In view of the coronavirus pandemic, we are making LIVE CLASSES and VIDEO CLASSES completely FREE to prevent interruption in studies
Mathematics of Finance and Elementary Probability

Approaches of Probability

The probability theory has many definitions – mathematical or classical, relative or empirical, and the theorem of total probability. In this article, we will talk about each of these definitions and look at some examples as well.

Suggested Videos

Play
Play
Play
Arrow
Arrow
ArrowArrow
Introduction to Probability
Events and its Algebra
Venn Diagrams & Probability
Slider

 

Classical or Mathematical Definition of Probability

Let’s say that an experiment can result in (m + n), equally likely, mutually exclusive, and exhaustive cases. Also, ‘m’ cases are favorable to the occurrence of an event ‘A’ and the remaining ‘n’ are against it. In such cases, the definition of the probability of occurrence of the event ‘A’ is the following ratio:

\( \frac {m}{m + n} \) = \( \frac {\text {Number of cases favorable to the occurrence of the event ‘A’}}{\text {Total number of equally likely, mutually exclusive, and exhaustive cases}} \)

The probability of the occurrence of the event ‘A’ is P(A). Further, P(A) always lies between 0 and 1. These are the limits of probability.

Instead of saying that the probability of the occurrence of the event ‘A’ is \( \frac {m}{m + n} \), we can say that “Odds are m to n in favor of event A or n to m against the event A.” Therefore,

Odds in favor of the event A = \( \frac {\text {No. of cases favorable to the occurrence of the event A}}{\text {No. of cases against the occurrence of the event A}} \) = \( \frac {\frac {m}{m + n}}{\frac {n}{m + n}} \) = \( \frac {m}{n} \)

Odds against the event A = \( \frac {\text {Number of cases against the occurrence of the event A}}{\text {Number of cases in favour of the occurrence of the event A}} \) = \( \frac {n}{m} \)

Note: The ratio \( \frac {m}{n} \) or \( \frac {n}{m} \) is always expressed in its lowest form (integers with no common factors).

  • If m = 0, or if the number of cases favorable to the occurrence of the event A = 0 then, P(A) = 0. In other words, event A is an impossible event.
  • If n = m, then P(A) = \( \frac {m}{m} \) = 1. This means that the event A is a certain or sure event.
  • If neither m = 0 nor n = 0, then the probability of occurrence of any event A is always less than 1. Therefore, the probability of occurrence the event satisfies the relation 0 < P < 1.

If the events are mutually exclusive and exhaustive, then the sum of their individual probabilities of occurrence = 1.

For example, if A, B, and C are three mutually exclusive events, then P(A) + P(B) + P(C) = 1. The probability of the occurrence of one particular event is the Marginal Probability of that event.

Choosing an object at random from N objects means that each object has the same probability \( \frac {1}{N} \) of being chosen.

probability theory

                                                                                                                                                   Source: Wikipedia

Empirical Probability or Relative Frequency Probability Theory

The Relative Frequency Probability Theory is as follows:

We can define the probability of an event as the relative frequency with which it occurs in an indefinitely large number of trials. Therefore, if an event occurs ‘a’ times out of ‘n’, then its relative frequency is \( \frac {a}{n} \).

Further, the value that \( \frac {a}{n} \) approaches when ‘n’ becomes infinity is the limit of the relative frequency.

Symbolically,

P(A) = \( \lim_{n \to ∞} \frac {a}{n} \)

However, in practice, we write the estimate of P(A) as follows:

P(A) = \( \frac {a}{n} \)

While the classical probability is normally encountered in problems dealing with games of chance. On the other hand, the empirical probability is the probability derived from past experience and is used in many practical problems.

Total Probability Theorem or the Addition Rule of Probability

If A and B are two events, then the probability that at least one of then occurs is P(A∪B). We also have,

P(A∪B) = P(A) + P(B) – P(A∩B)

If the two events are mutually exclusive,  then P(A∩B) = 0. In such cases, P(A∪B) = P(A) + P(B).

probability theory

Multiplication Rule

If A and B are two events, the probability of their joint or simultaneous occurrence is:

P(A∩B) = P(A) . P(A/B)

If the events are independent, then

  • P(A/B) = P(A)
  • P(B/A) = P(B)

 

probability theory

Therefore, we now have,

If the events are independent, then P(A∪B) = P(A) + P(B) – P(A∩B)

Also, P(A/B) is the conditional probability of the occurrence of the event A when event B has already occurred. Similarly, P(B/A) is the conditional probability of the occurrence of event B when event A has already occurred. If the events are independent, then the occurrence of A does not affect the occurrence of B.

∴ P(B/A) = P(B)

Also, P(A/B) = P(A)

Some Solved Illustrations

Illustration 1

If you toss two coins, then there are four possibilities – {HH, HT, TH, TT}.

If E1 is the event of getting one head and one tail, then P(E1) = \( \frac {2}{4} \) = \( \frac {1}{2} \)

If E2 is the event of getting two tails, then P(E2) = \( \frac {1}{4} \)

If E3 is the event of getting two heads, then P(E3) = \( \frac {1}{4} \)

If E4 is the event of getting a head or a tail, then P(E4) = \( \frac {4}{4} \) = 1 (Certain or Sure Event)

If E5 is the event of getting three heads or three tails or one head and two tails or one tail and two heads, then P(E5) = \( \frac {0}{4} \) = 0 (Impossible Event).

Illustration 2

The percentages of 100 teenage offenders are in the following six categories:

Categories Group A Group B Total
Non-offenders 28 42 70
First-time offenders 05 15 20
Repeat offenders 07 03 10
Total 40 60 100

If you choose a teenager at random, then the following events are defined:

  • The teenager is from Group A
  • The teenager is from Group B
  • The teenager is not an offender

Find the following probabilities:

  1. P(A)
  2. P(B)
  3. P(C)

Solution:

The probability that a teenager is from Group A is

P(A) = \( \frac {Number of teenagers in Group A}{Total Number of teenagers} \) = \( \frac {28 + 5 + 7}{100} \) = \( \frac {40}{100} \) = 0.4

The probability that a teenager is from Group B is

P(B) = \( \frac {Number of teenagers in Group B}{Total Number of teenagers} \) = \( \frac {45 + 15 + 3}{100} \) = \( \frac {60}{100} \) = 0.6

The probability that a teenager is from Group C is

P(C) = \( \frac {Number of teenagers who are not offenders}{Total Number of teenagers} \) = \( \frac {28 + 42}{100} \) = \( \frac {70}{100} \) = 0.7

Solved Question

Q1. An analyst develops the following table of joint probabilities relating the size of firm (measured in terms of the number of employees) and the type of firm.

Number of Employees Industry
Construction Manufacturing Retail
Under 20 0.2307 0.0993 0.5009
20 – 99 0.0189 0.0347 0.0876
100 or more 0.0019 0.0147 0.0113

If one is selected at random, find the probability of the following events:

  1. P(A) = The firm employs fewer than 20 employees
  2. P(B) = The firm is in the retail
  3. P(C) = A firm in the construction industry employs between 20 to 99 workers
  4. P(D) = A firm in the retail industry employed more than 20 workers.

Solution:

i. The firm employs fewer than 20 employees

P(A) = 0.2307 + 0.0993 + 0.5009 = 0.8303

ii. The firm is in the retail

P(B) =  0.5009 + 0.0876 + 0.0113 = 0.5998

iii. A firm in the construction industry employs between 20 to 99 workers

P(C) = \( \frac {0.0189}{0.2307 + 0.0189 + 0.0019} \) = \( \frac {0.0189}{0.2515} \) = 0.075

iv. A firm in the retail industry employed more than 20 workers

P(D) = \( \frac {0.876 + 0.0113}{0.5009 + 0.876 + 0.0113} \) = 0.296

 

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

Leave a Reply

avatar
  Subscribe  
Notify of

Stuck with a

Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.
toppr Code

chance to win a

study tour
to ISRO

Download the App

Watch lectures, practise questions and take tests on the go.

Get Question Papers of Last 10 Years

Which class are you in?
No thanks.