The document is a course on probability that covers basic concepts, problem-solving techniques, and formulae related to probability. It explains key terms such as sample space, events, and various properties of probability, including independent and conditional probabilities, as well as Bayes’ theorem. The course also includes practical examples and problems to enhance understanding and application of probability concepts.

1. MAKING PROBABILITY
EASY!!!
GAURAV SAHA

2. Salient features of this course
 Briefly covers the basic concepts and makes understanding
easy and really quick
 Makes you familiar with the problem solving techniques
 Makes you deal easily and quickly with the term
“PROBABILITY”
 Its really fun and lots more to learn here !!!

3. What are we going to do in this course???
Understanding the
word “PROBABILITY”
Step-
1 Deal with
some
formulaes
Step-
2
Solve
Examples
Step-
3

4. Lets get started
!!!!!!!ALL THE
BEST!!

5. DEFINITION
PROBABILITY is the extent to which something is likely to happen or be the case in an event.
It is the ratio of the desired outcomes to the total outcomes.
So, if n(e) = desired outcomes , t(e) = total outcomes from an event
then we can say P(e) = n(e)/t(e)

6. Some terms related to PROBABILITY
Sample Space
 In probability theory, the sample
space of an experiment or random
trial
is the set of all possible outcomes or
results of that experiment.
Event
 In probability theory, an event is a
set
of outcomes of an experiment (a
subset of the sample space) to which
a probability is assigned.

7. So lets take some EXAMPLES
 If I roll a dice, then total number of outcomes will be six.
The possible outcomes can be {1,2,3,4,5,6}
So, probability of getting a 1 while throwing a dice = 1/6 as desired number of
outcomes = n(e) = 1 and total number of possible outcomes = t(e) = 6.
So, according to the formulae P(e) = n(e)/t(e) , we get 1/6 .
Similarly, Probability of getting a 2 = Probability of getting a 3 = Probability of
getting a 4 = Probability of getting a 5 = Probability of getting a 6 = 1/6 .

8. LETS take a PRACTICE
If I toss a coin, then what is the probability that I will get a head?
Also calculate the probability to get a tail.

9. Answer to previous question
If I toss a coin then the possible outcomes are either head or tail.
So, total number of possible outcomes = t(e) = 2
Now, probability of getting a head = P(e) = n(e)/t(e) = ½
Similarly, Probability of getting a tail = ½.
Here, we see both of them add to 1.
Thus, our answer is correct.

10. Two or more events at the same time
In case of “AND” term
If the AND term is used while describing the outcomes of an event,
then we have to multiply the probabilities of the outcomes of the
event.

11. Practise problem
If I throw a disc and toss a coin at the same time, what is the
probability that I will get a 5 and heads??
 Analysing the problem:-
Two Simultaneous events– Throwing a disc, toss a coin
Here, the term AND is used in case of outcomes
So, we have to multiply the corresponding probabilities.
Probability of getting a 5 = 1/6
Probability of getting a heads = ½
So, required probability = P(5 and heads) = 1/6*1/2 =1/12

12. Two or more events at the same time
In case of “OR” term
If the OR term is used while describing the outcomes of an event,
then we have to add the probabilities of the outcomes of the
event.

13. Practise problem
If I throw a disc and toss a coin at the same time, what is the probability
that I will get a 5 or heads??
 Analysing the problem:-
Two Simultaneous events– Throwing a disc, toss a coin
Here, the term OR is used in case of outcomes
So, we have to add the corresponding probabilities.
Probability of getting a 5 = 1/6
Probability of getting a heads = ½
So, required probability = P(5 or heads) = 1/6+1/2 = 2/3

14. Discussion of Basic concepts from Set Theory
The probability that Events A and B both occur is the probability of
the intersection of A and B. The probability of the intersection of Events A and
B is denoted by P (A ∩ B).
The probability that either Event A or B occur is the probability of
the union of A and B. The probability of the union of Events A and
B is denoted by P(A U B).
The compliment of Event A is A
It is a set of all outcomes that are not in A.
When A and B have no outcomes in common, then they are said to be mutually exclusive
or mutually disjoint events.
c

15. ILLUSTRATION
 Suppose, A, B and C are three events.
A = {1,5,7,9}
B = {2,3,4,6,8}
C = {5,7,6}
So, (A U B) = {1,2,3,4,5,6,7,8,9}
(A ∩ B) = {0} as they have no elements in common. So, they are mutually
exclusive events.
A = {2,3,4,6,8}, elements not present in A.
Also, (A ∩ C) = {5,7}. So, they are not mutually exclusive.
c

16. PROPERTIES of PROBABILITY
Probability of an event A is symbolized by P(A). Probability of an event A is lies between 0 ≤ P(A) ≤ 1.
 The first property of probability is that the probability of the entire sample space is one. Symbolically we
write P (S ) = 1
 The second property of probability deals with mutually exclusive events. If E1 and E2 are mutually exclusive
events, meaning that they have an empty intersection, then P (E1 U E2 ) = P (E1) + P(E2). If they are not
mutually exclusive then
P (E1 U E2 ) = P (E1) + P (E2) - P (E1 ∩ E2 )
As long as this occurs, the probability of the union of the events is the same as the sum of the probabilities:
P (E1 U E2 U . . . U En ) = P (E1) + P (E2) + . . . + En

17. PROPERTIES of PROBABILITY
 According to third property, if we denote the complement of the event E by E C then,
E and E C have an empty intersection and are mutually exclusive.
Furthermore E U E C = S, the entire sample space.
Thus, we can say , 1 = P (S ) = P (E U E C) = P (E ) + P (E C) .
Rearranging above equation ,we get P (E ) = 1 - P (E C).
If A, B and C are not independent or mutually exclusive then the union probability is given by:
P(A∪B∪C) =P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) - P(A∩B∩C)

18. PROPERTIES of PROBABILITY
 INDEPENDENT EVENTS
Two events are said to be independent of each other when the
probability that one event occurs in no way affects the probability of the
other event occurring. An example of two independent events is as
follows; say you rolled a die and flipped a coin.
If A and B are two independent events, then P(A ∩ B) = P(A)*P(B)
Similarly, P(A ∩ B ∩ C) = P(A)*P(B)*P(C) provided A,B,C are independent set of
events.

19. CONDITIONAL PROBABILITY
 Definition: If E and F are two events associated with the same sample space of a random
experiment, the conditional probability of the event E given that F has occurred, i.e. P (E|F) is given
by P(E|F) = P (E ∩ F) / P(F) provided P(F) ≠ 0
Example
 If P(A) = 7 /13 , P(B) = 9 /13 and P(A ∩ B) = 4 /13 , evaluate P(A|B).
Solution : We have P(A|B) = P(A ∩ B) / P(B) = (4/13)/(9/13) = 4/9

20. Total Probability Theorem
Statement: Let {E1 , E2 ,...,En } be a partition of the sample space S, and suppose that each of
the events E1 , E2 ,..., En has nonzero probability of occurrence. Let A be any event associated with
S , then
P(A) = P(E1 ) P(A|E1 ) + P(E2 ) P(A|E2 ) + ... + P(En ) P(A|En ) = 𝒋=𝟏
𝒏
𝑷 𝑬𝒋 𝑷( A|Ej )

21. Problem
 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the
construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed
on
time if there is a strike. Determine the probability that the construction job will be completed on time.
 Solution Let A be the event that the construction job will be completed on time, and B be the event that there
will be a
strike.
So, We have to find P(A).
We have P(B) = 0.65,
Hence, P(no strike) = P(B′) = 1 − P(B) = 1 − 0.65 = 0.35
Also, P(A|B) = 0.32, P(A|B′) = 0.80
Since events B and B′ form a partition of the sample space S, therefore, by theorem of total probability, we have
P(A) = P(B) P(A|B) + P(B′) P(A|B′) = 0.65 × 0.32 + 0.35 × 0.8 = 0.208 + 0.28 = 0.488
Thus, the probability that the construction job will be completed in time is 0.488.

22. BAYES’ THEOREM
 If E1 , E2 ,..., En are n non-empty events which constitute a partition of
sample space S,
i.e. E1 , E2 ,..., En are pairwise disjoint and ( E1 ∪ E2 ∪ ... ∪ En ) = S and A is
any event
of non-zero probability, then
P(Ei|A)=
𝑷 𝑬𝒊 𝑷 𝑨 𝑬𝒊
𝒋=𝟏
𝒏 𝑷 𝑬𝒋 𝑷(𝑨|𝑬𝒋)
for any i = 1, 2, 3, ..., n
This is Bayes’ Theorem.

23. Example
 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at
random
from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.
Solution : Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II and A be the event of drawing
a red
ball.
Then P(E1 ) = P(E2 ) = 1/2 .
Also P(A|E1 ) = P(drawing a red ball from Bag I) = 3/7 and
P(A|E2 ) = P(drawing a red ball from Bag II) = 5/11
Now, the probability of drawing a ball from Bag II, being given that it is red, is P(E2 |A) By using Bayes' theorem, we
have
P(Ei|A)=
𝑷 𝑬𝒊 𝑷 𝑨 𝑬𝒊
𝒋=𝟏
𝒏 𝑷 𝑬𝒋 𝑷(𝑨|𝑬𝒋)
= (½ * 5/11)/(1/2 * 3/7 + ½ * 5/11) = 35/68

24. THANK YOU !!!!!
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