Theory of Probability

 

CAIIB Paper 1 (ABM) Module A Unit 6: Theory of Probability (New Syllabus) 

IIBF has released the New Syllabus Exam Pattern for CAIIB Exam 2023. Following the format of the current exam, CAIIB 2023 will have now four papers. The CAIIB Paper 1 (Advanced Bank Management) includes an important topic called “Theory of Probability”. Every candidate who are appearing for the CAIIB Certification Examination 2023 must understand each unit included in the syllabus.

In this article, we are going to cover all the necessary details of CAIIB Paper 1 (ABM) Module A (Statistics ) Unit 6 : Theory of Probability, Aspirants must go through this article to better understand the topic, Theory of Probability and practice using our Online Mock Test Series to strengthen their knowledge of Time Series. Unit 6 : Theory of Probability

Introduction To Probability

• Probability means chance/s or possibility of happening of an event. For example, suppose we want to plan for a picnic in a weekend.
• Before planning we may check the weather forecast and see what is the chance that there will be rain at that time, accordingly we may do the planning.
• Probability gives a numerical measure of this chance or possibility.
• Suppose it says that there is a 60% chance that rain may occur in this weekend, 60% or 0.6 is called the probability of raining. To understand the concept of probability first we have to understand the concepts of Factorial, Permutations and Combinations.

Factorial:

• In mathematics, Factorial is equal to the product of all positive integers which are less than or equal to a given positive integer. The Factorial of an integer is denoted by that integer and an exclamation point.

• Thus, factorial five is written as 5! which is equal to 1 × 2 × 3 × 4 × 5 = 120 
• The product of the first n natural numbers is called factorial n and is denoted by n!  =

                                      n × (n – 1) × (n – 2) × … × 2 × 1 

• The above formula can also be represented as  n! = n × (n – 1) … (n – r + 1) × (n – r)! 
• Where r < n  It may be noted that:

                                                    0! = 1, 1! = 1 

Permutations and Combinations 

• A permutation is the arrangement of objects in which order is the priority. The fundamental difference between permutation and combination is the order of objects, in permutation, the order of objects is very important, i.e., the arrangement must be in the stipulated order of the number of objects, taken only some or all at a time.
• The combination is the arrangement of objects in which order is irrelevant. The notation for permutation is P (n, r) or nPr, denoting the number of permutations of n things when r things are selected at a time.
• If there are three things a,b, and c then permutations of three things taken two at a time is denoted by  P (3, 2) or 3P2.
• It is given by (a, b), (a, c), (b, c), (b, a), (c, a), (c, b) = 6
• In general,

P (n, r) is the number of permutations when r things are selected at a time from n items.

The notation for combination is C(n, r) or nCr  which is the number of combinations or selections of n  things if only r things are selected.  If there are three things a, b and c then combination of these three things taken two at a time is denoted  by 3C2 and is given by (a, b), (a, c), (b, c) = 3

Example: Using 5 letter of word SHYAM, how many distinct word can be formed?

N= 5

R= 5

5P5 = 5!/ (5-5)! = 5*4*3*2/0! = 5*4*3*2/1 = 120

Note: Permutation and Combination are related to each other by formula P(n,r) = r! * C(n,r).

Example:  In how many ways 3 pencils can be selected from 5 pencils?

3 pens can be selected from 5 pens in 5C3 ways

5C3 =  5! / 3! 2!  ×  = 10 ways

Example: From a group of 7 boys and 6 girls, 3 boys and 4 girls is to be selected. In how many ways this can be done?

3 boys can be selected from 7 boys in 7C3 ways

= 7C3 =  7!  /3! 4!  ×

=  7* 6 *5 *4!  /3*2*4! = 35

4 girls can be selected from 6 girls in 6 C4 ways

=  6!  4! 2! =  6 *5 *4!  /4! *2  = 15

3 boys and 4 girls can be selected in 7C3 × 6C4 = 35 × 15 = 525 ways.

Random Experiment or Trial 

An operation or experiment conducted under identical conditions and which has a number of possible outcomes is called Random Experiment or Trial.

Example: 1. Tossing a coin 2. Throwing a dice 3. Selecting a card form a pack of cards

Sample Space and Sample Points 

The set of all possible outcomes of a random experiment is called sample space.

The elements of the sample space are called sample points. Sample space is denoted by S.

Example: 1. In an experiment of throwing a coin, S = {H,T]

2. In an experiment of throwing a dice, S = {1, 2, 3, 4, 5, 6}

The number of sample points in a sample space of random experiment is denoted by n (s).

For example,  (1) n (S) = 2, and

example (2) n (S) = 6

Event 

Any subset of the sample space S is called an event.

If S is a sample space and A is a subset of S (i.e.,  A⊂ S), then A is called an event.

Example: In an experiment of throwing dice where S = {1, 2, 3, 4, 5, 6}, the event of getting odd numbers is A = {1, 3, 5}

Types of Events 

Certain Event

• If sample points in an event are same as sample points in sample space of that random experiment, then the event is called a certain event.
• Example: Getting any number between 1 to 6 on a dice is a certain event.

 Impossible Events 

• An event which never occurs or which has no favourable outcomes is called an impossible event. In other words, the event corresponding to the set φ (null set) is called an impossible event.
• Example: Getting a number 7 on a dice is an impossible event.

Mutually Exclusive Events 

• Events are said to mutually exclusive if the happening of any of them restricts the happening of the others i.e., if no two or more of them can happen together or simultaneously in the same trial.
• Example: In tossing a coin event head and tail are mutually exclusive. Note: If A & B are mutually exclusive events of sample space S, then A ∩ B = φ.

Example: In tossing a coin event head and tail are mutually exclusive.  Note: If A & B are mutually exclusive events of sample space S, then A ∩ B = φ.

Equally Likely Events 

• Events are said to be equally likely if they have equal choice to occur. In other words, outcomes of a trial are said to be equally likely if taking into consideration all relevant evidences, there is no reason to prefer one with respect to other.
• Example: In throwing a dice all the six faces are equally likely to occur.

Exhaustive Events 

• If the sample points of the events taken together
• Note: If A & B are exhaustive events of sample space S, then A U B = S.
• Example: Random Experiment:
• Throwing a dice  S = {1,2,3,4,5,6},
• A= Event of odd numbers = {1, 3, 5}
• B= Event of odd numbers = {2, 4, 6},
• C= Event of multiple of 3 = {3, 6}
• Here A U B = {1, 2, 3, 4, 5, 6} = S,
• Here A and B are called exhaustive events
• But A U C = {1, 3, 5, 6} ≠ S,
• so A and C are not exhaustive events.

Complementary Event 

If A is an event in sample space S, then the non-occurrence event of A is called Complementary event of A.

Two events A and B are called complementary events, if A and B exhaustive as well as mutually exclusive events.

In other words, A and B are called complementary events if

A U B = S and A ∩ B = φ.

Example: Random Experiment: Throwing a dice, S = {1, 2, 3, 4, 5, 6}, A = {1, 2}, B = {3, 4, 5, 6} As A U B = S and A ∩ B = φ, A and b are complementary events.

Complementary event of A is denoted by Ac, A/ or A.

Mathematical Definition Of Probability

If the sample space S of a random experiment consists of n equally likely, exhaustive and mutually exclusive sample points and m of them are favourable to an event A, then the probability of event A is given by

Number of favourable items/ Total number of outcomes

Example: Two unbiased dice are thrown. Find the probability that: 

• Both the dice show same number.
• First dice shows 6.
• The total of the numbers on the dice is 8.

Solution  In a random throw of two dice, the total number of cases is given below:

A: Both the dice show same number

n(A)/n(S) = 6/36 = 1/6

B: First die show 6 n(B)/n(S) = 6/36 = 1/6

C: Total of the number

on the dice is 8 n(C)/n(S) = 5/36

Example: Two unbiased coins are tossed simultaneously. Find the probability of getting –

at least one tail,

majority of heads

S = {(H, H), (H, T), (T, H), (T, T)} 

n (S) = 4

(i)A: At least one tail,

P (A) = n (A)/ n (S) =  3 /4

(ii)B: Majority of heads

P (B) = n (B)/  n( S)  =  1/ 4

Addition Theorem 

Let A and B are two events (subsets of sample space S) and are not disjoint, then the probability of the occurrence of A or B or A and B both, in other words probability of occurrence of at least one of them is given by,

Example: Find the probability that a card drawn from a pack of cards will be a red or a picture card. 

Probability of selecting a red card = 26 = Event A

P(A) = 26/52 = 1/2

  Probability of getting picture card = 6 = Event B

  P(B) = 12/52 = 3/13

There are 6 red cards which are picture cards,

P (A∩B) = 6/52

P (A U B) = P (A) + P (B) – P (A∩B)

½ + 3/13 – 6/52 = 8/13

Conditional Probability

• The conditional probability of an event A is the probability that the event will occur given the knowledge that an event B has already occurred.

P (A/B).

• If the events A and B are such that the occurrence of A doesn’t depend upon occurrence of event B, (A and B are independent event), the conditional probability of event A given event B is simply the probability of event A, that is P (A).
• Similarly, probability of event B given that event A has already occurred is denoted by P (B/A).

P (B/A) = P (A ∩ B) / P(A)

Example: Consider a fair coin is tossed 3 times

S = (HHH, HHT, HTH, TTT, TTH, THT, THH, HTT) = 8

Event A = Atleast two tail appear

Event B – First coin show Head

P(A) = (TTT, TTH, THT, HTT) = 4/8 = ½

P(B) = (HHH, HHT, HTH, HTT) = 4/8 = ½

P (A ∩ B) = 1/8

P(A/B) = 1/ 8 / ½ = 1/4

Multiplication Theorem 

• If A and B are two events of a sample space S associated with an experiment, then the probability of simultaneous occurrence of events A and B is given by

P (A ∩ B) = P(A) P(B/A) = P(B) P(A/B)

Independent Events 

Two events A and B are independent of each other if the occurrence or non-occurrence of one does not affect the occurrence of the other.

P (A ∩ B) = P(A) P (B)

Example: Two balls are drawn from a bag one by one with 2 white and 3 black balls. What is the probability that the second ball is white?

Event W1 = first ball – White Ball

Event B1 = First Ball – Black Ball

Event W2 = Second Ball – White Ball

1st White Ball = 2/5 + ¼

2nd Black Ball = 3/5 + 2/4

P(W2) = P(W1) + P(W2/W1) + P(B1) + P(W2/B1)

2/5 + ¼ + 3/5 + 2/4= 2/5

Random Variable

• A random variable is a function that associates a real number with each element in the sample space.
• In other words, a random variable is a function X: S → R,
• where S is the sample space of the random experiment under consideration and R is the real number line.

Example. Consider the random experiment of tossing a coin two times and observing the result (a Head or a Tail) for each toss.

Let X denote the total number of heads obtained in the two tosses of the coin.

Example: Suppose that you play a certain lottery by buying one ticket per week. Let X be the number of weeks until you win a prize. X is a random variable.

Discrete Random Variable:

• If a random variable takes a finite number or countable infinite number of  possibilities, it is called a discrete random variable.
• Example: 1. Age in years 2. Number of arrivals in a clinic 3. Number of accidents

Continuous Random Variable:

• If a random variable takes infinite number of possibilities, it is called a continuous random variable.
• Example 1. Percentage of marks 2. Weight 6.5 PROBABILITY

Binomial Distribution

• Consider a random experiment consisting of n repeated independent trials with p the probability of success at each individual trial. Let the random variable X represent the number of successes in the n repeated trials.
• Then X follows a Binomial distribution.

The definition of this distribution is:

• A random variable X has a binomial distribution,

X ~ Binomial (n,  p), if the discrete density of X is given by: 

P[X=x] = f(x) = nCx px (1 – p)n–x,

x = 0, 1, 2…, n  = 0 otherwise

f(x) = nCx px qn–x;

x = 0, 1, 2…, n  = 0 otherwise where p + q = 1

P = the probability of success

n is the total number of trials.

Example: Toss a coin for 10 times and you wan to get head 4 times & probability of coming head is 0.5 calculate f(x)?

n = 10, x= 4 & p = 0.5

q = 1- 0.5 = 0.5

f(x) = nCx px qn–x

= 10C4* 0.510 o.510

Binomial Distribution Real Life Examples 

Many instances of binomial distributions can be found in real life. 

• If a new drug is introduced to cure a disease, it either cures the disease (it’s successful) or it doesn’t cure the disease (it’s a failure).
• If you purchase a lottery ticket, you’re either going to win money, or you aren’t. Basically, anything you can think of that can only be a success or a failure can be represented by a binomial distribution.
• Mean = np
• Variance = np (1 – p) = npq
• SD = √ variance
• Measure of Skewness = â1 = (1-2p)2/npq
• Measure of Kurtosis = â2 = 3 + [1 – 6pq /npq]
• Binomial Distribution is symmetric if p = q = 0.5
• If p < 0.5, distribution is positively skewed and
• if p > 0.5, distribution is negatively skewed.

Mode

M= (n+1) p

• If M is not an integer, mode is the integral part lying between M – 2 and M. 
• If M is an integer, there are two modes and thus the distribution is bimodal, and two modes are M – 1 and M.

Problem: If X follows Binomial distribution with n = 8, p = 1/2, then Find P [IX-4I ≤ 2]

Solution : P[-2<=(X-4)<=2]

P[2<= X<=6]

P[2<= X<=6] = P(X=2) +  P(X=3) +  P(X=4) +  P(X=5) +  P(X=6)

f(x) = nCx px qn–x = 8Cx 1/2x 1/28–x

8Cx 1/2x 1/28–x =

= 8C2 1/22 1/28–2 + 8C3 1/23 1/28–3  + 8C4 1/24 1/28–4 + 8C5 1/25 1/28–5  + 8C6 1/26 1/28–6

= (1/2)8 (8C2 +  8C3 +  8C4 +  8C5 +  8C6)

= 1/256 (128+ 56+ 70 +56 +28) = 119/128

Poisson Distribution

• The Poisson probability distribution was introduced by S. D. Poisson.
• A random variable X, taking on one of the values 0, 1, 2,…, is said to be a Poisson random variable with parameter λ, λ > 0, if its  probability mass function is given by

• The symbol e stands for a constant approximately equal to 2.7183. It is a famous constant in mathematics, named after the Swiss Mathematician L. Euler, and it is also the base of the so-called natural logarithm

Some examples of Poisson probability are: 

• The number of misprints on a page (or a group of pages) of a book.
• The number of people in a community living to 100 years of age
• The number of wrong telephone numbers that are dialed in a day.
• The number of transistors that fail on their first day of use.
• The number of customers entering a post office on a given day.
  • Mean = λ,
  • Variance = λ
  • Measure of Skewness = β1 = 1 /λ
  • Measure of Kurtosis β2 = 3 +1/λ
  • Mode: If l is not an integer mode is the integral part lying between λ–1 and λ
  • If λ s an integer, there are two modes and thus the distribution is bimodal and two modes are λ −1 , λ

Problems. Births in a hospital occur randomly at an average rate of 1.8 births per hour. 

• What is the probability of observing 4 births in a given hour at the hospital?
• What about the probability of observing more than or equal to 2 births in a given hour at the hospital?

Let X= No. of births in a given hour = 4 with Mean rate λ = 1.8

e-1.8 = .16529

f(x) = e-1.8 * 1.84 /4!

.16529 * 10.4976/ 24 = 0.0723

(ii) We want P(X ≥ 2) = P(X = 2) + P(X = 3) +…, i.e., an infinite number of probabilities to calculate

but P(X ≥ 2) = P(X = 2) + P(X = 3) +… = 1 − P(X < 2)

= 1 − (P(X = 0) + P(X = 1)

= 1 − (0.16529 + 0.29753) = 0.538

Normal Distribution

• A normal distribution is a distribution that occurs naturally in many situations where 50% of the data will fall to the left of the mean and 50% will fall to the right.
• For example, Height of the population, most of the people in a specific population are of average height. The number of people taller and shorter than the average height people is almost equal, and a very small number of people are either extremely tall or extremely short.
• Some other examples are distribution of Income in economy, distribution of marks in an exam, etc.
• A random variable X is said to follow Normal Distribution if its pdf is given by

Note:

• µ and σ2 are called parameters of Normal Distribution.
• If µ = 0 and σ2 = 1, then the Normal variable is called Standard Normal Variable. Generally, it is denoted by Z.

• The graph of Normal Distribution is bell shaped and symmetric.
• Quartile deviation is 0.6745σ
• Mean deviation is 0.7979σ

Mean = Median = Mode = µ

Problem: Normal population of 1000 employees has mean income Rs. 800 per day and variance 400, Find no.  of employees where income between [ P(Z= 1) = 0.3413, P(Z= 2) = 0.4772 & P(Z= 2.5) = 0.4938 P(Z= 5) = 0.5]

P (750 < x < 820)

P (x > 700)

P (x > 760)

n= 1000 , µ= 800 & σ2 = 400

Z = X- µ /σ

i)X = 750 = 750- 800/ 20 = 2.5 = 0.4938

X = 820 = 820-800 / 20 = 1 = 0.3413

= 0.4938 + 0.3413 = 0.8351 = 83.51%

ii)X = 700 = 700 – 800/ 20 = 5  = 0.5

0.5 + 0.5 = 1 = 1000 employees

n= 1000 , µ= 800 & σ2 = 400

Z = X- µ /σ

iii) X = 760 = 760-800/20 = 2 = 0.4772

0.4772 + 0.5 = 0.9772

Credit Risk

• We can apply probability concept and different formulas and laws of probability in different practical field.
• One very important application is Credit Risk.
• When lenders offer mortgages, credit cards, any type of loan to different customers, there could be a risk that the customer or borrower might not repay the loan.
• Similarly, if a company extends credit to a customer, there could be a risk that the customer might not pay their invoices.
• We are interested to calculate this risk of not repaying any due payment. This is called Credit Risk.
• Credit risk also represents the risk that a bond issuer may fail to make a payment when requested, or an insurance company will not be able to pay a claim.
• Thus, Credit Risk is the possibility or chance or probability of a loss occurring due to a borrower’s failure to repay a loan to the lender or to satisfy contractual obligations. It refers to a lender’s risk of having its cash flows interrupted when a borrower does not repay the loan taken from him.

There are three types of credit risks. 

Credit default Risk :

Credit default risk is the type of loss that is incurred by the lender either when the borrower is unable to repay the amount in full or when 90 days pass the due date of the loan repayment. This type of credit risk is generally observed in financial transactions that are based on credit like loans, securities, bonds or derivatives.

Concentration Risk:

Concentration risk is the type of risk that arises out of significant exposure to any individual or group because any adverse occurrence will have the potential to inflict large losses on the core operations of a bank. The concentration risk is usually associated with significant exposure to a single company or industry or individual.

Country risk 

• The risk of a government or central bank being unwilling or unable to meet its contractual obligations is called Country or Sovereign Risk.
• When a bank or financial institution or any other lender has an indication that the borrower may default the loan payment, he will be interested to calculate the expected loss in advance.
• The expected loss is based on the value of the loan (i.e., the exposure at default, EAD) multiplied by the probability, that the borrower will default (i.e., probability of default, PD).
• In addition, the lender takes into account that even when the default occurs, it might still get back some part of the loan.
• Hence, PD * EAD is further multiplied by the estimation of the part of the loan which will be lost in case that a default occurs (i.e., loss given default, LGD).

Expected loss = PD * EAD * (1 – LGD)

Problem: Let a credit of Rs. 2,000,000 was extended to a company one year ago. Determine the expected loss for the exposure if the company defaults completely, where the loss given default is 50%.

Probability of default, PD = 100

Loss given default, LGD = 50% 

Expected loss = 100% * Rs. 2,000,000 * (1 – 50%) 

= Rs. 1,000,000

Value At Risk (VaR)

• The concept of value at risk is associated with portfolio of an individual or an organisation.
• A portfolio is a collection of different kinds of assets owned by an individual or organisation to fulfil their financial objectives.
• One can include fixed deposit or any investment where he or she can earn a fixed interest, equity shares, mutual funds, debt funds, gold, property, derivatives, and more in his portfolio.
• In any type of investment where one can earn fixed interest are not risky, but risk is associated with the investments in Equity market, Mutual Funds, Gold, etc.
• Value at risk (VaR) is a financial metric that one can use to estimate the maximum risk of an investment over a specific period.

If the portfolio value is Rs. 30,000 and if 1-month average return and standard deviation is 10% and  12% respectively, calculate daily VaR at 95% confidence level. 

VAR at 95% confidence level

=  [Return of the portfolio –1.65 *σ][ Value of the portfolio]

= [0.1–1.65*0.12] *30000

= [0.1 –0.198] *30000 = –2940 = 9.8% of the portfolio