This document discusses probability theories and formulas. It defines probability as a branch of mathematics dealing with random phenomena and introduces key concepts like experiments, sample spaces, events, and equally likely probabilities. It provides examples of experiments like coin tosses, dice rolls, and drawing balls from an urn. The document also covers basic probability formulas like addition rule, complementary events, disjoint and independent events. It gives examples of calculating probabilities and introduces the conditional probability formula.

1. PROBABILITY
THEORIES
MARC JOSHUA D. GOROSPE, MA-EDUCATIONAL MGT.

2. PROBABILITY THEORY
•a branch of mathematics concerned with
the analysis of random phenomena. The
outcome of a random event cannot be
determined before it occurs, but it may
be any one of several possible outcomes.
The actual outcome is considered to be
determined by chance.

3. EXPERIMENTS, SAMPLE SPACE,
EVENTS, AND EQUALLY LIKELY
PROBABILITIES
•The fundamental ingredient of
probability theory is an experiment
that can be repeated, at least
hypothetically, under essentially
identical conditions and that may lead
to different outcomes on different
trials.

4. •The set of all possible outcomes of an
experiment is called a “sample space.” The
experiment of tossing a coin once results in
a sample space with two possible outcomes,
“heads” and “tails.” Tossing two dice has a
sample space with 36 possible outcomes, each
of which can be identified with an ordered pair
(i, j), where i and j assume one of the values 1,
2, 3, 4, 5, 6 and denote the faces showing on

5. •It is important to think of the dice as
identifiable (say by a difference in colour),
so that the outcome (1, 2) is different
from (2, 1). An “event” is a well-defined
subset of the sample space. For example,
the event “the sum of the faces showing
on the two dice equals six” consists of the
five outcomes (1, 5), (2, 4), (3, 3), (4, 2),

6. •A third example is to draw n balls
from an urn containing balls of
various colours. A generic
outcome to this experiment is
an n-tuple, where the ith entry
specifies the colour of the ball
obtained on the ith draw (i = 1,

7. •In spite of the simplicity of
this experiment, a thorough
understanding gives the
theoretical basis for opinion
polls and sample surveys.

8. •For example, individuals in a population favouring a
particular candidate in an election may be identified
with balls of a particular colour, those favouring a
different candidate may be identified with a different
colour, and so on.
CANDIDA
TE 1
CANDIDA
TE 2

9. •In contrast to the experiments
described above, many experiments
have infinitely many possible
outcomes. For example, one can toss
a coin until “heads” appears for the
first time.
•The number of possible tosses is n =

10. The probability of formula is used to
compute the probability of an event to
occur.
What is the probability that a certain
event occurs?
A probability is a chance of prediction.

11. Let’s say, X be the chances of happening
an event then at the same time (1-x) are
the chances for “not happening” of an
event.

12. Number of favourable outcome
P(A) = Total number of favourable
outcomes
or, P(A) = n(A) / n(S)
where,
•P(A) is the probability of an event
•n(A) is the number of favourable
outcomes
•N(S) is the total number of events in
the sample space

13. Basic Probability Formulas
Probability Range 0 ≤ P(A) ≤ 1
Rule of Addition P(A∪B) = P(A) + P(B) –
P(A∩B)
Rule of Complementary Events
P(A’) + P(A) = 1
Disjoint Events P(A∩B) = 0
Independent Events P(A∩B) = P(A) ⋅ P(B)
Conditional Probability P(A | B) = P(A∩B) / P(B)
Bayes Formula P(A | B) = P(B | A) ⋅ P(A) /
P(B)

14. Example1.
What is the probability that a card
taken from a standard deck, is an
Ace?
Total number of cards a
standard pack contains
= 52
Number of Ace cards in a
deck of cards = 4
the number of favourable
outcomes = 4
P(Ace) = (Number of
favourable outcomes) /
(Total number of
favourable outcomes)
P(Ace) = 4/52
= 1/13

15. Example 2.
Calculate the probability of
getting an odd number if a dice is
rolled.
Solutions:
Sample space (S) = {1, 2, 3, 4,
5, 6}
n(S) =
6
Let “E” be the event of getting
an odd number, E = {1, 3, 5}
n(E) =
3

16. Example 2.
Calculate the probability of getting an odd number
if a dice is rolled.
Solutions:
Sample space (S) = {1, 2, 3, 4, 5, 6}
n(S) = 6
Let “E” be the event of getting an odd
number, E = {1, 3, 5}
n(E) = 3
P(E) = (Number of outcomes
favorable)/(Total number of
outcomes)
= n(E)/n(S) = 3/6
= ½

17. Conditional Probability
Formula
Conditional probability formula gives
the measure of the probability of an
event given that another event has
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

18. FORMULA FOR CONDITIONAL
PROBABILITY
Conditional Probability
of A given B
P (A|B) = P(A ∩ B)⁄P(B)
Conditional Probability
of B given A
P (B|A) = P(B ∩ A) ⁄ P(A)

19. Question
1.
The probability that it is Friday and that a student is
absent is 0.03. Since there are 5 school days in a
week, the probability that it is Friday is 0.2. What is
the probability that a student is absent given that
today is Friday?
P (B|A) = P(A ∩ B) ⁄ P(A)
P(Absent | Friday)=
P (Absent and Friday) ⁄ P(Friday)
0.03/0.2 = 0.15 = 15 %

20. Question
1.
A teacher gave her students of the class two tests
namely maths and science. 25% of the students passed
both the tests and 40% of the students passed the
maths test. What percent of those who passed the
maths test also passed the science test?
Percentage of students who passed the maths
test = 40%
Percentage of students who passed both the tests
= 25%
Let A and B be the events of the number of students who
passed maths and science tests.

21. According to the
given,
P(A) = 40% = 0.40 P(A ⋂ B) = 25% = 0.25
Percent of students who passed the maths test
also passed the science test
= Condition probability of B given A
= P(B|A)
= P(A ⋂ B)/P(A)
= 0.25/0.40
= 0.625 = 62.5%