binomial distribution….
 Example 2: The exam has five questions and each question has
four multiple choice in which one of the choice is the correct
answer. If a student answers all the question by guess.
1. What is the probability that he will answer 3 out of 5 questions
correctly?
2. What is the probability that he will answer more than 3
questions correctly?
34

Continuous Probability
Distributions

Continuous probability Distributions
A continuous random variable has an infinite number of
possible values that can be represented by an interval on
the number line.
Proportion of patients with positive HIV
test result per day
0% 25%
12.5% 37.5% 62.5%
50% 75% 100%
87.5%
The proportion of patients with
positive HIV test result can be
any number between 0% and
100% inclusive.
The probability distribution of a continuous random
variable is called a continuous probability distribution.

Continuous Probability Distributions
 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.
37
f (x)
x
Uniform
Normal
Skewed

Properties of Normal 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

The Normal Distribution
 The formula that generates the normal probability distribution is:
39
Where, s = Population variance
µ = population mean
e =2.718…, π= 3.14…
2
)
(
2
1
2
1
)
( s


s




x
e
x
f
This is a bell shaped
curve with different
centers and spreads
depending on  and s

Properties of a Normal Distribution
1. The mean, median, and mode are equal.
2. The normal curve is bell-shaped and symmetric about
the mean.
3. The total area under the curve is equal to one.
4. The normal curve approaches, but never touches the x-
axis as it extends farther and farther away from the
mean.
5. Between μ  σ and μ + σ (in the center of the curve), the
graph curves downward. The graph curves upward to
the left of μ  σ and to the right of μ + σ. The points at
which the curve changes from curving upward to
curving downward are called the inflection points.

μ  3σ μ + σ
μ  2σ μ  σ μ μ + 2σ μ + 3σ
Inflection points
Total area = 1
x

The Family of Normal Distribution
The Line of symmetry for the curve indicates the mean of the
distribution, and the spread shows the magnitude of the standard
deviation
A normal distribution can have any mean and
any positive standard deviation.

The area under the curve
 The area under a curve can be obtained:
a. By taking the integral of an interval, (a, b)
b. By preparing a tables containing areas for each curve
However, both of these are not good solutions because:
i. Either it requires us to have some knowledge of calculus or
ii. Preparing tables for the infinite family of normal curves is
impossible
2
2
2
/
)
(
b
a
2
1
)
(
f(x)dx
=
b
&
a
b/n
curve
under the
Area
s

s





x
e
x
Where a b
f

3 1
2 1 0 2 3 z
The Standard Normal Distribution
Standardization solves the above two problems
Each data value of normally distributed random variable x
can be transformed into a z-score by using the formula:
The resulting distribution will be the standard normal with
a mean of 0 and a standard deviation of 1.
The horizontal scale
corresponds to z-scores.
- -
Value Mean
= = .
Standard deviation
x μ
z
σ
z-score = no. of σ-units above (positive z) or below (negative
z) a distribution mean μ

The Standard Normal Distribution
After the formula is used to transform an x-value into a z-
score, a Standard Normal Table can be used to find the
cumulative area under the curve.
The area that falls in the interval
under the unstandardized normal curve
(the x-values) is the same as the area
under the standard normal curve (within
the corresponding z-boundaries)
That means standardization preserves
area.
3 1
2 1 0 2 3
z
μ
X
x1, x2, x3
-x3, -x2, -x1

The Standard Normal Table
Properties of the Standard Normal Distribution
1. The cumulative area is close to 0 for z-scores close to z = 3.49.
2. The cumulative area increases as the z-scores increase.
3. The cumulative area for z = 0 is 0.5000.
4. The cumulative area is close to 1 for z-scores close to z = 3.49
z = 3.49
Cum. Area is close to 0.
z = 0
Cum. Area is 0.5000.
z = 3.49
Cum. Area is close to 1.
z
3 1
2 1 0 2 3

Example:
Find the cumulative area that corresponds to a z-score
of 2.71.
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964
2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974
2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981
Find the area by finding 2.7 in the left hand column, and
then moving across the row to the column under 0.01.
The area to the left of z = 2.71 is 0.9966.
Standard Normal Table

Example:
Find the cumulative area that corresponds to a z-score
of 0.25.
z .09 .08 .07 .06 .05 .04 .03 .02 .01 .00
3.4 .0002 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003
3.3 .0003 .0004 .0004 .0004 .0004 .0004 .0004 .0005 .0005 .0005
Find the area by finding 0.2 in the left hand column, and
then moving across the row to the column under 0.05.
The area to the left of z = 0.25 is 0.4013
0.3 .3483 .3520 .3557 .3594 .3632 .3669 .3707 .3745 .3783 .3821
0.2 .3859 .3897 .3936 .3974 .4013 .4052 .4090 .4129 .4168 .4207
0.1 .4247 .4286 .4325 .4364 .4404 .4443 .4483 .4522 .4562 .4602
0.0 .4641 .4681 .4724 .4761 .4801 .4840 .4880 .4920 .4960 .5000
Standard Normal Table

Guidelines for Finding Areas
Finding Areas Under the Standard Normal Curve
1. Sketch the standard normal curve and shade the
appropriate area under the curve.
2. Find the area by following the directions for each case
shown.
a. To find the area to the left of z, find the area that
corresponds to z in the Standard Normal Table.
1. Use the table to find
the area for the z-score.
2. The area to the
left of z = 1.23
is 0.8907.
1.23
0
z

b. To find the area to the right of z, use the Standard
Normal Table to find the area that corresponds to z.
Then subtract the area from 1.
3. Subtract to find the area to
the right of z = 1.23:
1  0.8907 = 0.1093.
1. Use the table to find
the area for the z-score.
2. The area to the
left of z = 1.23 is
0.8907.
1.23
0
z

c. To find the area between two z-scores, find the area
corresponding to each z-score in the Standard
Normal Table. Then subtract the smaller area from
the larger area.
4. Subtract to find the area of
the region between the two
z-scores:
0.8907  0.2266 = 0.6641.
1. Use the table to find the area for
the z-score.
3. The area to the left
of z = 0.75 is
0.2266.
2. The area to the
left of z = 1.23
is 0.8907.
1.23
0
z
0.75

Normal Distributions
and Probabilities

Probability and Normal Distributions
We know that the area under any normal curve is 1 unit
Therefore, we can link these areas with probability
i.e. if a random variable, x, is normally distributed, the
probability that x will fall in a given interval is the area
under the normal curve for that interval.
Or P(a  x  b) = area under the curve
between a and b.
There is no probability attached to any single value of x.
That is, P(x = a) = 0.

Same area
P(x < 15) = P(z < 1) = Shaded area under the curve
= 0.8413
15
μ =10
P(x < 15)
μ = 10
σ = 5
Normal Distribution
x
1
μ =0
μ = 0
σ = 1
Standard Normal Distribution
z
P(z < 1)

Example: The average 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.
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

Example:
b) Based on the above example, find the probability that
a pregnant woman weights greater than 85kg.
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
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

Example:
From the above example, find the probability that a randomly
selected pregnant woman weights between 60 and 80.
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
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

Finding z-Scores
Example:
In a certain population, the proportion of individuals with
uric acid level less than a certain limit is 36.7%.
a. Find the z-score that corresponds to this cut of point
b. If the distribution of uric acid level is normal with mean
and standard deviation of 8 and 2.5 units respectively, find
the value of the cut of point in its unit .
Find the z-score by locating 0.367 in the body of the Standard
Normal Table. Use the value closest to 0.367.
0.3 .3483 .3520 .3557 .3594 .3632 .3669 .3707 .3745 .3783 .3821
0.2 .3859 .3897 .3936 .3974 .4013 .4052 .4090 .4129 .4168 .4207
0.1 .4247 .4286 .4325 .4364 .4404 .4443 .4483 .4522 .4562 .4602
0.0 .4641 .4681 .4724 .4761 .4801 .4840 .4880 .4920 .4960 .5000
The z-score is 0.34.
z .09 .08 .07 .06 .05 .04 .03 .02 .01 .00

Finding a z-Score Given a Percentile
Example:
Find the z-score that corresponds to P75
The z-score that corresponds to P 75 is the same as the z-score
that corresponds to an area of 0.75.
The z-score is 0.67.
?
μ =0
z
0.67
Area = 0.75

Transforming a z-Score to an x-Score
To transform a standard z-score to a data value, x, in
a given population, use the formula
Example:
The monthly expenses for cigarette by smokers in a city are
normally distributed with a mean of 120Birr and a standard
deviation of 16 Birr. Find the monthly expense corresponding
to a z-score of 1.60.

x μ+ zσ.

x μ+ zσ
=120+1.60(16)
=145.6
We can conclude that an expense of 145.60 Birr for cigarette is
1.6 standard deviations above the mean.

Exercise
1. What proportion of sandwiches will weight above 289.2
grams?
2. What is the probability that a randomly selected sandwich
will weight between 250 and 289.2 grams?
3. What is the 75 percentile in gram for the distribution of
weight of sandwiches?
4. What is the probability that a randomly selected sandwich
will weight exactly 260 grams?
A population of sandwich has a mean
weight of 250 grams with standard
deviation of 20 grams. Based on this
information give a short answer to the
following questions.

Area between 0 and z
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
62
Table : Normal distribution

Lecture-3 Probability and probability distribution.ppt

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Lecture-3 Probability and probability distribution.ppt