1. Mathematics

2. Session
Probability - 1

3. Probability of an Event
Session Objectives
Experiment
Sample Space
Event
Types of Events
Probability of an Event
Class Exercise

4. Experiment
Experiment: An operation, which results in some well-defined
outcomes is called an experiment.
Random Experiment: If we conduct an experiment and we do not
know which of the possible outcome will occur this time, the
experiment is called a random experiment.
For example:
Tossing a coin
Throwing a die
3. Drawing a card from a well shuffled pack of cards

5. Sample Space
Sample Space: The sample space of an random experiment is the set
of all possible elementary outcomes. It is denoted by S.
For example:
When we toss a coin, the sample space
S = {H, T}
It is a random experiment, because when we toss a coin this time, we do
not know whether we shall get head or tail.
When we throw a die, the sample space
S = {1, 2, 3, 4, 5, 6}

6. Event
Event : A subset ‘E’ of a sample space is called an event. An event
is a combination of one or more of the possible outcomes of an
experiment.
For example:
In a single throw of a die, the event of getting a prime number is given
by E = {2, 3, 5} and the sample space S = {1, 2, 3, 4, 5, 6}.

7. For any three events A, B and C with sample space S.
Algebra of Events
For any three events A, B and C with sample space S.

8. Algebra of Events

9. Types of Events (Sure Event)
Sure Event: In the throw of a die, sample space S = {1, 2, 3, 4, 5, 6}

10. Types of Events (Impossible Event)
Impossible Event: In the throw of a die, sample space
S = {1, 2, 3, 4, 5, 6}.
Let E be the event of getting an ‘8’ on the die.
Clearly, no outcome can be a number 8.

11. Simple and Compound Event
Simple or Elementary Event
An event that contains only one element of the sample space is called
a simple or an elementary event.
Compound Event
A subset of sample space which contains more than one element is
called compound event or mixed event or composite event.
For example: In a simultaneous toss of two coins, the sample space is
S = {HT, TH, HH, TT}

12. Equally Likely Outcomes
The outcomes are said to be equally likely, if none of them is expected
to occur in preference to the other or the chances of occurrence of all
of them are same.
For example:
In throwing of a single die, each outcome is equally likely to happen.

13. Mutually Exclusive Events
Two or more events are said to be mutually exclusive if no two or more
of them can occur simultaneously in the same trial.
Facts:
Elementary events related to an experiment are always
mutually exclusive.
2. Compound events may or may not be mutually exclusive.

14. In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}
Example
In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}

15. Exhaustive Events Exhaustive Events
In a random experiment, two or more events are exhaustive if
their union is the sample space.

16. In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}
Example
In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}

17. Example-1
A die is thrown twice. Each time the number appearing on it is recorded. Describe the events :
A: both numbers are odd.
B: sum of numbers is less than 6.
C: both numbers are even.
Describe Which pairs of events are mutually exclusive.
Solution: A ={(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}
B = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
C ={(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}

18. Solution Cont.
{(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (3,1)
(3,2), (3,3), (3,5), (4,1), (5,1), (5,3), (5,5)}
{(1,1),(1,3),(3,1)}
{(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5)
(4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)}
A and C are mutually exclusive

19. Example –2 Three coins are tossed . Describe
two mutually exclusive events A and B.
three mutually exclusive exhaustive events A, B and C.
two events E and F which aren’t mutually exclusive.
Solution: Sample space for the toss of three coins is
S = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T),
(T, H, T), (T, T, H), (T, T, T)}
Let A:{Event of getting three heads} ={(H, H, H)}
and B:{Event of getting three tails}={(T, T, T)}
A and B are mutually exclusive

20. Solution (Cont.) Let C:{Event of getting one or two head}
={(H, H, T),(H, T, H),(T, H, H),(H, T, T),(T, H,T),(T,T,H)}
A, B and C are mutually exclusive exhaustive events.
Let E: {Event of getting two heads}
= {(H, H, T),(H, T, H),(T, H, H)} and
F: {Event of getting at least one tail}
= {(H, H,T), (H, T, H),(T, H, H), (H, T, T), (T, H, T), (T, T, H), (T,T,T)}
E and F aren’t mutually exclusive.

21. Probability of an Event
In a random experiment, the probability of happening of the event A
with sample space S is defined as

22. Probability of an Event (Cont.)
If m is the number of outcomes favourable to an event and n is the
total number of possible outcomes. Then,
Hence, the probability of an event always lies between 0 and 1.

23. Probability of an Event (Cont.)
Sure Event: An event A is said to be sure or certain if P(A) = 1
Impossible Event: An event A is said to be impossible if P(A) = 0
Odds in favour of occurrence of an event A are defined as m : n - m,
i.e. ratio of favorable outcomes to unfavorable ones.
Odds against occurrence of event A are defined as n - m : m , i.e.
ratio of unfavourable outcomes to favourable ones.

24. Example-3
What is the probability of getting at least two heads in a simultaneous
throw of three coins?
Solution: If three coins are tossed together possible outcomes are
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Number of these exhaustive outcomes, n(S) = 8
At least two heads can be obtained in the following ways
E = {HHH, HHT, HTH, THH}

25. Number of favourable outcomes, n(E) = 4
Solution Cont.
Number of favourable outcomes, n(E) = 4

26. Thus, the number of such outcomes = 5
Example-4
In a single throw of two dice what is the probability of getting
(i) 8 as the sum (ii) a total of 9 or 11
Solution: In throwing of a pair of dice, total number of outcomes in
sample space = 6 × 6 = 36
(i) To get 8 as the sum favourable outcomes are
(2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).
Thus, the number of such outcomes = 5

27. Thus, the number of such outcomes = 6
Solution Cont.
(ii) Favourable outcomes to the event of getting the sum as 9 or 11
are (3, 6), (4, 5), (5, 4), (6, 3), (5, 6) and (6, 5).
Thus, the number of such outcomes = 6

28. Example-5
The letters of the word ‘SOCIETY’ are placed at random in a row. What
is the probability that three vowels come together.
Solution: There are 7 letters in the word ‘SOCIETY’.
These 7 letters can be arranged in a row in 7! ways.
O, I, E are three vowels in the word ‘SOCIETY’.
Assuming these three vowels as one letter, we get 5 letters which
can be arranged in a row in 5! ways.

29. But three vowels O, I, E can be arranged in 3! ways.
Solution Cont.
But three vowels O, I, E can be arranged in 3! ways.
The total number of arrangements in which three vowels come
together is 5! × 3!.

30. Example-6
A five-digit number is formed by the digits 1, 2, 3, 4, 5 without
repetition. Find the probability that the number is divisible by 4.
Solution: Total number of ways in which a five digit number can be
formed by digits 1, 2, 3, 4, 5 = 5!
A number is divisible by 4 if the numbers formed by last two digits are
divisible by 4.

31. Solution Cont.
Thus for an outcome to be favorable, the last two digits can be
(1, 2), (2, 4), (3, 2), (5, 2).
The last two digits can have only these 4 arrangements. But the rest
of the three digits can be arranged in 3! ways.

32. Example-7
A bag contains 8 red and 5 white balls. Three balls are drawn at random.
Find the probability that one ball is red and two balls are white.
Solution: Total number of balls = = 13
n(S) = number of ways of selecting 3 out of 13 balls
Let A be the event of selecting one red and 2 white balls out of 8 red
and 5 white balls.

33. Solution Cont.

34. Example-8
A bag contains 50 tickets numbers 1, 2, 3, …50 of which five are
drawn at random and arranged in ascending order of magnitude

35. Solution Cont.

36. Solution: Out of 9 students 4 students can be selected in
Example-9
Out of 9 outstanding students in a college, there are 4 boys and 5 girls.
A team of four students is to be selected for a quiz programme. Find the
probability that two are boys and two are girls.
Solution: Out of 9 students 4 students can be selected in
Total number of events
There are 4 boys and 5 girls out of which 2 boys and 2 girls can be
selected in

37. Favourable number of events
Solution Cont.
Favourable number of events

38. Example-10
Four cards are drawn at random from a pack 52 playing cards.
Find the probability of getting
all the four cards of the same suit (CBSE 1993)
(ii) all the four cards of the same number (CBSE 1993)
Solution:
(i) There are four suits: club, spade, heart and diamond, each of 13 cards.
Therefore, the total number of ways of getting all the four cards of
the same suit

39. Favourable number of events = 13
Solution Cont.
(ii) Four cards of the same number:
(1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), …(13, 13, 13, 13).
Favourable number of events = 13

40. Example –11
Two dice are thrown. Find the odds in favour of getting the sum to be
(i) 4 (ii) 5 (iii) what are the odds against getting the sum to be six.
Solution: The sample space when two dice are thrown is
S = {(1, 1), (1, 2), ... (1, 6), (2, 1), (2, 2), ... (2, 6),
(3, 1), (3, 2), ... (3, 6), (4, 1), (4, 2), ... (4, 6),
(5, 1), (5, 2), ... (5, 6), (6, 1), (6, 2), ... (6, 6)}
n(S) = 36

41. Solution (Cont.)
Let A be the event of getting the sum on the pair of dice to be 4
A = {(1, 3), (2, 2), (3, 1)}
Let B be the event of getting the sum on the pair of dice to be 5.
B = {(1, 4), (2, 3) (3, 2) (4, 1)}

42. Solution (Cont.)
Let C be the event of getting the sum to be six on the pair of dice
C = {(1, 5), (2, 4) (3, 3) (4, 2), (5, 1)}

43. Thank you