Probability

pIFFERENT APPROACHES
Thereare following three types of
à
approaches to theory of probability:
Experimental approach, empirical approach or observed required approach.
(i) Classical approach.
(iii) Axiomatic approach.
Inthis chapter, we will study only the

IMPORTANT TERMS AND

DEFINITIONS
Experiment
An activity which gives some well-defined outcomes is called experiment "Tossing a coin" gives either head (H) or tail (1), S
an experiment.
Random experiment
It is an experiment, in which we know about all the possible outcomes but not sure about a specific outcome. For example, in
browing a dice, possible out comes are 1, 2, 3, 4, 5, 6 but we are not sure that the no on dice is '4.
Event
number 1 or 2 or 3 or 4 or 5 or 6 is
Thepossible outcomes of a trial are called events., e.g.. when a die is rolled, showing the
an event.

Sure event
called a sure event and its empirical probability
When all the outcomes of a random experiment favour an event, the event is
is 1.

Impossible event
event is called an impossible event and its empirical probability
When no outcome ofa random experiment favours an event, the
is 0.

Favourable events
an event are called favourable cases to that event.
The cases, which ensure the occurrence of

Sample space
an experiment is called sample space.
The collection of all possible outcomes in
Some special sample spaces
A
die is thrown once
S={1,2,3,4, 5, 6};n(S)=6
A coin is tossed once
S={H,T; n(S) =2
A coin is tossed twice
S={HH, HT,TH,TT}; n(S) =4=2?
or
simultaneously
Two coins are tossed

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Mathematics Class IX
A coin is tossed three [HHH, HHT, HTH,THH)
times
S=
|TTT,TTH, THT,HTT sMS)=
8=2
Three coins are tossed
simultaneously
[1,1),(1,2), (1,3), (1,4), (1,5), (1,6)
(2, 1), (2,2), (2,3), (2,4), (2, 5),(2,6)
Two dice are thrown together (3,1), (3, 2), (3,3), (3,4),(3, 5), (3, 6) n(s) = 36 =62
Or |(4,1),(4,2),(4, 3),(4,4),(4,5), (4,6)|
A die is thrown twice (5,1),(5,2), (5,3), (5,4),(5,5), (5,6)
(6,1),(6,2), (6,3), (6,4),(6,5),(6,6)

EMPIRICAL PROBABILITY
The empirical probability is a type of probability which quantifies the occurrence of some event on the basis
of empirical evidence collected by the experiment. Empirical probability is determined by finding the ratio of number
of outcomes to an event to the number of trials in the actual experiment, not theoretically. Let us suppose that E be an
event defined in some sample space, then the empirical probability of E is calculated by dividing number of
favorable outcomes and total number of times the experiment is performed. The formula for empirical probability
is given
Number of favourable cases
Probability of an event E = P(E)= Total number of cases
The probability of an event can be minimum 0 or maximum li.e. it lies between 0and 1.
There are broadly two types of probabilities; (i) empirical probability and (ii) theoretical probability.
Empirical probability is the calculation of probability of an event based on the practical experiment.
NOTE:

P(E)+P (not E) = 10s P(E) S1Sum of the probabilities of all the outcomes of random experiment is 1.
ILLUSTRATION 1: A
coin is tossed 500 times with the following frequencies of two outcomes: Head 240 times, tail 260 times.
Find the probability of occurrence of each of these events.
Solution: It is given that the coin is tossed 500 times.
.:. Total number of cases = 500

The events of getting a head and of geting a tail are denoted by Aand Brespectively. Then.
Number of cases in which the event A happens = 240
and, Number of cases in which the event B happens = 260
Probability of event getting head
P(A) =
Number of cases in which the event A happens 240
Total number of cases =0.48
S00
Probability of event getting tail
Number of cases in which the event
P(B) =
Total number of cases
happens 260
=0.52
500

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ILLUSTRATION 2: Two coins are tossed simultaneously 1000 times with the following frequencies of different outcomes:
Two heads: 210 times
One head: 550 times
No head: 240 times
occurrence of each of these events.
Eind the probability of
two heads, B= Getting one head
Solution: Let, A = Getting
C= Getting no head.
cases=1000
Total number of
happens =210
Number of cases in which event A
happens = 550
Number of cases in which event B
C happens = 240
Number of cases in which event Number of cases in which the event A happens 210
=0.21
P(A) = Total number of cases 1000

Number of cases in which the event B happens 550

=0.55
P(B) = Total number of cases 1000
240
Number of cases in which the event C happens =0.24
P(C) = Total number of cases 1000

times out of 40 balls he played. Find the probability that he

ILLUSTRATION 3: In a cricket match, a batsman hits a boundary 8
didn't hit a boundary.
hit a boundary.
Solution: Let A denote the event that the batsman did not
We have, Total number of cases = 40
Number of cases in which the eventAhappened = 40- 8=32
32 4
P(A)= -=0.8
40 5

250 consecutive days, its weather forecost were

ILLUSTRATION 4: The record of a weather station shows that out of the past
correct 175 times. What is the probability that on a given day.
). it was correct?
(ii) it was not correct?
Solution: We have,
lotal number of days for which the weather forecast was made = 250
Number of days for which the forecast was correct = 175
Number of days for which the forecast was not correct = 250 --175 =75
Therefore,
(1) Probability that the forecast was correct on a given day
Number of days for which the forecast was correct
Number of days for which the forecast was made
175
= =0.7
250
(1) Probabilitythat the forecast was not correct on a given ay

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Number of days for which the forecast was not correct
Number of days for which the forecast was made
75
= 0.3
250

ILLUSTRATION 5: There are 40 students in aclass and their results is presented as below :
Result Pass Fail
Number of Students 30 10
Ifa student is chosen at random out of the class. find the probability that the student has passed the examination.
Solution: Total number of students = 40
Cases which favour a student to pass = 30
30
The probability of the required event, i.., the student has passed the examination = 40 = 0.75

ILLUSTRATION6: A coin is tossed 150 times and the outcomes are recorded. The frequency distribution of theoutcomesH
(i.e. head) and T (i.e., tail) is given below:
Outcome H T

Frequency 85 65
Find the value of P(H), i.e., probability of getting a head in a single case.
Solution: Total number of cases = 150
Cases which favour the outcome H= 85
85
P(H) =
150
=0.567 (approx.)
ILLUSTRATION 7: 400students of class Xof aschool appeared in atest of 100 marks in the subject of social studies and the
data about the marks secured is as below :

Marks secured 0-25 26-50 51-75 Above 75 Total no. of students

Number of Students 50 220 100 30 400
If the result card of a student be picked up at random, what is the probability that the student has secured more than 50
marks'
Solution: Totalnumber of students, i.e., the total frequency =400
The totalnumber of students who secured more than 50 marks = 100 +30 = 130
130
’Probability that the marks secured are more than 50 = =0.325.
400
ILLUSTRATION8: A die is thrown 1000 times with the following frequencies for the
outcomes 1.2.3.4, 5, and 6 as give
below:
Outcomes 2 3 4 6
Frequency 179 150 157 149 175 190
Solution: PE) =Probability of gettingoutcome 1
Frequency of 1
Total number of times the die is
thrown

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 Probability
179
= 0.179
1000
P(E,)= Probability of getting outcome 2
Frequency of 2 150
=0.15
Total number of times the die is thrown 1000

Similarly, we have, 149 175

157 =0.149, P(E;) =-1000-=0.175 and
PE;)= 1000 =0.157, P(E,)= 1000
190 =0.19
P(E;) = 1000
following data were recorded
families with 2children were selected randomly, and the
MUSTRATION9: 1000
0 1 2
Number of boys in a family
Number of families
140 560300
one boy (v) at
probability that it has (i) No boy (ii) one boy (ii) 2 boys (iv) at least
the
Ifa family is chosen at random, find
most one boy
two children = 1000
Solution: Total number of families with
boy child = 140
() Number of families having no 140 =0.14
have a boy child =
at random does not
’ Probability that a family chosen 1000
boy child = 560
(ii) Number of families having one 560
boy child = =0.56
has one
Probability that afamily chosen at random 1000
(ii) Number of families having two boys children =300
300
=0.3
Probability that a family chosen at random has two boys children = 1000
’
560+ 300 =860
(iv) Number of families having at least one boy child=
860
has at least one boy child 1000-=0.86
’ Probability that a family chosen at random
child= 140 + 560 =700
(V) Number of families having at most one boy
700
has at most one boy child = -=0.7
’ Probability that a family chosen at random 1000
tests are given below:
LLUSTRATION 10: The percentage of marks obtained by astudent in the monthly unit
I III IV V
Unit
|Percentage of marks obtained 58 64 76 62 85

Find the probability that the student gets:

(() a first division i.e. atleast 60% marks
(i) marks between 70% and 80%
(iii) a distinction i.e.. 75% or above.
(iv) less than 65% marks.
Dolution: Total number of unit tests held =5
) Number of unit tests in which the student gets a first class i.e. atleast 60% marks =4
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’ Probability that the student gets a first class ==0.8

5

(11) Number of unit tests in which the student gets marks between 70% and 80% = 1
1
’ Probability that a student gets marks between 70% and 80% = =0.2
5

(ii1) Number of unit tests in which the student gets distinction = 2

2
Probability that the student gets distinction = =0.4

(iv) Number of unit tests in which the student gets less than 65% marks =3
3
Probability that a student gets less than 65% marks = :0.6

ILLOSTRATION 11: On the page of a telephone directory, there were 200 telephone numbers. The frequency distribution of
their unit place digit is given in thetable below:
1 2 3 5 6
Digit
26 22 22 20 10 14 28 16 20
Frequency 22
A number ischosen at random, find the probability that the digit at its unit's place is:
(i) 6 (ii) a non-zero multiple of 3
(1iii) a non-zero even number (iv) an odd number.
Solution: We have,
Total number of selected telephone numbers = 200
()) Itis given that the digit 6occurs 14 times at unit's place.
14
’ Probability that the digit at unit's place is 6 = -=0.07
200
(ii) Anon-zero multiple of 3i.e.3, 6 and 9.
Number of telephone numbers in which unit's digit is either 3 or 6 or 9 = 22 + 14+ 20 = 56
56
:. Probability of getting a telephone numbers having a multiple of 3 at unit's place = =0.28
200
(ii) Number of telephone numbers having an even number (2 or 4 or 6 or 8) at unit's place
= 22 +20 +14 + 16= 72
.:. Probability of getting a telephone numbers having an even number at unit's place = 72 =0.36
200
(iv) Number of telephone numbers having an odd digit (1 or 3 or 5 or 7 or 9) at unit's place
= 26 +22 + 10 + 28 + 20 = 106

.. Probability of getting a telephone numbers having an odd number at unit's place = 106
=0.53
200
ILUSTRATION 12: Two similar coins were tossed simultaneously 1000timesand the frequency
on each toss is as below: distribution of heads obtained
Number of heads 1 2
Frequency 200 500 300

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of the following:
Bind the probabilities
getting one head.
(i) Probability of
Probability of getting two heads.
(ii)
Probability of getting at least one head.
(i) heads.
Probability of getting less than two
(iv) two heads.
Probability of getting not more than
(v)
of cases = 1000
Solution: Total number event of getting one head =
500.
favouring the
() Total number of cases
500
head) =0.5
so, P (one 1000
300
(ii) P(two head) =0.3
1000
500+ 300 800 =0.8
=P(1 head or 2heads) = 1000
(iji) Patleast one head) 1000
200 + 500 700
=0.7
= P (0head or 1head) = 1000
(iv) P(less than two heads) 1000
200+ 500 + 300 1000 1
than two heads) = 1000
(v) P(not more 1000 data
survey in the filed about the life of these batteries. The
manufacturing car batteries made a
ILLUSTRATION13: A factory
obtained is as under : Total number of batteries
36 to 48 More than 48
Less than 24 24 to 36
Life time (in months) 200 1000
40 220 540
Frequency or the number of batteries
in acar, what is the probability that
Abattery of this company is put
more than 36 months?
() the battery will last for
than 48 months ?
(i) the battery will last for less
months?
() the battery will last for 36 to 48
number of cases made = 1000
Solution: Total frequency or the total 540 + 200 = 740
which last for more than 36 months =
() The total number of batteries
740
than 36 months) = 0,74
Now, P (battery will last for more 1000
800
which last for less than 48 months = 40 + 220 + 540 =
(11) The total number of batteries
800
=0.80
than 48 months) = 1000
So, P (battery will last for less
which last for 36 to 48 months = 540
) The total number of batteries
540
(battery will last for 36 to 48 months) = 1000 =0.54
So, P
under
14: Fify seeds were selected at random from each of Sbags A, B, C, D, E of seeds, and were kept
0STRATION of seeds which had germinated in each
to germination. After 20 days, the number
equally favourable
Standardized conditions follow:
collection were counted and recorded as
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Complete Foundation Guide for lIT-JEE_Mathematics Class IX
Bag A D
Number of seeds germinated 40 48 42 39 41
What is the probability of germination of
() more than 40 seeds in a bag?
(ii) 49 seeds in a bag?
(iii) more than 35 seeds in bag?
Solution: (i) Number of bags in which more than 40 seeds out of 50 seeds germinated = 3 (these are B, C and E)
Total number of bags =5
3
So, P (more than 40 seeds in a bag germinated) ==0.60
5

(i) Number of bags in which 49 seeds germinated = 0

0
So, P(49seeds germinated in a bag) ==0
5
(ii) Number of bags in which more than 35 seeds out 50 seeds germinated=5
Total number of bags = 5

So, P (more than 35 seeds in a bag germinated) ==1

5

ILLUSTRATION 15: An insurance company selected 2000 drivers at random in a particular city to find a relationship between
age and accidents. The data obtained are given in the following table:
Number of accidents in one year
Age group of drivers (in years) 2 3 more than 3
18-29 440 160 110 61 35
30- 50 s05 125 60 22 18
Above 50 360 45 35 15

Findthe probability of the following events for a driver selected at random from the city:
(i) being 18 29 years of age and having exactly 3 accidents in one year
(ii) being 30 - 50 years of age and having one or more accidents in a year
(iii) having no accident in one year
Solution: (i) The number of drivers in the age group 18-29 having exactly 3 accidents = 61
Total number of cases = 2000
61
So, P (driver in age group 18--29 having exactly 3 accidents in one year) =0.0305
2000
(ii) The number of drivers having in the age group 30-50 and having one or more than one accidents in one year =125 +
60 + 22+18 =225
225
P(driver inage group 30-50 having one or more accidents in one year) =: =0.1125
2000
(iii) The number of drivers having no accident in one year = 440 + 505 + 360 = 1305
1305
So, P (driver having no accident) =0,6525
2000

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ILLUSTRATION 16: 100 plants each were planted in 100 sschhools
Probability
during Van Mahotsava. After one month, the
survived was recorded as in data below:
that
number of plants
Less than 25 26-50 51 6061-70
Number ofplants survived More than 70 Total number of
schools
Number of schools (frequency)15 20 30 30 5 100
When a school is
selected at random for inspection, what is the probability that:
school?
() More than 25 plants survived in the
(i) Lessthan 61 plants survived in
the school?
school?
(it) 61 to 70plants survived in the
planted = 100
Calution: Total frequency or the total number of school in which plants were
(I) Number of schools in which more than 25 plants survived = 20+ 30 + 30 + 5 - 85
85
=0.85
P (more than 25 plants survived in the school) 100

(ii) Number of schoois in which less than 61 plants survived) = 15 + 20 +30=65

65
P(Less than 61 plants survived) = = 0.65
100
30
(ii) P(61 to 70 plants survived) 100
=0.30

CONCEPT TEST 1
1. A
coin is tossed 1000 times with the following frequencies :
Head 455, Tail 545
Find the probability for each event.
2. Two coins are tossed simuitaneously 500 times, and we get
Two heads: 105 times
One head: 275 times
No head: 120 times
Find the probability of occurrence of each of these events.
3. Refer to the following table:
Marks Number of students Cumulative frequency
0-20 7 7

20 - 30 10 17
30-40 10 27
40- 50 20 47
50 - 60 20 67
60 - 70 15 82
70 and above 8 90

Total 90

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