The document defines several types of probability: 1) Classical/mathematical probability which defines the probability of an event as the number of favorable outcomes divided by the total number of possible outcomes for experiments with equally likely outcomes. 2) Relative/frequency probability which is based on repeating an experiment many times and defining probability as the limit of the ratio of favorable outcomes to total outcomes as the number of repetitions approaches infinity. 3) Subjective probability which is a personal judgment of likelihood not based on formal calculations. It also discusses concepts like independent and mutually exclusive events, combinations, and properties of probability like additivity.

1. Various Definitions
of Probability
RAJU DAVID.C, MSc, MBA, PGDCA, BCC
Faculty, Department of Psychology
Rajagiri College of Social Science
Kalamassery, Ernakulam

2. Mathematical / Classical / a priori
definition of probability
Consider a fair six-sided die numbered 1,2,3,4,5,6 on its faces.
The sample space has 6 elementary outcomes. S = {1,2,3,4,5,6}
Then probability of getting the face 4 in a single roll of the die is 1/6.
In a coin tossing experiment, the probability of getting tail in a single trial is ½ as the
sample space contains 2 outcomes. S = {head, tail}
Definition
If an event can occur in N mutually exclusive, exhaustive and equally
likely ways, and if m of these possess a trait A,
Then, probability of the occurrence of the event A is equal to m/N, which is
denoted by P(A) = m/N

3. Problems with the Classical Method
The classical method requires a game to be broken down into
equally likely outcomes. It is not always possible to do this.
It is not always clear when possibilities are equally likely.
Another method, known as the frequency method had also been
used by Pascal and Fermat to verify results obtained by the classical
method.

4. Relative frequency / a posteriori probability
Relative frequency or a posteriori probability depends on
repeatability of some process/ability to count number of repetitions
and number of times that the event of interest occurs.
Definition
If some process is repeated a large number of times, say n, and if
some resulting event with the characteristic E occurs m times, then
the relative frequency of occurrence of E is = m/n, will be
approximately equal to the probability of E.
i.e., P(E) = m/n [where m/n is an estimate of P(E)]

5. SUBJECTIVE PROBABILITY
Subjective probability is a type of probability derived from an individual's personal
judgment or own experience about whether a specific outcome is likely to occur. It
contains no formal calculations and only reflects the subject's opinions and past
experience.
• It is often the only means available for making probabilistic estimates.
• Frequently used in making business decisions.
• Different people often arrive at different subjective probabilities
Examples
• Estimating the yield of crop based on the past experience of a farmer.
• you have an 80% chance of your best friend calling today, because her car broke down
yesterday and she’ll probably need a ride.
• You think you have a 50/50 chance of getting the job you applied for, because the
other applicant is also very qualified.

6. Elementary properties of probability or
AXIOMS of probability
Three properties:
1. The probability of an event must be between 0 and 1. The smallest possible probability
value is 0 (IMPOSSIBLE EVENT). You cannot have a negative probability. The largest possible
probability value is 1 (CERTAIN EVENT). You cannot have a probability greater than 1.
2. If events in a set are mutually exclusive and collectively exhaustive, the sum of their
probabilities must add up to 1.
3. If two events A and B are mutually exclusive, the probability of either event A or event B
occurring, which is same as the probability of (AUB), is the sum of their separate probabilities.
i.e., P(AUB) = P(A) + P(B)

7. COMBINATIONS
Suppose we have N items of m are to be selected. This is termed as “a combination of m from N” and
is denoted by Ncm. For example 2 girls can be selected from 5 girls in 5c2 different ways. Let us
consider the girls are G1, G2, G3, G4 & G5. We can select 2 girls as G1G2, G1G3, G1G4, G1G5, G2G3,
G2G4, G2G5, G3G4, G3G5 & G4G5. There are 10 different combinations. (Note that G1G2 & G2G1 are
the same combination). The number of different combinations can be found out using the formula
5c2 = (5x4)/(1x2) = 10, Similarly 5c3 = (5x4x3)/(1x2x3) = 10
5c1 = 5/1 = 5 5c4 = (5x4x3x2)/(1x2x3x4) = 5
5c0 = 1 5c5 = (5x4x3x2x1)/(1x2x3x4x5) = 1
From the above example, it can be seen that the value of zero-combination (Nc0) and maximum
combination NcN is always 1.
In general,
Nc0 = NcN, Nc1 = Nc(N-1), Nc2 = Nc(N-2), Nc3 = Nc(N-3) and so on

8. COMBINATIONS FROM COMBINED GROUP
Suppose we have 2 different groups of sizes N and M from which we have to
form a combined group of (r+s) members in such a way that r from the 1st group
and s from the 2nd . This can be done in Ncr x Mcs different ways.
For example, suppose there are 3 boys B1,B2,B3 and 4 girls G1,G2,G3,G4
We have to select 2 persons which include a boy and a girl.
The possible combinations are
B1G1, B1G2, B1G3, B1G4, B2G1, B2G2, B2G3, B2G4, B3G1, B3G2, B3G3 and B3G4.
i.e., 12 different combinations.
This number of different combinations can be found out using the formula
3c1 x 4c1 = (3/1) x (4/1) = 4x3 = 12

9. Examples
1. How many different ways 3 Heart cards can be drawn from a pack of cards?
A pack of cards contains 52 cards. Among these 52 cards, there are 13 hearts cards.
We want to draw 3 hearts. The 3 cards drawn should be heart which means the cards should
definitely be drawn from the 13 hearts cards only. This can be done in 13c3 different ways.
So, the answer is 13c3 = (13x12x11)/(1x2x3) = 858 different ways.
2. How many different ways 2 boys and 3 girls from a class contains 10 boys and 12 girls?
2 boys must be drawn from 10 boys and 3 girls from 12 girls.
A combination of 2 boys and 3 girls can be taken from a combined group of 10 boys
and 12 girls in 10c2 x 12c3 different ways.
So, the required answer is 10c2 x12c3 = [(10x9)/(1x2)] x [(12x11x10)/(1x2x3)]
= 45 x 660 = 29700 ways

10. 1. How many different ways 2 letters can be taken from the word “YESTERDAY”?
2. A bag contains 5 red balls and 4 white balls. How many different ways 3 balls
can be drawn at random of which there should be exactly 2 white balls?
3. In how many different ways two A’s can be drawn from the word
“MALAYALAM”?
4. Find the number of different cases that a king and a queen can be drawn
from a pack of cards?
5. How many different ways 2 spades and 3 clubs cards can be drawn from a
deck of cards?
EXERCISES

11. INDEPENDENCE OF EVENTS
Two events A & B are said to be mutually independent if the probability of their joint
occurrence is equal to the product of their individual probabilities.
i.e., if two events are independence, they should not be mutually exclusive. Then
P(AՈB) = P(A) X P(B)
Three events are said to be totally independent,
1. if they are pairwise independent
i.e., P(AՈB) = P(A) X P(B),
P(AՈC) = P(A) X P(C),
P(BՈC) = P(B) X P(C), and
2. P(AՈBՈC) = P(A) X P(B) X P(C)

12. INDEPENDENT EVENTS contd…
Note:
The terms ‘independent’ and ‘mutually exclusive’ do not mean
the same thing.
If A and B are independent and event A occurs, the occurrence of B
is not affected.
If A and B are independent, AՈB is a valid event and exists.
If A and B are mutually exclusive, however, and event A occurs,
event B cannot occur.

13. ADDITION THEOREM OF PROBABILITY
If A and B are any two events then, the probability of happening of at least one of the events is
defined as P(AUB) = P(A) + P(B) - P(A∩B).
Proof:
Since events are nothing but sets,
From set theory, we have
n(AUB) = n(A) + n(B) - n(A∩B),
(where n(A), n(B), n(AUB), n(A∩B) indicates number of outcomes favourable to respective events)
Dividing the above equation by n(S), (where S is the sample space)
n(AUB)/n(S) = n(A)/n(S) + n(B)/n(S) - n(A∩B)/n(S)
Then by the definition of probability,
P(AUB) = n(AUB)/n(S), P(A) = n(A)/n(S), P(B) = n(B)/n(S) & P(A∩B) = n(A∩B)/ n(S)
Therefore,
P(AUB) = P(A) + P(B) - P(A∩B).
A B

14. Example:
Q1. If the probability of solving a problem by two students George and James
are 1/2 and 1/3 respectively then what is the probability of the problem to be
solved?
Solution:
Let A and B be the probabilities of solving the problem by George and James
respectively.
Then P(A)=1/2 and P(B)=1/3.
The problem will be solved if it is solved at least by one of them also.
So, we need to find P(AUB).
By addition theorem on probability, we have
P(AUB) = P(A) + P(B)- P(A∩B).
P(AUB) = 1/2 + 1/3 – 1/2 * 1/3 = 1/2 +1/3 - 1/6 = (3+2-1)/6 = 4/6 = 2/3
•

15. Addition theorem for 3 events
A, B, C are 3 events (not necessarily be mutually exclusive) then
P(AUBUC) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) + P(A∩B∩C)
This can also be written as S
P(AUBUC) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)
A
B
C

16. Exercise
Q2.What is the probability that a leap year contains 53 Wednesdays?
Ans. A leap year contains 366 days. A week contains 7 different week days and 52
weeks in an year takes 52 x 7 = 364 days. So 52 weeks covers all weekdays 52 times.
The remaining 2 days of the leap year may 2 consecutive week days, viz.,
SUN-MON, MON-TUE, TUE-WED, WED-THU, THU-FRI, FRI-SAT & SAT-SUN
which constitute the sample space. Then n(S) = 7
Let A denotes the event of occurring Wednesday.
Here wed can occur in two combinations, tue-wed & wed-thu. i.e., n(A) = 2
Therefore The prob. of getting 53 Wednesdays in a leap year = P(A) =n(A)/n(S) =2/7

17. Questions
Q3. Consider 2 fair dice are rolled together.
a) Write down the sample space
b) What is the probability that the sum of faces is 8?
c) What is the probability that each die shows same number?
d) What is the probability that the sum of faces is an odd number?
e) What is the probability that the first die always shows less number than the second?
f) What is the probability that the sum of faces is a square number?
g) What is the probability that the sum should at least 4?
h) What is the complement probability of the sum is utmost 10?

18. A deck of cards

19. EXERCISES
Q4. Find the probability of drawing an spade card from a pack of cards.
An:- A pack of cards contain 4 types of cards, viz., SPADE, CLUBS, HEARTS & DIAMONDS.
Spade & Clubs are BLACK in colour where as Hearts & Diamonds are RED
Each item has 13 different cards - Ace(1),2,3,4,5,6,7,8,9,10,Jack,King & Queen.
So, a deck of cards contains 13 x 4 = 52 cards.
Let A denote the event that drawing a spade card form a pack of cards.
We can draw a spade card from 13 spade cards in 13C1 = 13/1 = 13 ways.
i.e., n(A) = 13, the number of favourable cases.
A card can be drawn from a pack of 52 cards in 52C1 different ways. Total number of possible cases = Total
number of cases in the sample space S, i.e., n(S) = 52/1 = 52
The required probability P(A) = No. of favourable cases/ Total No. of possible cases = n(A)/n(S)
= 13/52 = ¼ = 0.25 or 25%

20. ASSIGNMENT -
Q1. A bundle of 100 tickets numbered serially from 1 to 100 is well shuffled a ticket
is drawn at random. What is the probability that
a) It is an even number?
b) It is a multiple of 10?
c) It is a perfect square?
d) It is a multiple of 12 or a perfect cube?
e) It is a square or a cube?
f) It is an odd number greater than 75?
h) It is a prime number less than 25?

21. ASSIGNMENT
Q2. A bag contains 5 white, 4 red and 3 black balls. 3 balls are drawn from the
bag at random. Find the probability that
a) there are 2 white and a black balls
b) There is no white ball
c) balls have all the 3 colours
d) none of them are red or black
e) Exactly one white
g) at least one red
H) balls are same colour

22. Meaning of certain terms used in probability
1. Prob. of ATLEAST ONE = Total prob. – Prob. of getting NONE
e.g. Prob of getting at least one head in a tossing of 2 coins
Sample space S = {HH,HT,TH,TT}
At least one head = {HH, HT,TH}
So prob of at least one head = ¾
But total prob = 1, prob of no head = ¼
Total prob – prob of none = 1 – ¼ = c

23. Meaning of certain terms used in probability
PROB. OF ATMOST N items = P(x≤N) = TOTAL PROB – P(x≥N)
= P(none) + P(1) + P(2) + P(3) + … + P(N)
= 1 – Prob(x≥N)
E.g. Prob of getting at most 5 in a die rolling experiment
Sample space S = {1,2,3,4,5,6}
The event A = {1,2,3,4,5} and so required Prob = n(A)/n(S) = 5/6
Total prob = 1, Prob(x≥5) = Prob(x=6) = 1/6
1 - Prob(x≥5) = 1 – 1/6 = 5/6
i.e., P(x≤N) = 1 – Prob(x≥N)