Statistical computing 1

Statistical Computing
M.V. Padmavati
BIT, Durg
Venkata Padmavathi Metta

UNIT I: Probability Theory
• Sample Spaces- Events - Axioms – Counting
• Conditional Probability and Bayes’ Theorem
• The Binomial Theorem
• Random variable and distributions : Mean and
Variance of a Random variable
• Binomial-Poisson-Exponential and Normal
distributions.
• Curve Fitting and Principles of Least Squares-
Regression and correlation.

Random Experiment
• The experiment whose set of all outcomes are
known but not possible to predict which
outcome will occur.
• Example 1. Tossing a coin
2. Casting an un biased die.

Probability
• How likely something is to happen.
• Many events can't be predicted with total
certainty.
• The best we can say is how likely they are to
happen, using the idea of probability.

Sample space
• Consider a random experiment.
• The set of all the possible outcomes is called the
sample space of the experiment and is usually
denoted by S.
• Examples:
1. Experiment of tossing a coin
S1={H,T}
2. Selecting a digit
S2={0,1,2,3,4,5,6,7,8,9}

Sample space
Example 3. Tossing a die.
The sample space is S = {1, 2, 3, 4, 5, 6}.
Example 4. Tossing a die twice.
The sample space is S = {(i, j) : i, j = 1, 2, . . . , 6}
={(1,1),(1,2),(1,3),……….(6,6)} which contains 36
elements

Event
• One or more outcomes of an experiment
• Any subset E of the sample space S is called an
event
• Examples
1. Tossing a coin. The sample space is S = {H, T}.
E = {T} is an event, which can be described in
words as “Getting Tail”.
2. Tossing a die. The sample space is S = {1, 2, 3, 4,
5, 6}.
E = {2, 4, 6} is an event, which can be described in
words as “Getting even number”.

Example 3. Tossing a coin twice. The sample space is S = {HH, HT, TH, TT}.
Consider the event “The first toss results in a Heads”.
What is the value of E?
E = {HH, HT}
Example 4. Tossing a die twice. The sample space is S = {(i, j) : i, j = 1, 2,
. . . , 6}, which contains 36 elements. Consider the event “Getting 10 as
sum”. What is the value of E?
E={(4,6),(5,5),(6,4)}
Event

Probability
• Probability of an event E happening
= Number of favorable outcomes/ Total number
of outcomes
• Denoted as P(E)= #(E)/#(S)

Probability
• Tossing a Coin is an event
• When a coin is tossed, there are two possible
outcomes: So sample space S={H,T}
• heads (H) or
• tails (T)
• We say that the probability of the coin
landing H is ½
• And the probability of the coin landing T is ½

Probability
1. Tossing a coin. The sample space is S = {H, T}.
E = {T} is an event, which can be described in
words as “Getting Tail”.
P(E)=1/2
2. Tossing a die. The sample space is S = {1, 2, 3,
4, 5, 6}.
E= {2, 4, 6} is an event, which can be
described in words as “Getting even number”.
P(E)=3/6=1/2

Example 3. Tossing a coin twice. The sample space is S = {HH, HT, TH, TT}.
Consider the event “The first toss results in a Heads”.
What is the value of E?
E = {HH, HT}
Example 4. Tossing a die twice. The sample space is S = {(i, j) : i, j = 1, 2,
. . . , 6}, which contains 36 elements. Consider the event “Getting 10 as
sum”. What is the value of E?
E={(4,6),(5,5),(6,4)}
Probability
P(E)=2/4=1/2
P(E)=3/36=1/12

Axioms of Probability
• Consider an experiment with sample space S.
• For any event E, we refer to P(E) as the probability of E.
– Axiom 1: For any event E, P(E)≥0.
– Axiom 2: Probability of the sample space S is P(S)=1.
– Axiom 3: for any sequence of events E1, E2, . . . which are
mutually disjoint (mutually exclusive),
P(E1∪E2∪E3⋯)=P(E1)+P(E2)+P(E3)+⋯

Axioms of Probability
• The smallest value for P(E) is zero and if P(E)=0,
then the event A will never happen.
• The second axiom states that the probability of
the whole sample space is equal to one,
i.e., 100 percent.
– Axiom 1: For any event E, P(E)≥0.
– Axiom 2: Probability of the sample space S is P(S)=1.

In a presidential election, there are four candidates. Call them A, B, C, and D.
Based on our polling analysis, we estimate that A has a 20 percent chance of
winning the election, while B has a 40 percent chance of winning. What is the
probability that A or B win the election?
• Notice that the events that {A wins},{B wins}, {C wins}, and {D
wins} are disjoint since more than one of them cannot occur at the
same time.
• For example, if A wins, then B cannot win.
• From the third axiom of probability, the probability of the union of
two disjoint events is the summation of individual probabilities.
Therefore,
P(A wins or B wins)= P({A wins}∪{B wins}
= P({A wins})+P({B wins})
= 0.2+0.4
= 0.6

Implied Axioms
• For any event E, P(Ec)=1−P(E).
• The probability of the empty set is zero,
i.e., P(∅)=0.
• For any event E, 0 ≤ P(E)≤1.
• The probability of the certain event E,
i.e., P(E)=1.

Questions
• A bag contains three white and five black balls.
What is the chance that a ball drawn at random
will be black. Ans: 5/8
• Three coins are tossed. What is the probability
that all are heads?
S={hhh,htt,hth,thh,……….}-8 elements
E={hhh} Ans: 1/8
• Die is thrown. Find the probability that 1,2 or 3
turns up? Ans: ½
• One card is drawn from pack of cards (52 cards).
What is the probability that it is (1) diamond (2)
not a diamond Ans: ¼,3/4

Questions
• A man draws at random 3 balls from a bag
containing 6 red and 5 white balls. What is the
chance of getting the balls all red.
• A man draws at random 3 balls from a bag
containing 6 blue and 4 red balls. What is the
chance of getting two blue and one red.
Ans. 6/11*5/10*4/9=4/33 or 6C3 / 11C3
6C2 * 4C1/10C3

Counting
• We start with the fundamental principle of
counting. Suppose that two experiments are to
be performed.
• Experiment 1 can have n1 possible outcomes
and
• for each outcome of experiment 1, experiment
2 has n2 possible outcomes
• Then together there are n1* n2 possible
outcomes.

Counting- Generalization
• Generalize the fundamental principle of
counting to k experiments.
• Assume that we have a collection of n objects
and we wish to make an ordered arrangement
of k of these objects.
• Using the generalized multiplication principle,
the number of possible outcomes is
n * (n- 1)*……*(n- k + 1).
We will write this as nPk.

Example 1
• For a group of n individuals, one is chosen to
become the president and a second is chosen
to become the treasurer.
• By the multiplication principle, if this position
are held by different individuals, then this task
can be accomplished in n x (n- 1) possible
ways

Inclusion-Exclusion Principle
• P(A∪B)=P(A)+P(B)−P(A∩B)
• If A and B are mutually exclusive then
P(A∪B)=P(A)+P(B)

Conditional probability
• It is a measure of the probability of
an event occurring, given that another event
has already occurred.
• If the event of interest is A and the event B is
known or assumed to have occurred,
• The conditional probability of A given B, or
"the probability of A under the condition B", is
usually written as P(A|B)

Conditional Probability
• The probability that any given person has a
Covid(A) after testing is only 5%.
• But if we know or assume that the person is
having Flu(B), then they are much more likely
to be Covid positive.
• For example, the conditional probability that
someone with Flu symptoms is Covid +ve is
75%, in which case we would have
that P(Covid) = 5% and P(Covid|Flu) = 75%.

• P(A|B) may or may not be equal to P(A)
• If P(A|B) = P(A), then events A and B are said
to be independent.
• P(A|B) typically differs from P(B|A).
• P(Covid/Flu) is different from P(Flu/Covid)

The probability P(A) = 0.30 + 0.10 + 0.12 = 0.52
The conditional probability
• P(A|B1) = 1,
• P(A|B2) = 0.12 ÷ (0.12 + 0.04) = 0.75
• P(A|B3) = 0.

Independent events vs. mutually
exclusive events
If statistically independent If mutually exclusive
P(AB)= P(A) 0
P(BA)= P(B) 0
P(AՈB)= P(A)P(B) 0
The concepts of mutually independent events and mutually exclusive
events are separate and distinct.

Example 1
• (i) If P(A)=2/5, P(B) = 1/3 and P(A  B)= ½, find
P(A/B) and P (B/A)
• (ii) if A and B two independent events and
P(A)=2/3, P(B) =3/5, find P(AB).
Ans: (i) 7/10, 7/12
Ans: (ii) 13/15

Example 2
In an examination, 30% of the students have failed in
Mathematics, 20% of the students have failed in Chemistry
and 10% have failed in both Mathematics and Chemistry. A
student is chosen at random.
(i) What is the probability that the student has failed in
Mathematics if it is known that he has failed in Chemistry?
(ii) What is the probability that the student has failed either
in Mathematics or in Chemistry?
Ans: (i) 0.5 Ans: (ii) 0.4

Example 3
The probability that a man will be alive in 25 years is 3/5, and the
probability that his wife will be alive in 25 years is 2/3. Find the
probability that:
(i) both will be alive,
(ii) only wife will be alive,
(iii) only man will be alive,
(iv) atleast one of them will be alive.
Ans: (i) 2/5 Ans: (ii) 4/15
Ans: (iii) 1/5 Ans: (iv) 13/15

Example 4
A problem in Statistics is given to three students, A, B, C, whose
chances of solving it are ½ , 1/3 and 1/4 respectively, if they try
it independently, what is the probability that the problem will be
solved?
Ans: 1/4

Example 5
Two sets of candidates are competing for the positions
on the Board of Directors of a company. The
probabilities that the first and second sets will win are
0.6 and 0.4 respectively. If the first set wins, the
probability of introducing a new product is 0.8 and the
corresponding probability, if the second set wins, is 0.3.
What is the probability that the new product will be
introduced?
Ans: 0.6

In Exton School, 60% of the boys play baseball, and 24% of the
boys play baseball and football.
What percent of those that play baseball also play football?
Ans: 2/5
Example 6

45% of the children in a school have a dog, 30% have a
cat, and 18% have a dog and a cat.
What percent of those who have a dog also have a cat?
• Ans : 40%
Example 7

Among the examinees in an examination 30%, 35% and 45%
failed in Statistics, in Mathematics and in at least one of the
subjects respectively, An examinee is selected at random. Find
the probabilities that
(i) he failed in Mathematics only,
(ii) he passed in Statistics if it is known that he failed in
Mathematics.
Example 1: Bayes’ Theorem

Example-Conditional Probability
• Roll a fair die.
• Let A be the event that the outcome is an odd
number, i.e., A={1,3,5}
• Let B be the event that the outcome is less
than or equal to 3, i.e., B={1,2,3}.
• What is the probability of A, P(A)?
• What is the probability of A given B, P(A|B)?

Bayes’ Theorem Statement
An event A can occur only if any one of the set of
exhaustive and mutually exclusive events B1, B2,
…,Bn occurs.
The probabilities P(B1),P(B1), …., P(Bn) and the
conditional probabilities P(A/Bi); i=1,2,…,n for A to
occur are known.
Then the conditional probabilities P(Bi/A) when A
has occurred is given by
P(Bi│A) = P(Bi)*P(A│Bi)/ 𝑘=1
𝑛
P(Bk)P(A|Bk)

Bayes’ Theorem Proof

There are two identical boxes containing respectively 4
white and 3 red balls and 3 white and 7 red balls. A box is
chosen at random and a ball is drawn from it. If the ball is
white, what is the probability that it is from the first box?

In a bolt factory, machines M1, M2, M3 manufacture
respectively 25, 35 and 40 percent of the total output. Of
their output 5, 4 and 2 percent respectively, are defective
bolts, one bolt is drawn at random from the product and is
found to be defective. What is the probability that it is
manufactured in the machine M2?

There are two identical boxes containing respectively 5
white and 3 red balls and 4 white and 6 red balls. A box is
chosen at random and a ball is drawn from it. If the ball is
white, what is the probability that it is from the first box?

In a bolt factory, machines X, Y and Z manufacture
respectively 20, 35 and 45 percent of the total output. Of
their output 8, 6 and 5 percent respectively, are defective
bolts, one bolt is drawn at random from the product and is
found to be defective. What is the probability that it is
manufactured in the machine Z?

Example
• Suppose we have the following information:
– There is a 60 percent chance that it will rain today.
– There is a 50 percent chance that it will rain tomorrow.
– There is a 30 percent chance that it does not rain either
day.
• Find the following probabilities:
– The probability that it will rain today or tomorrow.
– The probability that it will rain today and tomorrow.
– The probability that it will rain today but not tomorrow.
– The probability that it either will rain today or tomorrow,
but not both.
0.7, 0.4, 0.2, 0.1

Random Variable
• Random variable and distributions : Mean and
Variance of a Random variable

Random Variable
• A variable whose numerical value is determined by the
outcome( or result) of a random experiment is called
random variable or a chance variable
Example: Consider a random experiment of tossing 3 coins. The sample space is
S={HHH,HHT,HTH,HTT,THT,THH,THT,TTH,TTT}
Let X denote a random variable and it denotes the number of
heads obtained then
X(HHH)=3, X(HHT)=2, X(TTT)=0

Types of Random Variable
• A random variable can be discrete or continuous
Discrete Random variable Continuous Random Variable
A discrete variable is a variable
whose value is obtained by counting.
A Continuous variable is a variable
whose value is obtained by
measuring.
 number of students present
 number of red marbles in a jar
 number of heads when flipping
three coins
 height of students in class
 weight of students in class
 time it takes to get to school
If a random variable can take only
finite set of values (Discrete Random
Variable), then its probability
distribution is called as Probability
Mass Function or PMF
If a random variable can take
continuous set of values (Continuous
Random Variable), then its probability
distribution is called as Probability
Density Function or PDF

Random Variable

Probability distribution of a discrete random variable
• The probability distribution of a random variable X tells
what the possible values of X are and how probabilities
are assigned to those values
• Example: Let X represent the sum of two dice.
• Then the probability distribution of X is as follows:
X 2 3 4 5 6 7 8 9 10 11 12
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
If a random variable can take only finite set of values (Discrete Random Variable),
then its probability distribution is called as Probability Mass Function or PMF.

Probability distribution of a discrete random variable
• The probability distribution of discrete
random variable X is as follows:
X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2+k
(i) Determine K (ii) find P(X<6) (iii) what will be P(X>=6) and P(0<X<5)?

Expected Mean of a discrete Probability distribution
• The probability distribution of discrete
random variable X is as follows:
X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2+k
(i) Determine K (ii) find P(X<6) (iii) what will be P(X>=6) and P(0<X<5)?

Expectation or Expected Mean of a discrete random
probability distribution

Variance of a discrete random probability distribution
The variance of a random variable X is calculated as
Variance= E(X2)-(E(X))2
E(X) is the expected Mean.
Standard Deviation

A random variable has following probability distribution
X 4 5 6 8
P(X) 0.1 0.3 0.4 0.2
Find expectation (or mean), variance and standard deviation of the random
Example 1: Mean, variance and standard deviation of a random variable

A bag contains 5 white and 7 black balls. Find the expectation
of a man who is allowed to draw two balls from the bag and
who is to receive one rupee for each black ball and two
rupees for each white ball.
Example 2: Mean, variance and standard deviation of a random variable

• It is a discrete probability distribution and is
defined by the probability function
Binomial distribution or Bernoulli Distribution
f(x)=nCx px qn-x
Which gives the probability of x successes in a
series of n independent trials (x<=n), where p is
the constant probability of success in a single trail
and q=1-p Venkata Padmavathi Metta

Conditions of Binomial distributions
Binomial distribution holds under the following
conditions
– Each observation or trial is independent. In other words, none of
your trials have an effect on the probability of the next trial.
– The probability of success (tails, heads, fail or pass) is exactly the
same from one trial to another.
– There are only two possible outcomes i.e success or failure for
each trial.
– Trials are repeated under identical conditions for a fixed number
of times say n

Q. A coin is tossed 10 times. What is the probability of getting
exactly 6 heads?
• The number of trials (n) is 10
• The Probability of success (“getting a head”) is
p=0.5 (So q=1-p = 0.5)
x = 6
• P(x=6) = 10C6 * 0.5^6 * 0.5^4
= 210 * 0.015625 * 0.0625
= 0.205078125
f(x)=nCx px qn-x

Six coins are tossed. What is the probability of getting exactly 3
tails?
f(x)=nCx px qn-x

Suppose that 80% of adults with allergies report symptomatic relief with a
specific medication. If the medication is given to 10 new patients with
allergies, what is the probability that it is effective in exactly seven?
1. The outcome is relief from symptoms (yes or no), and
here we will call a reported relief from symptoms a
'success.‘
2. The probability of success for each person is p=0.8. So
q=1-p=0.2
We know that:
Number of observations is n=10
Number of successes or events of interest is x=7
p=0.80
The probability of 7 successes is= 10C7(0.8)7(0.2)3

Continuous Random Variables
• A continuous random variable is not defined at specific values. Instead, it is
defined over an interval of values, and is represented by the area under a
curve (in advanced mathematics, this is known as an integral).
• The probability of observing any single value is equal to 0, since the number of
values which may be assumed by the random variable is infinite.
• Suppose a random variable X may take all values over an interval of real
numbers. Then the probability that X is in the set of outcomes A, P(A), is
defined to be the area above A and under a curve. The curve, which
represents a function p(x), must satisfy the following:
1: The curve has no negative values p(x) > 0 for all x
2: The total area under the curve is equal to 1.
A curve meeting these requirements is known as a density curve.

Statistical computing 1

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