STATISTICS 183
8. Aclass teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent. Number of 6 - 10] 10 - 14] 14 - 20] 20 -28 | 28 - 38)38 - 40
days
Number of 11 10 7 4 4 3 1
students
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate. . Literacy rate (in %) 45-55 | 55-65 | 65 - 75 | 75 - 85 | 85 - 95
ES
13.3 Mode of Grouped Data - : 2
Recall from Class IX, a mode is that value among the observations which occurs most
often, that is, the value of the observation having the maximum frequency. Further, we
discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining
a mode of grouped data. It is possible that more than one value may have the same
maximum frequency. In such situations, the data is said to be multimodal. Though
grouped data can also be multimodal, we shall restrict ourselves to problems having a
single mode only. ae ) ;
Let us first recall how we found the mode for ungrouped data through the following
example. .


maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only. ae ) ; Let us first recall how we found the mode for ungrouped data through the following example. . Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows:
2. .6 '4 5 0 2 1 3 2 3
Find the mode of the data. Solution : Let us form the frequency distribution table of the given data as follows:
Number of 1 2 3 4 5
wickets
Number of 1 1 3 2 1 1 1
matches
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184 MATHEMATICS Clearly, 2 is the number of wickets taken by the bowler in the maximum number
(i.e., 3) of matches. So, the mode of this data is 2. In a grouped frequency distribution, it is not possible to determine the mode by
looking at the frequencies. Here, we can only locate a class with the maximum
frequency, called the modal class. The mode is a value inside the modal class, and is
given by the formula: Mode = (ae lsh
2h fo- hr
where /= lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
J, = frequency of the modal class, °
J, = frequency of the class preceding the modal class, '
f, = frequency of the class succeeding the modal class. Let us consider the following examples to illustrate the use of this formula. Example 5: A survey conducted on 20 households in a locality by a group of students
resulted in the following frequency table for the number of family members in a
household: Number of 7 2 2 1
families Find the mode of this data. Solution : Here the maximum class frequency is 8, and the class corresponding to this
frequency is 3 - 5. So, the modal class is 3 - 5.


for the number of family members in a household: Number of 7 2 2 1 families Find the mode of this data. Solution : Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 - 5. So, the modal class is 3 - 5. Now modal class = 3 -5, lower limit (/) of modal class = 3, class size (h) = 2 frequency (f,) of the modal class = 8, frequency (f,) of class preceding the modal class = 7, frequency (f,) of class succeeding the modal class = 2. Now, let us substitute these values in the formula :
Reprint 2025-26


STATISTICS 185
Mode = (ae lsh
2h ~ fi 07 f 2
= 3+ _ 8-7 x2=347=3,286
2x8-7-2 7 Therefore, the mode of the data above is 3.286. Example 6 : The marks distribution of 30 students in a mathematics examination are
given in Table 13.3 of Example |. Find the mode of this data. Also compare and
interpret the mode and the mean. Solution : Refer to Table 13.3 of Example 1. Since the maximum number of students
(i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore, the lower limit (/) of the modal class=40, the class size (h) = 15, . - _ the frequency (f,) of modal class = 7, _ ° the frequency (f,) of the class preceding the modal class = 3, the frequency (f,) of the class succeeding the modal class = 6. Now, using the formula: - _ , Cc) Mode = 1 + [she )m
a ° 2fi- fo -h t Mode = 40+ -=3- |x 15 252
we ge ode = 4-6-3 =
So, the mode marks is 52. . Now, from Example 1, you know that the mean marks is 62. So, the maximum number of students obtained 52 marks, while on an average a
student obtained 62 marks.


Mode = 40+ -=3- |x 15 252 we ge ode = 4-6-3 = So, the mode marks is 52. . Now, from Example 1, you know that the mean marks is 62. So, the maximum number of students obtained 52 marks, while on an average a student obtained 62 marks. Remarks : °
1. In Example 6, the mode is less than the mean. But for some other problems it may
be equal or more than the mean also.
2. It depends upon the demand of the situation whether we are interested in finding the
average marks obtained by the students or the average of the marks obtained by most Reprint 2025-26


186 MATHEMATICS
of the students. In the first situation, the mean is required and in the second situation,
the mode is required. Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations
assigned to the groups. Ask each group to find the mode of the data. They should also
compare this with the mean, and interpret the meaning of both. Remark : The mode can also be calculated for grouped data with unequal class sizes. However, we shall not be discussing it. EXERCISE 13.2 : 1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years) 15-25 | 25-35 | 35-45 | 45-55 | 55-65
namberafpatenss | 6 | [| a | ow | s | Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency. - . ° 2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components : . , Lifetimes (in hours) 20-40 | 40-60 | 60-80 | 80-100) 100-120
\ "
Determine the modal lifetimes of the components. 3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families.


components : . , Lifetimes (in hours) 20-40 | 40-60 | 60-80 | 80-100) 100-120 \ " Determine the modal lifetimes of the components. 3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure : Expenditure (in) Number of families
1000 - 1500 24
1500 - 2000 40
2000 - 2500 33
2500 - 3000 28
3000 - 3500 30
3500 - 4000 22
4000 - 4500 16
4500 - 5000 7
Reprint 2025-26


STATISTICS 187
4. The following distribution gives the state-wise teacher-student ratio in higher
secondary schools of India. Find the mode and mean of this data. Interpret the two
measures. Number of students per teacher Number of states /U.T.
15 - 20 3
20 - 25 8
25 - 30 9
30-35 10
35-40 3
40 - 45 0
45 - 50 0
50-55 2
5. The given distribution shows the number of runs scored by some top batsmen of the
world in one-day international cricket matches. © ° :
3000 - 4000 4
4000 - 5000 18
5000 - 6000 9
6000 - 7000 7
7000 - 8000 6
8000 - 9000 3
9000 - 10000 1
10000 - 11000 1
Find the mode of the data. 6. A student noted the number of cars passing through a spot on a road for 100
periods each of 3 minutes and summarised it in the table given below. Find the mode
of the data : 10-20] 20-30 | 30-40] 40-50 | 50-60] 60-70] 70 - 80
wos [7 t*letel[~[ofsfs
Reprint 2025-26


188 MATHEMATICS
13.4 Median of Grouped Data
As you have studied in Class IX, the median is a measure of central tendency which
gives the value of the middle-most observation in the data. Recall that for finding the
median of ungrouped data, we first arrange the data values of the observations in
n+l ; ;
ascending order. Then, if n is odd, the median is the 3 th observation. And, ifn
is even, then the median will be the average of the stn and the (3 + 7 th observations. Suppose, we have to find the median of the following data, which gives the
marks, out of 50, obtained by 100 students in a test : . :
First, we arrange the marks in ascending order and prepare a frequency table as
follows : ;
Table 13.9. Marks obtained Number of students
(Frequency)
20 6
25 20
28 24
29 28
33 15
38 4
42 2
43 1
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STATISTICS 189
Here n = 100, which is even. The median will be the average of the " th and the
2
G + i observations, i.e., the 50th and 51st observations. To find these
observations, we proceed as follows:
Table 13.10
Marks obtained Number of students
20 6 .
upto 25 6+20= 26 ;
upto 28 26 + 24=50
upto 29 50 + 28 =78
upto 33 78 +15=93
upto 38 93+4=97
upto 42 97+2=99
upto 43 99+ 1=100
Now we add another column depicting this information to the frequency table
above and name it as cumulative frequency column.
\ ) Table 13.11
Marks obtained Number of students Cumulative frequency
20 6 6
25 20 26
28 24 50
29 28 78
33 15 93
38 4 97
42 2 99
43 1 100
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190 MATHEMATICS
From the table above, we see that:
50th observaton is 28 (Why?)
51st observation is 29
28 + 29
So, Median = > = 28.5
Remark : The part of Table 13.11 consisting Column | and Column 3 is known as
Cumulative Frequency Table. The median marks 28.5 conveys the information that
about 50% students obtained marks less than 28.5 and another 50% students obtained
marks more than 28.5. Now, let us see how to obtain the median of grouped data, through the following
situation. . : . Consider a grouped frequency distribution of marks obtained, out of 100, by 53
students, in a certain examination, as follows: - " . Table 13.12 -
0-10 5
10 - 20 3
20 - 30 4
30 - 40 3
40 - 50 3
50 - 60 4
60 - 70 7
70 - 80 9
, 80 - 90 7
90 - 100 8
From the table above, try to answer the following questions:
How many students have scored marks less than 10? The answer is clearly 5. Reprint 2025-26


STATISTICS 191 How many students have scored less than 20 marks? Observe that the number
of students who have scored less than 20 include the number of students who have
scored marks from 0 - 10 as well as the number of students who have scored marks
from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, ie., 8. We say that the cumulative frequency of the class 10-20 is 8. Similarly, we can compute the cumulative frequencies of the other classes, i.e.,
the number of students with marks less than 30, less than 40, . . ., less than 100. We
give them in Table 13.13 given below: Table 13.13 :
Marks obtained Number of students
(Cumulative frequency) Less than 10 5 Less than 20 5+3=8 Less than 30 8+4=12 Less than 40 124+3=15 Less than 50 15+3=18 Less than 60 18+4=22 Less than 70 22+7=29 Less than 80 29+9=38 Less than 90 384+7=45 Less than 100 45+8=53 The distribution given above is called the cumulative frequency distribution of
the less than type. Here 10, 20, 30, .. . 100, are the upper limits of the respective
class intervals. ~ .


18+4=22 Less than 70 22+7=29 Less than 80 29+9=38 Less than 90 384+7=45 Less than 100 45+8=53 The distribution given above is called the cumulative frequency distribution of the less than type. Here 10, 20, 30, .. . 100, are the upper limits of the respective class intervals. ~ . We can similarly make the table for the number of students with scores, more
than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From
Table 13.12, we observe that all 53 students have scored marks more than or equal to
0. Since there are 5 students scoring marks in the interval 0 - 10, this means that there
are 53 - 5 = 48 students getting more than or equal to 10 marks. Continuing in the
same manner, we get the number of students scoring 20 or above as 48 - 3 = 45, 30 or
above as 45 - 4 = 41, and so on, as shown in Table 13.14. Reprint 2025-26


192 MATHEMATICS
Table 13.14
Marks obtained Number of students
(Cumulative frequency)
More than or equal to 0 53
More than or equal to 10 53-5 =48
More than or equal to 20 48 --3 =45
More than or equal to 30 45-4=41
More than or equal to 40 41-3 =38
More than or equal to 50 38 --3 =35 . More than or equal to 60 35-4=31
More than or equal to 70 31-7=24
More than or equal to 80 24-9=15
More than or equal to 90 15-7= 8
The table above is called a cumulative frequency distribution of the more
than type. Here 0, 10, 20, .. ., 90 give the lower limits of the respective class intervals. Now, to find the median of grouped data, we can make use of any of these
cumulative frequency distributions.


8 The table above is called a cumulative frequency distribution of the more than type. Here 0, 10, 20, .. ., 90 give the lower limits of the respective class intervals. Now, to find the median of grouped data, we can make use of any of these cumulative frequency distributions. Let us combine Tables 13.12 and 13.13 to get Table 13.15 given below:
so Table 13.15
Number of students (/) Cumulative frequency (cf)
0-10 5 5
10 - 20 3 8
20 - 30 4 12
30 - 40 3 15
40 - 50 3 18
50 - 60 4 22
60 - 70 7 29
70 - 80 9 38
80 - 90 7 45
90 - 100 8 53
Now in a grouped data, we may not be able to find the middle observation by
looking at the cumulative frequencies as the middle observation will be some value in
Reprint 2025-26


STATISTICS 193
a class interval. It is, therefore, necessary to find the value inside a class that divides
the whole distribution into two halves. But which class should this be?
n
To find this class, we find the cumulative frequencies of all the classes and 2°
We now locate the class whose cumulative frequency is greater than (and nearest to)
n n
> This is called the median class. In the distribution above, n = 53. So, 7 = 26.5. Now 60 - 70 is the class whose cumulative frequency 29 is greater than (and nearest
to) ~ i.e, 26.5 |
0) 5» Le., 26.5. Therefore, 60 - 70 is the median class. .


is called the median class. In the distribution above, n = 53. So, 7 = 26.5. Now 60 - 70 is the class whose cumulative frequency 29 is greater than (and nearest to) ~ i.e, 26.5 | 0) 5» Le., 26.5. Therefore, 60 - 70 is the median class. . After finding the median class, we use the following formula for calculating the
median. . a, .
* cf oe 2
Median= / + | 2-- xh, -
where != lower limit of median class, .
n= number of observations,
f°? \
cf = cumulative frequency of class preceding the median class,
\ a
f= frequency of median class,
h= class size (assuming class size to be equal).
. . n
Substituting the values 5 26.5, 1= 60, cf =22,f=7,h=10
in the formula above, we get
26.5 - 22
Median= 60 + (25-22) x10
: 60 45
= + -_
7
= 66.4
So, about half the students have scored marks less than 66.4, and the other half have
scored marks more than 66.4. Reprint 2025-26


194 MATHEMATICS
Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school
was conducted and the following data was obtained:
Height (in cm) Number of girls Less than 140 4 Less than 145 11 Less than 150 29 Less than 155 40 : Less than 160 46 , Less than 165 51 ,
Find the median height. . oe
Solution : To calculate the median height, we need to find the class intervals and their
corresponding frequencies. sO . . The given distribution being of the less than type, 140, 145, 150,..., 165 give the
upper limits of the corresponding class intervals. So, the classes should be below 140,
140 - 145, 145 - 150, ..., 160 - 165. Observe that from the given distribution, we find
that there are 4 girls with height less than 140, i.e., the frequency of class interval
below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with
height less than 140. Therefore, the number of girls with height in the interval
140 - 145 is 11 - 4 = 7. Similarly, the frequency of 145 - 150 is 29 - 11 = 18, for
150 - 155, it is 40 - 29 = 11, and so on.


height less than 140. Therefore, the number of girls with height in the interval 140 - 145 is 11 - 4 = 7. Similarly, the frequency of 145 - 150 is 29 - 11 = 18, for 150 - 155, it is 40 - 29 = 11, and so on. So, our frequency distribution table with the
given cumulative frequencies becomes: ao Table 13.16 Class intervals Cumulative frequency Below 140 4 4
140 - 145 7 11
145 - 150 18 29
150 - 155 11 40
155 - 160 6 46
160 - 165 5 51 Reprint 2025-26


STATISTICS 195
n 51 . . oa,
Now n=5l. So, 5 25.5 . This observation lies in the class 145 - 150. Then,
l (the lower limit) = 145,
cf (the cumulative frequency of the class preceding 145 - 150) = 11,
f (the frequency of the median class 145 - 150) = 18,
h (the class size) = 5.
5 - cf :
Using the formula, Median = / + Ff x h | we have
Median = 145+ (ase SoU) a .
18 - .
145 725 149.03 . . TOG TN
So, the median height of the girls is 149.03 cm. -
This means that the height of about 50% of the girls is less than this height, and
50% are taller than this height... - °
Example 8 : The median of the following data is 525. Find the values of x and y, if the
total frequency is 100. | J .
0- 100 2
100 - 200 5
200 - 300 x
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 y
700 - 800 9
800 - 900 7
900 - 1000 4
Reprint 2025-26


196 MATHEMATICS
Solution :
Class intervals Cumulative frequency
0 - 100 2 2 100 - 200 5 7 200 - 300 x 7T+x 300 - 400 12 19+x 400 - 500 17 36+x 500 - 600 20 564+x 600 - 700 y S64x4+y 700 - 800 9 654+x4+y 800 - 900 7 T2Z+x+y 900 - 1000 4 J64+x+y
It is given that n = 100 oe -
So, 76+x+y=100, ie, x+y=24 : (1)
The median is 525, which lies in the class 500 - 600
So, /=500, f=20, cf=36+x, h=100 os Cc) . . re cf
Using the formula : - Median= /+ a h, we get
oo 525 = 500 + (92362) 109
- 20
L.e., 525-500 = (14-x) x5
L.e., : 25= 70 --5x
ie, : 5x = 70-25=45
So, x=9
Therefore, from (1), we get 9+y= 24
L.e., y= 15
Reprint 2025-26


STATISTICS 197 Now, that you have studied about all the three measures of central tendency, let
us discuss which measure would be best suited for a particular requirement. The mean is the most frequently used measure of central tendency because it
takes into account all the observations, and lies between the extremes, i.c., the largest
and the smallest observations of the entire data. It also enables us to compare two or
more distributions. For example, by comparing the average (mean) results of students
of different schools of a particular examination, we can conclude which school has a
better performance. " However, extreme values in the data affect the mean. For example, the mean of
classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20,
21, 18, then the mean will certainly not reflect the way the data behaves. So, in such
cases, the mean is not a good representative of the data. _ . In problems where individual observations are not important, and we wish to find
out a 'typical' observation, the median is more appropriate, e.g., finding the typical
productivity rate of workers, average wage in a country, etc.


such cases, the mean is not a good representative of the data. _ . In problems where individual observations are not important, and we wish to find out a 'typical' observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations
where extreme values may be there. So, rather than the mean, we take the median as
a better measure of central tendency. . _ aa
In situations which require establishing the most frequent value or most popular
~~
item, the mode is the best choice, e.g., to find the most popular T. V. programme being
watched, the consumer item in greatest demand, the colour of the vehicle used by
most of the people, etc. Remarks : . .
1. There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean
2. The median of grouped data with unequal class sizes can also be calculated. However,
we shall not discuss it here. Reprint 2025-26


198 MATHEMATICS
EXERCISE 13.3
1. The following frequency distribution gives the monthly consumption of electricity of
68 consumers of a locality. Find the median, mean and mode of the data and compare
them.
65 - 85 4
85 - 105 5
105 - 125 13 |
125 - 145 20
145 - 165 14
165 - 185 8
185 - 205 4
2. Ifthe median of the distribution given below is 28.5, find the values of x and y.
0- 10 5
10- 20 x
20- 30 20
30 - 40 15
40 - 50 y
. 50- 60 5
ee
3. A life insurance agent found the following data for distribution of ages of 100 policy
holders. Calculate the median age, if policies are given only to persons having age 18
years onwards but less than 60 year. Reprint 2025-26


STATISTICS
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89 ,
Below 50 92
Below 55 98
Below 60 100
4, The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and
the data obtained is represented in the following table :
118 - 126 3
127-135 5
136 - 144 9
145 - 153 12
154 - 162 5
163-171 4
. 172-180 2
Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median,
since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, ..., 171.5 - 180.5.)
Reprint 2025-26


200 MATHEMATICS 5. The following table gives the distribution of the life time of 400 neon lamps : Life time (in hours) Number of lamps 1500 - 2000 14 2000 - 2500 56 2500 - 3000 60 3000 - 3500 86
3500 - 4000 74 4000 - 4500 62
4500 - 5000 48 Find the median life time of a lamp. . ", , 6. 100 surnames were randomly picked up from a local telephone directory and the
frequency distribution of the number of letters in the English alphabets in the surnames
was obtained as follows: se : Number of letters 10-13 | 13-16 | 16-19
female |» |= fw ls]. Determine the median number of letters in the surnames. Find the mean number of
letters in the surnames? Also, find the modal size of the surnames. ~~ ae . 7. The distribution below gives the weights of 30 students of a class. Find the median
weight of the students. Weight (in kg) 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 woman] 2 [> [ele lols le 13.5 Summary
In this chapter, you have studied the following points:
1.


gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg) 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 woman] 2 [> [ele lols le 13.5 Summary In this chapter, you have studied the following points: 1. The mean for grouped data can be found by :
Dfex,
(i) the direct method: x = Xi
Yi
= Lid;
(ii) the assumed mean method : x = a + Sp
Reprint 2025-26


STATISTICS 201
Zhu,
(iii) the step deviation method: X =a+ (2) xh,
with the assumption that the frequency of a class is centred at its mid-point, called its
class mark.
2. The mode for grouped data can be found by using the formula:
Mode = / (fhe
2h -fo- hr
where symbols have their usual meanings.
3. The cumulative frequency of a class is the frequency obtained by adding the frequencies
of all the classes preceding the given class. .
4. The median for grouped data is formed by using the formula: .
7 _ of Be
Median = 7 + 2 xh, _
'| Ff
where symbols have their usual meanings.
. A NOTETO THE READER
For calculating mode and median for grouped data, it should be
ensured that the class intervals are continuous before applying the
formulae. Same condition also apply for construction of an ogive. Further, in case of ogives, the scale may not be the same on both the axes. Reprint 2025-26


at
Se
eat
ant be
oll
[] ape a
1062CH15
The theory of probabilities and the theory of errors now constitute
a formidable body of great mathematical interest and of great
practical importance. - , °
Bo ~ -R.S. Woodward
14.1 Probability - A Theoretical Approach
Let us consider the following situation: - . Suppose a coin is tossed at random,
When we speak of a coin, we assume it to be 'fair', that is, it is symmetrical so
that there is no reason for it to come down more often on one side than the other. We call this property of the coin as being 'unbiased'. By the phrase 'random toss',
we mean that the coin is allowed to fall freely without any bias or interference. We know, in advance, that the coin can only land in one of two possible ways -
either head up or tail up (we dismiss the possibility of its 'landing' on its edge, which
may be possible, for example, if it falls on sand). We can reasonably assume that each
outcome, head or tail, is as likely to occur as the other. We refer to this by saying that
the outcomes head and tail, are equally likely. For another example of equally likely outcomes, suppose we throw a die
once.


it falls on sand). We can reasonably assume that each outcome, head or tail, is as likely to occur as the other. We refer to this by saying that the outcomes head and tail, are equally likely. For another example of equally likely outcomes, suppose we throw a die once. For us, a die will always mean a fair die. What are the possible outcomes? They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So
the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6. Reprint 2025-26


PROBABILITY 203 Are the outcomes of every experiment equally likely? Let us see. Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball
without looking into the bag. What are the outcomes? Are the outcomes - a red ball
and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you
would agree that you are more likely to get a red ball than a blue ball. So, the outcomes
(a red ball or a blue ball) are not equally likely. However, the outcome of drawing a
ball of any colour from the bag is equally likely. So, all experiments do not necessarily
have equally likely outcomes. However, in this chapter, from now on, we will assume that all the experiments
have equally likely outcomes. In Class IX, we defined the experimental or empirical probability P(E) of an
event E as ; : P(E) Number of trials in which the event happened
7 Total number of trials = - The empirical interpretation of probability can be applied to every event associated
with an experiment which can be repeated a large number of times. The requirement
of repeating an experiment has some limitations, as it may be very expensive or
unfeasible in many situations.


Total number of trials = - The empirical interpretation of probability can be applied to every event associated with an experiment which can be repeated a large number of times. The requirement of repeating an experiment has some limitations, as it may be very expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die throwing
experiments. But how about repeating the experiment of launching a satellite in order
to compute the empirical probability of its failure during launching, or the repetition of
the phenomenon of an earthquake to compute the empirical probability of a multistoreyed building getting destroyed in an earthquake? % wo . In experiments where we are prepared to make certain assumptions, the repetition
of an experiment can be avoided, as the assumptions help in directly calculating the
exact (theoretical) probability. The assumption of equally likely outcomes (which is
valid in many experiments, as in the two examples above, of a coin and of a die) is one
such assumption that leads us to the following definition of probability of an event. The theoretical probability (also called classical probability) of an event E,
written as P(E), is defined as


valid in many experiments, as in the two examples above, of a coin and of a die) is one such assumption that leads us to the following definition of probability of an event. The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as P(E) Number of outcomes favourable to E
Se on - 0000000000
Number of all possible outcomes of the experiment
where we assume that the outcomes of the experiment are equally likely. We will briefly refer to theoretical probability as probability. This definition of probability was given by Pierre Simon Laplace in 1795. Reprint 2025-26


204 MATHEMATICS Probability theory had its origin in the 16th century when an Italian physician and mathematician J.Cardan wrote the first book on the subject, The Book on Games of Chance. 2 prt ! Since its inception, the study of probability has attracted 2 'f 7 the attention of great mathematicians. James Bernoulli ee (1654 - 1705), A. de Moivre (1667 - 1754), and . { , | Pierre Simon Laplace are among those who made significant y 1, y contributions to this field. Laplace's Theorie Analytique clef Me des Probabilités, 1812, is considered to be the greatest ee contribution by a single person to the theory of probability. Pierre Simon Laplace In recent years, probability has been used extensively in (1749 - 1827) many areas such as biology, economics, genetics, physics, os sociology etc. . ° Let us find the probability for some of the events associated with experiments
where the equally likely assumption holds. oe -
Example 1 : Find the probability of getting a head when a coin is tossed once. Also
find the probability of getting a tail. . . Solution : In the experiment of tossing a coin once, the number of possible outcomes
is two - Head (H) and Tail (T). Let E be the event 'getting a head'.


getting a head when a coin is tossed once. Also find the probability of getting a tail. . . Solution : In the experiment of tossing a coin once, the number of possible outcomes is two - Head (H) and Tail (T). Let E be the event 'getting a head'. The number of
outcomes favourable to E, (.e., of getting a head) is 1. Therefore,
fo :
P(E) =P th ; _ Number of outcomes favourable toE 1
- a Number of all possible outcomes 2
Similarly, if Fis the event 'getting a tail', then
1
- P(F) = P(tail) = > (Why ?) Example 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls being
of the same size. Kritika takes out a ball from the bag without looking into it. What is
the probability that she takes out the (i) yellow ball? (ii) red ball? (iti) blue ball? Solution : Kritika takes out a ball from the bag without looking into it. So, it is equally
likely that she takes out any one of them. Reprint 2025-26


PROBABILITY 205 Let Y be the event 'the ball taken out is yellow', B be the event 'the ball taken
out is blue', and R be the event 'the ball taken out is red'. Now, the number of possible outcomes = 3,
(i) The number of outcomes favourable to the event Y = 1. Ss PCY) * 0, =3
a . 1 - 1 Similarly, (ii) PCR) = 3 and (iii) P(B) = 3
Remarks :
1. An event having only one outcome of the experiment is called an elementary
event. In Example 1, both the events E and F are elementary events. Similarly, in
Example 2, all the three events, Y, B and R are elementary events. |
2. In Example 1, we note that : P(E) + P(F) = 1 a, . In Example 2, we note that : P(Y) + PCR) + P(B) = 1 . . Observe that the sum of the probabilities of all the elementary events of
an experiment is |. This is true in general also. -
Example 3 : Suppose we throw a die once. (i) What is the probability of getting a
number greater than 4 ?


= 1 . . Observe that the sum of the probabilities of all the elementary events of an experiment is |. This is true in general also. - Example 3 : Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or
equal to 4? °
Solution : (i) Here, let E be the event 'getting a number greater than 4'. The number
of possible outcomes is six : 1,2, 3, 4, 5 and 6, and the outcomes favourable to E are 5
and 6, Therefore, the number of outcomes favourable to E is 2. So, 2 1
. P(E) = P(number greater than 4) = 673
(ii) Let F be the event ' getting a number less than or equal to 4'. Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4. So, the number of outcomes favourable to F is 4. Therefe P(F 42
erefore, = 6-3
Reprint 2025-26


206 MATHEMATICS
Are the events E and F in the example above elementary events? No, they are
not because the event E has 2 outcomes and the event F has 4 outcomes. Remarks : From Example 1, we note that
P(E) + P(F . + . 1 (1)
+ = 7tT=
(E) +P) = 5 +5
where E is the event 'getting a head' and F is the event 'getting a tail'. From (i) and (ii) of Example 3, we also get
1 2 :
P(E) + P(F) = ~ +> =1 (2)
3 3
where E is the event 'getting a number >4' and F is the event 'getting a number < 4'. Note that getting a number not greater than 4 is same as getting a number less
than or equal to 4, and vice versa. . a. In (1) and (2) above, is F not the same as 'not E'? Yes, it is. We denote the event
'not E' by E. oe . . So, P(E) + P(not E) = 1 :
1.¢., P(E) + P(E) = 1, which givesus P(E) =1-- P(E). In general, it is true that for an event E,
PE)=1-PE)
_ | ']
The event E, representing 'not E', is called the complement of the event E.


oe . . So, P(E) + P(not E) = 1 : 1.¢., P(E) + P(E) = 1, which givesus P(E) =1-- P(E). In general, it is true that for an event E, PE)=1-PE) _ | '] The event E, representing 'not E', is called the complement of the event E. We also say that E and E are complementary events. Before proceeding further, let us try to find the answers to the following questions:
(i) What is the probability of getting a number 8 in a single throw of a die? (ti) What is the probability of getting a number less than 7 in a single throw of a die? Let us answer (i) : ,
We know that there are only six possible outcomes in a single throw of a die. These
outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no
outcome favourable to 8, i.e., the number of such outcomes is zero. In other words,
getting 8 in a single throw of a die, is impossible.
: 0
So, P(getting 8) = 67 0
Reprint 2025-26


PROBABILITY 207 That is, the probability of an event which is impossible to occur is 0. Such an
event is called an impossible event. Let us answer (ii) : Since every face of a die is marked with a number less than 7, it is sure that we
will always get a number less than 7 when it is thrown once. So, the number of
favourable outcomes is the same as the number of all possible outcomes, which is 6. 6 Therefore, P(E) = P(getting a number less than 7) = 6 =1
So, the probability of an event which is sure (or certain) to occur is 1. Such an event
is called a sure event or a certain event. ,
Note : From the definition of the probability P(E), we see that the numerator (number
of outcomes favourable to the event E) is always less than or equal to the denominator
(the number of all possible outcomes). Therefore, oe OSP(E)S1 - Now, let us take an example related to playing cards. Have you seen a deck of
playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards eachspades (4), hearts (¥), diamonds (#) and clubs (#). Clubs and spades are of black
colour, while hearts and diamonds are of red colour.


related to playing cards. Have you seen a deck of playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards eachspades (4), hearts (¥), diamonds (#) and clubs (#). Clubs and spades are of black colour, while hearts and diamonds are of red colour. The cards in each suit are ace,
king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face
cards. " oN
Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Calculate the
probability that the card will _ . (i) be an ace,
(ii) not be an ace. Solution : Well-shuffling ensures equally likely outcomes. (i) There are 4 aces in a deck. Let E be the event 'the card is an ace'. The number of outcomes favourable to E = 4 The number of possible outcomes = 52 (Why ?) Therek pe = 4-4 erefore, (E) = 52 13
(ii) Let F be the event 'card drawn is not an ace'. The number of outcomes favourable to the event F = 52-4 =48 (Why?)
Reprint 2025-26


208 MATHEMATICS
The number of possible outcomes = 52
Theref P(F 8
erefore, (F) = 52. 13
Remark : Note that F is nothing but E. Therefore, we can also calculate P(F) as
follows: P(F) = P(E) = 1 -- P(E) = 1 1."
ollows: P(F) = P(E) = 1- P(E) = BB
Example 5 : Two players, Sangeeta and Reshma, play a tennis match. It is known
that the probability of Sangeeta winning the match is 0.62. What is the probability of
Reshma winning the match? Solution ;: Let S and R denote the events that Sangeeta wins the match and Reshma
wins the match, respectively. , OO
The probability of Sangeeta's winning = P(S) = 0.62 (given) . The probability of Reshma's winning = P(R) = 1 - P(S) -
~ [As the events R and S are complementary]
= 1-0.62 = 0.38
Example 6 : Savita and Hamida are friends, What is the probability that both will
have (i) different birthdays? (Gi) the same birthday? (ignoring a leap year).
pos
Solution ; Out of the two friends, one girl, say, Savita's birthday can be any day of the
year. Now, Hamida's birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely.


the same birthday? (ignoring a leap year). pos Solution ; Out of the two friends, one girl, say, Savita's birthday can be any day of the year. Now, Hamida's birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely. Gi) If Hamida's birthday is different from Savita's, the number of favourable outcomes
for her birthday is 365 - 1 = 364
tea -- 364
So, P (Hamida's birthday is different from Savita's birthday) = 365
(ii) P(Savita and Hamida have the same birthday)
: = 1--P (both have different birthdays)
: 364 _
= |--- [Using P(E) = 1 - P(E)]
365 g P(E)
I
365
Reprint 2025-26


PROBABILITY 209
Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15
are boys. The class teacher has to select one student as a class representative. She
writes the name of each student on a separate card, the cards being identical. Then
she puts cards in a bag and stirs them thoroughly. She then draws one card from the
bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy? Solution : There are 40 students, and only one name card has to be chosen.


a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy? Solution : There are 40 students, and only one name card has to be chosen. Gi) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25 (Why?)
; ; ; a)
Therefore, P (card with name of a girl) = P(Girl) = 40 = 8
(i) The number of outcomes favourable for a card with the name of a boy = 15 (Why?)
15.3 |
Therefore, P(card with name of a boy) = P(Boy) = 40 = 8
Note ; We can also determine P(Boy), by taking : "
. : 5 3
P(Boy) = 1 - P(not Boy) = 1 - P(Girl) = 1- 8 = 8
Example 8: A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn
at random from the box, what is the probability that it will be
& * \
(i) white? (ii) blue? (iii) red? Se" .


1 - P(Girl) = 1- 8 = 8 Example 8: A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be & * \ (i) white? (ii) blue? (iii) red? Se" . Solution : Saying that a marble is drawn at random is a short way of saying that all the
marbles are equally likely to be drawn. Therefore, the
number of possible outcomes = 3 +2+4=9 (Why?)
Let W denote the event 'the marble is white', B denote the event 'the marble is blue'
and R denote the event 'marble is red'. (i) The number of outcomes favourable to the event W = 2
Ss ° PCW) 2
0, / = 9
a . 31 - 4
Similarly, Gi) P(B)= - ==> and (iii) P(R)= -
9 3 9
Note that P(W) + P(B) + P(R) = 1. Reprint 2025-26


210 MATHEMATICS
Example 9 : Harpreet tosses two different coins simultaneously (say, one is of 71
and other of ~ 2), What is the probability that she gets at least one head? Solution : We write H for 'head' and T for 'tail'. When two coins are tossed
simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all
equally likely. Here (H, H) means head up on the first coin (say on ® 1) and head up
on the second coin (¥ 2), Similarly (H, T) means head up on the first coin and tail up on
the second coin and so on. The outcomes favourable to the event E, 'at least one head' are (H, H), (H, T)
and (T, H). (Why?) °
So, the number of outcomes favourable to E is 3. 3 a
Therefore, P(E) = 4 .
. 3 ;
i.e., the probability that Harpreet gets at least one head is 4°. Note : You can also find P(E) as follows: _ -
_ . =. 1
P(E)= 1-P()=1- I = 3 (sinc P(E) = Pino head) = i)
4 4 | 4 Did you observe that in all the examples discussed so far, the number of possible
outcomes in each experiment was finite? If not, check it now.


as follows: _ - _ . =. 1 P(E)= 1-P()=1- I = 3 (sinc P(E) = Pino head) = i) 4 4 | 4 Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite? If not, check it now. There are many experiments in which the outcome is any number between two
given numbers, or in which the outcome is every point within a circle or rectangle, etc. Can you now count the number of all possible outcomes? As you know, this is not
possible since there are infinitely many numbers between two given numbers, or there
are infinitely many points within a circle. So, the definition of (theoretical) probability
which you have learnt so far cannot be applied in the present form. What is the way
out? To answer this, let us consider the following example : Example 10* : In a musical chair game, the person playing the music has been
advised to stop playing the music at any time within 2 minutes after she starts
playing. What is the probability that the music will stop within the first half-minute
after starting? Solution : Here the possible outcomes are all the numbers between 0 and 2.


playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting? Solution : Here the possible outcomes are all the numbers between 0 and 2. This is
the portion of the number line from 0 to 2 (see Fig. 14.1).
tt
0 4 1 2
Fig. 14.1
* Not from the examination point of view. Reprint 2025-26


PROBABILITY 211
Let E be the event that "the music is stopped within the first half-minute'. The outcomes favourable to E are points on the number line from 0 to 5 .
1 1
The distance from 0 to 2 is 2, while the distance from 0 to 3 is 2°
Since all the outcomes are equally likely, we can argue that, of the total distance
1
of 2, the distance favourable to the event E is 2°
1
So, P(E) = Distance favourable to the event E _ 2_ lo. ;
Total distance in which outcomes canlie 2 4
Can we now extend the idea of Example 10 for finding the probability as the ratio of
the favourable area to the total area? . . ,
Example 11* : A missing helicopter is reported to have crashed somewhere in the
rectangular region shown in Fig. 14.2.


canlie 2 4 Can we now extend the idea of Example 10 for finding the probability as the ratio of the favourable area to the total area? . . , Example 11* : A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the
lake shown in the figure? .
-__----- 6km ----->
i a a ee
& we 2 " | cya Ne ee Nee See =
Ww ° e ° Coreen =
+ a 2S, rt ye aN t
LMS AD NES
ET Nas Nhe NUN ES &
ite OL I
MEL ies SMS OLN: IN . |
A. 9g .- >
: Fig. 14.2
Solution : The helicopter is equally likely to crash anywhere in the region. Area of the entire region where the helicopter can crash
= (4.5 x 9) km? = 40.5 km?
* Not from the examination point of view. Reprint 2025-26


212 MATHEMATICS
Area of the lake = (2.5 x 3) km? = 7.5 km? Therefore, P (helicopt hed in the lake) 1 2B 3
erefore, elicopter crashed in the lake) = 70> = 705 = 53
Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor
defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which
are good, but Sujatha, another trader, will only reject the shirts which have major
defects. One shirt is drawn at random from the carton. What is the probability that
(i) itis acceptable to Jimmy? (ii) it is acceptable to Sujatha? "
Solution : One shirt is drawn at random from the carton of 100 shirts. Therefore,
there are 100 equally likely outcomes. _ _ -
(i) The number of outcomes favourable (ie, acceptable) to Jimmy = 88 (Why?)
. '. 88
Therefore, P (shirt is acceptable to Jimmy) = 00 > 0.88
(i) The number of outcomes favourable to Sujatha =88+8=96 (Why?)
fo . 96
So, P (shirt is acceptable to Sujatha) = --~ = 0.96
ae ~ 100
Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write
down all the possible outcomes.


0.88 (i) The number of outcomes favourable to Sujatha =88+8=96 (Why?) fo . 96 So, P (shirt is acceptable to Sujatha) = --~ = 0.96 ae ~ 100 Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers
appearing on the top of the dice is
(i) 8? - Gi) 13? , (iii) less than or equal to 12? Solution : When the blue die shows '1', the grey die could show any one of the
numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows '2', '3', '4', 'S' or
'6'. The possible outcomes of the experiment are listed in the table below; the first
number in each ordered pair is the number appearing on the blue die and the second
number is that on the grey die. Reprint 2025-26


PROBABILITY 213
1 2 3 4 5 6
1 G,I) (4,2) (13) (4) (5) (1,6
<4> 2 2,1) @2) (@,3) @4) (2D (@,6
3 B,D G2) G3) G.4-"G3,5), 6)
4 (4,1) (4,2) (4,374,4)-4,5) (4,6)
6 (6,1) ((6,2)6,3) (6,4) (6,5) (6,6)
Fig. 14.3 a
Note that the pair (1, 4) is different from (4, 1). (Why?) _
So, the number of possible outcomes = 6 x 6 = 36, °
(i) The outcomes favourable to the event 'the sum of the two numbers is 8' denoted
by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see Fig. 14.3)
i.e., the number of outcomes favourable to E = 5. H C) P(E) = -
ence, Ke (E) =. 36
(ii) As you can see from Fig. 14.3, there is no outcome favourable to the event F,
'the sum of two numbers is 13',
" 0
8 , P(F) = -=0
So, : (P= 36
(iii) As you can see from Fig. 14.3, all the outcomes are favourable to the event G,
'sum of two numbers < 12',
S . P(G) = 26. 1
0, ( ) ~ 36
Reprint 2025-26


214 MATHEMATICS
EXERCISE 14.1
1. Complete the following statements:
(i) Probability of an event E + Probability of the event 'notE' =. (i) The probability of an event that cannot happen is ________ . Such an event is
called. (iii) The probability of an event that is certain to happen is ___ . Such an event
is called . (iv) The sum of the probabilities of all the elementary events of an experiment is
(v) The probability of an event is greater than or equal to _____ and less than or
equalto__ .
2. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. ~
(i) A player attempts to shoot a basketball. She/he shoots or misses the shot. (ui) A trial is made to answer a true-false question. The answer is right or wrong. (iv) A baby is born. Itis a boy ora girl. _ 3. Why is tossing a coin considered to be a fair way of deciding which team should get the
ball at the beginning of a football game? 4. Which of the following cannot be the probability of an event? » LN |
(A) 3 (B) -1.5 (C) 15% (D) 0.7 5.


Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? 4. Which of the following cannot be the probability of an event? » LN | (A) 3 (B) -1.5 (C) 15% (D) 0.7 5. If P(E) = 0.05, what is the probability of 'not E'? 6. A bag contains lemon flavoured candies only. Malini takes out one candy without
looking into the bag. What is the probability that she takes out (ij) an orange flavoured candy? (i) a lemon flavoured candy? 7. It is given that in a group of 3 students, the probability of 2 students not having the
same birthday is 0.992. What is the probability that the 2 students have the same
birthday? 8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (1) red? (ii) not red? 9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken
out of the box at random. What is the probability that the marble taken out will be
(i)red? (ii) white? (iii) not green? Reprint 2025-26


PROBABILITY 215
10. A piggy bank contains hundred 50p coins, fifty = 1 coins, twenty % 2 coins and ten & 5
coins. If it is equally likely that one of the coins will fall out when the bank is turned
upside down, what is the probability that the coin (i) will be a50 p coin ? (ii) will not be a5 coin? 11. Gopi buys a fish from a shop for his aquarium. The
shopkeeper takes out one fish at random from a ra
tank containing 5 male fish and 8 female fish (see (Nap, 85
Fig. 14.4). What is the probability that the fish taken i was tisamalefish? ° (Ves aecOy out is a male fish? ys 3 12. A game of chance consists of spinning an arrow
which comes to rest pointing at one of the numbers Fig. 14.4
1,2, 3,4, 5, 6, 7, 8 (see Fig. 14.5 ), and these are equally .
likely outcomes. What is the probability that it will °
Oe , fi yr) @) 8? (ii) an odd number? , ° (6 / \ 3
(ui) a number greater than 2? 5 : : . + 9 * ° (iv) anumber less than 9? . . Fig. 14.5 13. A die is thrown once.


probability that it will ° Oe , fi yr) @) 8? (ii) an odd number? , ° (6 / \ 3 (ui) a number greater than 2? 5 : : . + 9 * ° (iv) anumber less than 9? . . Fig. 14.5 13. A die is thrown once. Find the probability of getting
(i) a prime number; (ii) anumber lying between 2 and 6; (iii) an odd number. 14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) aking of red colour - (ii) a face card (iit) a red face card
(iv) the jack of hearts \) (v) a spade (vi) the queen of diamonds 15. Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their
face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (i) If the queen is drawn and put aside, what is the probability that the second card
picked up is (a) an ace? (b) a queen? 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just
look at a pen and tell whether or not it is defective. One pen is taken out at random from
this lot.


second card picked up is (a) an ace? (b) a queen? 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one. 17. @ A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (tt) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb
is drawn at random from the rest. What is the probability that this bulb is not
defective ? 18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random
from the box, find the probability that it bears (i) a two-digit number (ii) a perfect
square number (iii) anumber divisible by 5. Reprint 2025-26


216 MATHEMATICS 19. Achild has a die whose six faces show the letters as given below: [>|
The die is thrown once. What is the probability of getting (ij)A? (i) D?
20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is
the probability that it will land inside the circle with diameter 1m?
3m
OO] "
Fig. 14.6 - OO 21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri
will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one
pen at random and gives it to her. What is the probability that . (@) She will buy it ? , : . (i) She will not buy it? .
22. Refer to Example 13. (i) Complete the following table:
Event:
'Sum on2dice' | 2 3 4 5 7 10} | 2
Probabili 4+ = 1
robability 36 36 36
(i) A student argues that 'there are 11 possible outcomes 2, 3, 4,5, 6, 7, 8,9, 10, 11 and
os ]
12. Therefore, each of them has a probability Th Do you agree with this argument? Justify your answer. 23.


| 2 Probabili 4+ = 1 robability 36 36 36 (i) A student argues that 'there are 11 possible outcomes 2, 3, 4,5, 6, 7, 8,9, 10, 11 and os ] 12. Therefore, each of them has a probability Th Do you agree with this argument? Justify your answer. 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses
otherwise. Calculate the probability that Hanif will lose the game. 24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? Gi) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the
same experiment]
* Not from the examination point of view. Reprint 2025-26


PROBABILITY 217
25. Which of the following arguments are correct and which are not correct? Give reasons
for your answer. Gi) If two coins are tossed simultaneously there are three possible outcomes-two
heads, two tails or one of each. Therefore, for each of these outcomes, the
1
probability is 3
(i) Ifa die is thrown, there are two possible outcomes-an odd number or an even
1
number. Therefore, the probability of getting an odd number is 5°
14.2 Summary
In this chapter, you have studied the following points :
1. The theoretical (classical) probability of an event E, written as P(E), is defined as
PE) Number of outcomes favourable to E
~ Number of all possible outcomes of the experiment
where we assume that the outcomes of the experiment are equally likely.
2. The probability of a sure event (or certain event) is l.
3. The probability of an impossible eventis 0. -
4, The probability of an event E is a number P(E) such that
fo O<PE)<1 ©
5. An event having only one outcome is called an elementary event. The sum of the
probabilities of all the elementary events of an experiment is 1.
6. For any event E, P (E) + P(B) = 1, where £ stands for 'not E'.


P(E) such that fo O<PE)<1 © 5. An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1. 6. For any event E, P (E) + P(B) = 1, where £ stands for 'not E'. E and £ are called
complementary events. ,
_ A NOTE TO THE READER
The experimental or empirical probability of an event is based on
what has actually happened while the theoretical probability of the
event attempts to predict what will happen on the basis of certain
assumptions. As the number of trials in an experiment, go on
increasing we may expect the experimental and theoretical
probabilities to be nearly the same. Reprint 2025-26