Download Probability of an Event

Mathematics
Session
Probability - 1
Session Objectives
 Experiment
 Sample Space
 Event
 Types of Events
 Probability of an Event
 Class Exercise
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:
1.
Tossing a coin
2.
Throwing a die
3.
Drawing a card from a well shuffled pack of cards
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:
1.
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.
2.
When we throw a die, the sample space
S = {1, 2, 3, 4, 5, 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}.
E  S
Algebra of Events
For any three events A, B and C with sample space S.
1
A  B = either A or B or both
2
A  B = both A and B
3
 4
A = not A
A - B = A but not B
5  A  B' = A  B
Algebra of Events
6  A  B' = A  B
7
A  B  C =  A  B  C
8
A  B  C =  A  B  C
9
A  B  C =  A  B   A  C
10
A  B  C =  A  B   A  C
Types of Events (Sure Event)
Sure Event: In the throw of a die, sample space S = {1, 2, 3, 4, 5, 6}
S  S S represents an event.
Each outcome of the experiment is a member of S.
 S is called a sure event or certain event.
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.
E =  is an impossible event.
Complement of an event E or EC =not E=S-E
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}
Event E1 ={TT} of getting both tails is an elementary event.
Event E2 ={HT, TH, TT} of getting at least 1 tail is a compound event.
Equally Likely Outcomes
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.
Mutually Exclusive Events
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.
Two events E1 and E2 are mutually exclusive, if E1  E2 = .
Facts:
1. Elementary events related to an experiment are always
mutually exclusive.
2.
Compound events may or may not be mutually exclusive.
Example
In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}
Event E1 = event of getting a number less than 3
and E2 = event of getting a number more than 4, then
E1 = 1, 2 and E2 = 5, 6  E1  E2 = 
E1 and E2 are mutually exclusive events.
Exhaustive Events
Exhaustive Events
In a random experiment, two or more events are exhaustive if
their union is the sample space.
i.e. in a random experiment, events E1, E2 , ...En
with sample space S are exhaustive if
E1  E2  ...  En = S
Example
In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}
E1 : 1, 2 - the event of occurrence of '1' or '2'.
E2 : 2, 3, 4 - the event of occurrence of '2' or '3' or '4'.
E3 : 3, 4, 5, 6 - the event of occurrence of a number  3.
E1  E3  S, E1  E2  E3  S but E1  E2  S
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 A  B, A  B, A  C, A  C. 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)}
Solution Cont.
A  B= {(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)}
A  B  {(1,1),(1,3),(3,1)}
A  C  {(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C 
A and C are mutually exclusive
Example –2
Three coins are tossed . Describe
(1) two mutually exclusive events A and B.
(2) three mutually exclusive exhaustive events A, B and C.
(3) 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
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.
Probability of an Event
In a random experiment, the probability of happening of the event A
with sample space S is defined as
P A =
Number of outcomes favourable to A n  A 
=
Total number of possible outcomes
n S 
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,
P A =
m
n
Clearly, 0  m  n
0
m
1
n
 0  P  A  1
Hence, the probability of an event always lies between 0 and 1.
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.
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}
Solution Cont.
Number of favourable outcomes, n(E) = 4
Required probability =
n E 4 1
= =
n S 8 2
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
Solution Cont.
Required probability 
Number of favourable outcomes 5

Number of total outcomes
36
(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
 Re quired probability 
6 1

36 6
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.
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!.
Required probability =
5!×3! 1
=
7!
7
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!
Exhaustive number of cases=5!=120
A number is divisible by 4 if the numbers formed by last two digits are
divisible by 4.
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.
Number of favourable outcome=3!×4
Required probability =
4×3!
4  3!
1


5!
5  4  3! 5
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 = 8 + 5 = 13
n(S) = number of ways of selecting 3 out of 13 balls
=
13
C3 =
13×12×11
= 286
3×2×1
Let A be the event of selecting one red and 2 white balls out of 8 red
and 5 white balls.
Solution Cont.
 n  A  = 8C1 ×5 C2 = 8×10 = 80
P getting one red and two white balls
=P  A  =
n A 
n  S
=
80
40
=
286 143
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
 x1 <x2 <x3 <x5 .
Find the probability that x3  30.
Solution: Five tickets out of 50 can be drawn in
 Total number of elementary events = 50 C5
x1 < x2 < x3 < x5 and x3 =30
 x1, x2 <30 and this may happen in
29
C2 ways.
CBSE 2002
50
C5 ways.
Solution Cont.
x4 , x5 >30 and this may happen in
20
C2 ways.
 Favourable number of elementary events = 29C2 × 29 C2
Required probability =
29
C2 × 29 C2
551
=
50
C5
15134
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
9
C4 ways
Total number of events  9 C4
There are 4 boys and 5 girls out of which 2 boys and 2 girls can be
selected in = 4 C2 × 5 C2
Solution Cont.
Favourable number of events = 4 C2 × 5 C2
Required probability =
4
C2 × 5C2
9
C4
=
10
21
Example-10
Four cards are drawn at random from a pack 52 playing cards.
Find the probability of getting
(i) all the four cards of the same suit
(CBSE 1993)
(ii)
(CBSE 1993)
all the four cards of the same number
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 
13
C4 + 13 C4 + 13 C4 + 13 C4  4

13
C4

Solution Cont.
 Required probability =
4

13
52
C4
C4
=
198
20825
(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
Required probability =
13
13
=
52
C4 270725
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
Solution (Cont.)
(i) Let A be the event of getting the sum on the pair of dice to be 4
A = {(1, 3), (2, 2), (3, 1)}
Odds in favour of event A =
3
3
1
=
=
36 - 3 33 11
(ii) 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)}
Odds in favour of event B =
4
4
1
=
=
36 - 4 32 8
Solution (Cont.)
(iii) 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)}
Odds against event C =
36 -5 31
=
5
5
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