mathamatics xam idea

4. The class mark of the class 15.5 – 20.5 is

(a) 15.5 (b) 20.5 (c) 18 (d) 5

5. In the formula, Mode = l + f 2f1 f1 – fo f2 p× h, f2 is
– f0 –

(a) frequency of the modal class

(b) frequency of the second class

(c) frequency of the class preceding the modal class

(d) frequency of the class succeeding the modal class

6. Consider the following distribution: [NCERT Exemplar]

Marks obtained Number of students

Less than 10 5
Less than 20 12
Less than 30 22
Less than 40 29
Less than 50 38
Less than 60 47

The frequency of the class 50 – 60 is

(a) 9 (b) 10 (c) 38 (d) 47

7. For the following distribution: [NCERT Exemplar]

Class 0–8 8–16 16–24 24–32 32–40
Frequency 12 26 10 9 15

The sum of upper limits of the median class and modal class is

(a) 24 (b) 40 (c) 32 (d) 16

8. The runs scored by a batsman in 35 different matches are given below: [NCERT Exemplar]

Runs scored 0–15 15–30 30–45 45–60 60–75 75–90
Frequency 5 7 4 8 8 3

The number of matches in which the batsman scored less than 60 runs are

(a) 16 (b) 24 (c) 8 (d) 19

9. Consider the following frequency distribution of the height of 60 students of a class:
[NCERT Exemplar]

Height (in cm) Number of students

150 – 155 15
155 – 160 13
160 – 165 10
165 – 170 8
170 – 175 9
175 – 180 5

The sum of the lower limit of the modal class and upper limit of the median class is

(a) 310 (b) 315 (c) 320 (d) 330

Statistics 393

10. The class marks of the class 18–22 is

(a) 4 (b) 18 (c) 22 (d) 20

11. / fi = 18, / fi xi = 2p + 24 and mean of any distribution is 2, then p is equal to

(a) 3 (b) 4 (c) 8 (d) 6

12. In the formula x = a + hf / fi ui p, for finding the mean of a grouped frequency distribution,
ui = / fi

(a) xi + a (b) h(xi – a) (c) xi – a (d) a – xi
h h h

13. The abscissa of the point of intersection of the less than type and of the more than type

cummulative frequency curves of a grouped data gives its

(a) mean (b) median (c) mode (d) all of these

14. The arithmetic mean of 1, 2, 3 ...... n is

(a) n+1 (b) n (c) n –1 (d) n +1
2 2 2 2

15. For the following distribution [NCERT Exemplar]

Class 0–5 5–10 10–15 15–20 20–25
Frequency
10 15 12 20 9

The sum of lower limits of the median class and modal class is

(a) 15 (b) 25 (c) 30 (d) 35

16. Consider the following frequency distribution : [NCERT Exemplar]

Class 0–5 6–11 12–17 18–23 24–29
Frequency
13 10 15 8 11

The upper limit of the median class is

(a) 17 (b) 17.5 (c) 18 (d) 18.5

17. For the following distribution : [NCERT Exemplar]

Marks Number of students

Below 10 3
Below 20 12
Below 30 27
Below 40 57
Below 50 75
Below 60 80

The modal class is (b) 20 – 30 (c) 30 – 40 (d) 50 – 60
(a) 10 – 20

18. Consider the data: [NCERT Exemplar]

Class 65–85 85–105 105–125 125–145 145–165 165–185 185–205

Frequency 4 5 13 20 14 7 4

The difference of the upper limit of the median class and the lower limit of the modal class is

(a) 0 (b) 19 (c) 20 (d) 38

394 Xam idea Mathematics–X

19. The time in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:
[NCERT Exemplar]

Class 13.8 – 14 14 – 14.2 14.2 – 14.4 14.4 – 14.6 14.6 – 14.8 14.8 – 15
20
Frequency 2 4 5 71 48

The number of atheletes who completed the race in less then 14.6 seconds is:

(a) 11 (b) 71 (c) 82 (d) 130

20. Consider the following distribution : [NCERT Exemplar]

Marks obtained Number of students
More than or equal to 0 63
More than or equal to 10 58
More than or equal to 20 55
More than or equal to 30 51
More than or equal to 40 48
More than or equal to 50 42

The frequency of the class 30 – 40 is (d) 51

(a) 3 (b) 4 (c) 48

Answers

1. (c) 2. (d) 3. (b) 4. (c) 5. (d) 6. (a)
7. (c) 8. (b) 9. (b) 10. (d) 11. (d) 12. (c)
13. (b) 14. (a) 15. (b) 16. (b) 17. (c) 18. (c)

19. (c) 20. (a)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. Average of a data is called _____________ .
2. _____________ is mid value of class interval.
3. _____________ is graphical representation of cumulative frequency distribution.
4. _____________ is the value of the observation having the maximum frequency.
5. Mode = 3 _____________ – 2 _____________ .
6. In ungrouped data the data must be arranged in proper manner to find _____________ .
7. The median of grouped data obtained graphically as the _____________ coordinate of the point

of intersection of more and less than ogive.
8. If n is even then the median of ungrouped data is _____________ observation.
9. In step deviation method to find the mean ui is _____________ .
10. In less than ogive we take cumulative frequency and corresponding _____________ of the class

intervals.

Answers

1. mean 2. Class mark 3. Ogive 4. Mode 5. Median, Mean

6. median 7. x 8. 1  n th +  n + 1th  9. xi –A
10. upper limit 2 2   2  h


Statistics 395

Very Short Answer Questions [1 mark]

1. Find the class mark of the class 10 – 25.

Sol. Class mark = Upper limit + Lower limit = 10 + 25 = 35 =17.5
2 2 2

2. Find the mean of the first five natural numbers.

Sol. Mean = x1 + x2 + x3 + x4 + x5 = 1+ 2 + 3 + 4+5 = 15 =3
5 5 5

3. A data has 13 observations arranged in descending order. Which observation represents the
median of data?

Sol. Total no. of observations = 13, which is odd.

\ The median will be c n +1 th term = c 13 +1 th = c 14 th
2 2 2
m m m = 7th

i.e., 7th term will be the median.

4. If the mode of a distribution is 8 and its mean is also 8, then find median.

Sol. Mode = 8; Mean = 8; Median = ?

Relation among mean, median and mode is 3 median = mode + 2 mean

3 × median = 8 + 2 × 8.

Median = 8 +16 = 24 =8
3 3

5. In an arranged series of an even number of 2n terms which term is median?
Sol. No. of terms = 2n which are even.

;a n th +a n +1 th
2 2
k kE

\ The median term will be 2

<c 2n th + c 2n th nth + (n +1) th
2 2 2
m +1m F

= 2 = = G [Put n = 2n]

i.e., the mean of nth and (n + 1)th term will be the median.

6. What does the abscissa of the point of intersection of the less than type and of the more than

type cumulative frequency curves of a grouped data represent?
Sol. The abscissa of the point of intersection of the less than type and of the more than type cumulative

frequency curves of a grouped data gives its median.
7. Name the graphical representation from which the mode of a frequency distribution is

obtained.
Sol. The mode of frequency distribution is determined graphically from Histogram.

396 Xam idea Mathematics–X

8. A student draws a cumulative frequency curve for the marks obtained by 60 students of a class
as shown in fig 14.2. Find the median marks obtained by the students of the class.

Fig. 14.2

Sol. Here n = 60 \ n = 30
2

Corresponding to 30 on y-axis, the marks on x-axis is 40.

\ Median marks = 40.

9. Write the modal class for the following frequency distribution:

Class Interval 10–20 20–30 30–40 40–50 50–60 60–70
19 48
Frequency 33 38 65 52
[2 marks]
Sol. Maximum frequency, i.e., 65 corresponds to the class 30 – 40
\ Modal class is 30 – 40.

Short Answer Questions-I

1. If xi's are the mid-points of the class intervals of a grouped data. fi’s are the corresponding

frequencies and x is the mean, then find / fi ^xi – xrh .
/ fi xi
Sol. We know mean (xr) = / fi

\ / fi xi = xr / fi     ...(i)

Now the value of / fi ^xi – xrh= / fi xi – / fi xr

   = / fi xr – / fi xr = 0 . (Using (i))

2. Consider the following frequency distribution.

Class 0–5 6–11 12–17 18–23 24–29
Frequency 13 10 15 8 11

Find the upper limit of median class.

Statistics 397

Sol. Classes are not continuous, hence make them continuous by adding 0.5 to the upper limits and
subtracting 0.5 from the lower limits.

Class Interval Frequency Cumulative frequency

0–5.5 13 13
5.5–11.5 10 23
11.5–17.5 15 38
17.5–23.5 08 46
23.5–29.5 11 57
S f = 57
Total

Note :– Class interval can’t be negative hence the first CI is starting from 0.

Now to find median class we calculate Rf = 57 = 28.5
2 2

\ Median class = 11.5 – 17.5.

So, the upper limit is 17.5

3. Find the median class of the following distribution:

Class 0–10 10–20 20–30 30–40 40–50 50–60 60–70
Frequency 4 4 8 10 12 8 4

Sol. First we find the cumulative frequency

Classes Frequency Cumulative frequency
0–10 4 4
10–20 4 8
20–30 8 16
30–40 10 26
40–50 12 38
50–60 8 46
60–70 4 50
Total 50

Here, n = 50 = 25
2 2

\ Median class = 30 – 40.

4. Find the class marks of classes 15.5–18.5 and 50–75.

Sol. Class marks = upper limit + lower limit
2

Class marks of 15.5 – 18.5 = 18.5 +15.5 = 34 =17
2 2

Class marks of 50 – 75 = 75+ 50 = 125 = 62.5 .
2 2

398 Xam idea Mathematics–X

Short Answer Questions-II [3 marks]

1. If the mean of the following distribution is 6, find the value of p. 10 p+5
x246 1 2
f 323

Sol. Calculation of mean

xi fi fi xi
2 3 6
4 2 8
6 3 18
10 1 10
p+5 2 2p + 10
Total Sfi = 11 Sfixi = 2p + 52

We have, Sfi = 11, Sfixi = 2p + 52, X = 6 10 15
78
\ Mean ^ X h= Rfi xi
Rfi

⇒   6 = 2p+ 52 ⇒ 66 = 2p + 52
11

⇒ 2p = 14 ⇒ p=7

2. Find the mean of the following distribution:

x469
f 5 10 10

Sol. Calculation of arithmetic mean

xi fi fi xi
4 5 20
6 10 60
9 10 90
10 7 70
15 8 120
Total Sfi = 40 Sfixi = 360

\ Mean (X) = Σfi xi = 360 =9
Σfi 40

3. The following data gives the information on the observed lifetimes (in hours) of 225 electrical

components: [NCERT]

Lifetime (in hours) 0–20 20–40 40–60 60–80 80–100 100–120
61 38 29
Frequency 10 35 52

Determine the modal lifetimes of the components.

OR
The following table gives the number of participants in a yoga camp.

Statistics 399

Age (in years) 20–30 30–40 40–50 50–60 60–70

No. of participants 8 40 58 90 83

Find the modal age of the participants. [CBSE 2019 (30/5/1)]

Sol. Here, the maximum class frequency is 61 and the class corresponding to this frequency is 60–80.
So, the modal class is 60–80.

Here, l = 60, h = 20, f1 = 61, f0 = 52, f2 = 38

\ Mode =l+ f1 – f0 f2 # h = 60 + 2# 61 – 52 38 # 20 = 60 + 9 90 # 20
2f1 – f0 – 61 – 52 – 122 –

= 60 + 9 × 20 = 60 + 45 = 60 + 5.625 = 65.625
32 8

Hence, modal lifetime of the components is 65.625 hours.

OR
Similar solution as above only values are changed.

Ans: 58.205

4. 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
8 6 6 3 2
Number of students 2 3
[NCERT]


Sol. Calculation of median

Weight (in kg) Number of students (f) Cumulative frequency (cf)
40–45 2 2
45–50 3 5
50–55 8 13
55–60 6 19
60–65 6 25
65–70 3 28
70–75 2 30
Total
S fi = 30

We have, S fi = n = 30     ⇒     n =15
2

The cumulative frequency just greater than n =15 is 19, and the corresponding class is 55 – 60.
2
\ 55 – 60 is the median class.

Now, we have n =15 , l = 55, cf = 13, f = 6, h = 5
2

\ Median = l+f n – cf p#h
2 f

= 55 +c 15 – 13 m # 5= 55+ 2 # 5=55+1.67=56.67
6 6

Hence, median weight is 56.67 kg.

400 Xam idea Mathematics–X

5. The lengths of 40 leaves of a plant are measured correctly to the nearest millimetre, and the
data obtained is represented in the following table:

Length (in mm) 118–126 127–135 136–144 145–153 154–162 163–171 172–180

Number of leaves 3 5 9 12 5 4 2

Find the median length of the leaves. [NCERT]

Sol. Here, the classes are not in inclusive form. So, we first convert them in inclusive form by

subtracting h from the lower limit and adding h to the upper limit of each class, where h is the
2 2

difference between the lower limit of a class and the upper limit of preceding class.

Now, we have

Class Interval Number of leaves Cumulative frequency

117.5–126.5 3 3

126.5–135.5 5 8

135.5–144.5 9 17

144.5–153.5 12 29

153.5–162.5 5 34

162.5–171.5 4 38

171.5–180.5 2 40

Total Sfi = 40

We have, n = 40 ⇒ n = 20
2
n
And, the cumulative frequency just greater than 2 is 29 and corresponding class is 144.5 – 153.5.

So median class is 144.5 – 153.5.

Here, we have n = 20, l = 144.5, h = 9, f = 12, cf = 17
2
n
f 2 – cf p# = + 20 – 17
f 12
\ Median = l + h 144.5 c m#9

= 144.5 + 3 × 9 = 144.5 + 9 = 144.5 + 2.25 = 146.75 mm.
12 4

Hence, the median length of the leaves is 146.75 mm.

Long Answer Questions [5 marks]

1. 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 85–90

Number of cities 3 10 11 8 3

OR [CBSE 2019 (30/5/2)]
Find the mean of the following frequency distribution.

Class 0–20 20–40 40–60 60–80 80–100

Frequency 17 28 32 24 19

Sol. Here, we use step deviation method to find mean.
Let assumed mean A = 70 and class size h = 10

Statistics 401

So, ui = xi – 70
10

Now, we have

Literacy rate (in %) Frequency Class mark ui = xi – 70 fiui
10

45–55 3 50 – 2 – 6

55–65 10 60 – 1 – 10

65–75 11 70 0 0

75–85 8 80 1 8

85–95 3 90 2 6

Total Sfi = 35 Sfiui = –2

\ Mean (X) = A + h × Σfi ui = 70 + 10 × (−2) = 70 − 0.57 = 69.43%
Σfi 35

OR

Solution is same as above only values are changed.

Ans: 50

2. The following distribution shows the daily pocket allowance of children of a locality. The

mean pocket allowance is ` 18. Find the missing frequency f. [NCERT]

Daily pocket allowance (in `) 11–13 13–15 15–17 17–19 19–21 21–23 23–25

Number of children 7 6 9 13 f 5 4

OR
The arithmetic mean of the following frequency distribution is 53. Find the value of k.

Class 0–20 20–40 40–60 60–80 80–100

Frequency 12 15 32 k 13

[CBSE 2019 (30/1/2)]

Sol. Let the assumed mean A = 16 and class size h = 2, here we apply step deviation method.

So, ui = xi – A = xi – 16
h 2

Now, we have,

Class Interval Frequency Class mark ui = xi – 16 fiui
2
– 14
11–13 7 12 –2 –6
14
13–15 6 16 –1 0
15–17 9 18 13
20 0 2f
22 15
17–19 13 24 1 16
19–21 f Sfiui = 2f + 24
2

21–23 5 3
23–25 4
4

Total Sfi = f + 44

We have, mean (X) = 18, A = 16 and h = 2

402 Xam idea Mathematics–X

\ (X) = A + h × Rfi ui
Rfi

2f + 24 ⇒ 2f + 24
18 = 16 + 2 × e f + 44 o 2 = 2 × e f + 44 o

⇒ 1 = 2f + 24 ⇒ f + 44 = 2f + 24
f + 44

⇒ f = 44 – 24
⇒ f = 20

Hence, the missing frequency is 20.

OR
Similar solution as above. Only values are changed.

Ans : k = 28

3. The distribution below gives the marks of 100 students of a class.

Marks 0–5 5–10 10–15 15–20 20–25 25–30 30–35 35–40

Number of students 4 6 10 10 25 22 18 5

Draw a less than type and a more than type ogive from the given data. Hence, obtain the
median marks from the graph.

Sol.

Marks Cumulative frequency Marks Cumulative frequency
Less than 5 4 More than 0 100
Less than 10 10 More than 5 96
Less than 15 20 More than 10 90
Less than 20 30 More than 15 80
Less than 25 55 More than 20 70
Less than 30 77 More than 25 45
Less than 35 95 More than 30 23
Less than 40 More than 35 5
100

Fig. 14.3

Hence, median marks = 24

Statistics 403

4. For the following frequency distribution, draw a cumulative frequency curve (ogive) of 'more

than type' and hence obtain the median value. [CBSE 2019(30/5/1)]

Class 0–10 10–20 20–30 30–40 40–50 50–60 60–70
Frequency 5 15 20 23 17 11 9

Sol. We have cumulative frequency table,

Class Cumulative frequency

More than 0 100

More than 10 95

More than 20 80

More than 30 60

More than 40 37

More than 50 20

More than 60 9

N = 100

N = 50
2

Fig 14.4

a N2 = 50
` Median = 34.34 = 34.3

5. A survey was conducted by a group of students as a part of their environment awareness
programme, in which they collected the following data regarding the number of plants in
20 houses in a locality. Find the mean number of plants per house.

Number of plants 0–2 2–4 4–6 6–8 8–10 10–12 12–14

Number of houses 1 2 1 5 6 23
[NCERT]
Which method did you use for finding the mean and why?

404 Xam idea Mathematics–X

Sol. Calculation of mean number of plants per house.

Number of plants Number of houses (fi) Class mark (xi) fi xi
0–2 1 1 1

2–4 2 3 6

4–6 1 5 5

6–8 5 7 35

8–10 6 9 54

10–12 2 11 22

12–14 3 13 39

Total Sfi = 20 Sfixi = 162

Hence, Mean (X) = Σfi xi = 162 = 8.1
Σfi 20

Here, we used direct method to find mean because numerical values of xi and fi are small.

6. 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 years.

Age (in years) Number of policy holders Age (in years) Number of policy holders
Below 20 2 Below 45 89
Below 25 6 Below 50 92
Below 30 24 Below 55 98
Below 35 45 Below 60 100
Below 40 78

Sol. We are given the cumulative frequency distribution. So, we first construct a frequency table from
the given cumulative frequency distribution and then we will make necessary computations to
compute median.

Class interval Frequency (fi) Cumulative frequency (cf)
15–20 2 2

20–25 4 6

25–30 18 24

30–35 21 45

35–40 33 78

40–45 11 89

45–50 3 92

50–55 6 98

55–60 2 100

Total Sfi = 100

Here, n = 100 ⇒ n = 50
2
n
And, cumulative frequency just greater than 2 = 50 is 78 and the corresponding class is 35 – 40. So
35 – 40 is the median class.

Statistics 405

\ n = 50, l = 35, cf = 45, f = 33, h = 5
2

\ Median = l + KJKKKLKK n – cf OOONOPOO # h
2 f

= 35 + c 50 – 45 m# 5 = 35 + 5 # 5
33 33

= 35 + 25 = 35 + 0.76 = 35.76
33

Hence, the median age is 35.76 years.

7. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in `) 100–120 120–140 140–160 160–180 180–200

Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution, and draw

its ogive. [NCERT]

Sol. Converting given distribution to a less than type cumulative frequency distribution, we have,

Daily income (in `) Cumulative frequency

Less than 120 12

Less than 140 12 + 14 = 26

Less than 160 26 + 8 = 34

Less than 180 34 + 6 = 40

Less than 200 40 + 10 = 50

Now, let us plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34),
(180, 40), (200, 50) on a graph paper and join them by a freehand smooth curve.

Fig. 14.5
Thus, obtained curve is called the less than type ogive.

406 Xam idea Mathematics–X

HOTS [Higher Order Thinking Skills]

1. The mean of the following frequency table is 50. But the frequencies f1 and f2 in class 20–40
and 60–80 respectively are missing. Find the missing frequencies.

Classes 0–20 20–40 40–60 60–80 80–100 Total

Frequency 17 f1 32 f2 19 120

Sol. Let the assumed mean A = 50 and h = 20.

Calculation of mean

Class interval Mid-values (xi) Frequency (fi) ui = xi – 50 fi ui
0–20 10 17 20 –34

–2

20–40 30 f1 –1 –f1
40–60 50 32 0 0

60–80 70 f2 1 f2
80–100 90 19 2 38

Total Sfi = 68 + f1 + f2 Sfiui = 4 – f1 + f2

We have, Sfi = 120 (Given)
...(i)
⇒ 68 + f1 + f2 = 120

⇒ f1 + f2 = 52
Now, mean = 50

⇒ Xr = A+ h f Rfi ui p ⇒ 50 = 50 + 20 × ) 4 – f1 + f2 3
⇒ Rfi 120

50 = 50 + 4 – f1 + f2 ⇒ 0= 4– f1 + f2
6 6

f1 – f2 = 4 ...(ii)

From equations (i) and (ii), we get

f1 + f2 = 52

f1 – f2 = 4

2f1 = 56

⇒ f1 = 28

Putting the value of f1 in equation (i), we get

28 + f2 = 52 ⇒ f2 = 24

Hence, the missing frequencies f1 is 28 and f2 is 24.

2. If the median of the distribution given below is 28.5, find the values of x and y.

Class interval 0–10 10–20 20–30 30–40 40–50 50–60 Total
60
Frequency 5 x 20 15 y 5

Statistics 407

OR

Find the values of frequencies x and y in the following frequency distribution table, if N = 100

and median is 32. [CBSE 2019(30/5/1)]

Mark 0–10 10–20 20–30 30–40 40–50 50–60 Total

No. of students 10 x 25 30 y 10 100

Sol. Here, median = 28.5 and n = 60
Now, we have

Class interval Frequency (fi) Cumulative frequency (cf)
0–10 5 5

10–20 x 5+x

20–30 20 25 + x

30–40 15 40 + x

40–50 y 40 + x + y

50–60 5 45 + x + y

Total Sfi = 60

Since the median is given to be 28.5, thus the median class is 20 – 30.

\ n = 30, l = 20, h = 10, cf = 5 + x and f = 20
2

\ Median = l+f n – cf p#h ⇒ 28.5 = 20 + < 30 – (5+ x) F #10
2 f 20

⇒ 28.5 = 20 + 25 – x #10
20

⇒ 28.5 = 20 + 25 – x ⇒ 57 = 40 + 25 – x
2 x = 65 – 57 = 8

⇒ 57 = 65 – x ⇒ ( x = 8)
y=7
Also, n = Sfi = 60
⇒ 45 + x + y = 60

⇒ 45 + 8 + y = 60

\ y = 60 – 53 ⇒

Hence, x = 8 and y = 7.

OR

Similar solution as above. Only values are changed.

Ans : x = 9, y = 16

408 Xam idea Mathematics–X

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) The abscissae of the point of intersection of the less than type and of the more than type
cumulative frequency curve of a grouped data gives its

(a) Mean (b) Median (c) Mode (d) All of these

(ii) Which of the following is not a measure of central tendency?

(a) Mean (b) Range (c) Mode (d) Median

(iii) The runs scored by a batsman in 35 different matches are given below:

Runs Scored 0–15 15–30 30–45 45–60 60–75 75–90

Frequency 5 7 4 8 8 3

Number of matches in which the batsman scored less than 60 runs are

(a) 16 (b) 24 (c) 8 (d) 19

(iv) If the mean of data is 27 and mode is 45, the median is

(a) 30 (b) 27 (c) 32 (d) 33

(v) In the formula x = a + h> / fi ui H for finding the mean of grouped frequency distribution,
ui is equal to / fi

(a) xi + a (b) h (xi – a) (c) xi – a (d) a – xi
h h h

2. Fill in the blanks.

(i) An ogive is useful in determining _____________ .

(ii) The arithmetic mean of 1, 2, 3 . . . n is _____________ .

(iii) In the formula, mode = l + > 2f1 f1 – f0 f2 H # h, f2 is _____________.
– f0 –

(iv) Mode of a data is its _____________ value.

(v) _____________ of a data cannot be determined graphically.

QQ Very Short Answer Questions : [1 mark each]
3. Consider the following frequency distribution.

Class 0–10 10–20 20–30 30–40 40–50 50–60

Frequency 3 9 15 30 18 5

Determine the modal class.
4. Consider the following frequency distribution:

Class 0–5 6–11 12–17 18–23 24–29
15
Frequency 13 10 8 11

Find the upper limit of the median class. [NCERT Exemplar]

5. For the following distribution, find the modal class.

Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60
27 57 75 80
Number of students 3 12 [NCERT Exemplar]

Statistics 409

6. In the formula X = A+ hf Rfi ui p for finding the mean of grouped frequency distribution, what
Rfi
is the value of ui? [NCERT Exemplar]

7. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:

Class 13.8–14 14–14.2 14.2–14.4 14.4–14.6 14.6–14.8 14.8–15
Frequency 2 4 5 71 48 20

Find the number of athletes who completed the race in less then 14.6 seconds. [NCERT Exemplar]

8. In the following distribution:

Monthly income More than More than More than More than More than More than
range (in ™) ™10000 ™13000 ™16000 ™19000 ™22000 ™25000
Number of 100 85 69 50 33 15
families

Find the number of families having income range ™16000 – 19000.
9. Consider the following distribution:

Marks Obtained Less than 10 Less than 20 Less than 30 Less than 40 Less than 50

Numbers of students 02 10 15 30 40

Find the number of students having marks in range 30–40.

10. Which graphical representation helps in determining the median of a given frequency distribution?
11. What is the empirical relation between mean, median and mode?

12. Which measure of central tendency is given by the x-coordinate of the point of intersection of the
‘more than’ ogive and ‘less than’ ogive?

13. What is the value of the median of the data represented by the following graph of less than ogive
and more than ogive?

Fig. 14.6

410 Xam idea Mathematics–X

QQ Short Answer Questions-I: [2 marks each]
14. Find the mode of the following frequency distribution:

Class Interval 25–30 30–35 35–40 40–45 45–50 50–55
42
Frequency 25 34 50 38 14

15. Write the median class of the following distribution: [CBSE 2019, (30/2/1)]

Classes 0–10 10–20 20–30 30–40 40–50 50–60 60–70

Frequency 14 6 8 20 15 12 9

16. The mean, median and mode of grouped data are always different. State True or False and
justify your answer.

17. Find the class marks of classes 15–35 and 20–40.

18. If the mean of the following distribution is 2.6, then find the value of k.

xi 1 2 3 4 5
fi k 5 8 1 2

QQ Short Answer Questions-II: [3 marks each]

19. The table below shows the salaries of 280 persons:

Salary (In thousand `) Number of Persons

5 – 10 49

10 – 15 133

15 – 20 63

20 – 25 15

25 – 30 6

30 – 35 7

35 – 40 4

40 – 45 2

45 – 50 1

Calculate the median salary of the data. [CBSE 2018, (30/1)]

20. By changing the following frequency distribution ‘to less than type’ distribution, draw its ogive.

Classes 0–15 15–30 30–45 45–60 60–75

Frequency 6 8 10 64

[CBSE 2018, (C) (30/1)]

21. Find the mode of the following frequency distribution.

Class 0–10 10–20 20–30 30–40 40–50 50–60 60–70

Frequency 8 10 10 16 12 6 7

[CBSE 2019, (30/1/1)]

22. A class 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 days 0–6 6–12 12–18 18–24 24–30 30–36 36–42
Number of students 10 11 7 4 4 3 1

[CBSE 2019, (30/3/1)]

Statistics 411

23. The weights of tea in 70 packets is given in the following table:

Weight (In g.) Number of Packets

200 – 201 12
201 – 202 26
202 – 203 20

203 – 204 9
204 – 205 2

205 – 206 1

Find the modal weight. [CBSE 2019, (C) (30/1/1)]
24. Find the value of p, if the mean of the following distribution is 20.

x 15 17 19 20+p 23
4 5 6
f2 3

25. For the following distribution, calculate mean:

Classes 25–29 30–34 35–39 40–44 45–49 50–54 55–59
Frequency 14 22 16 6 5 3 4

26. Find the mean age of 100 residents of a town from the following data:

Age equal and above (in years) 0 10 20 30 40 50 60 70

No. of Persons 100 90 75 50 25 15 5 0

27. The table below shows the daily expenditure on food of 25 households in a locality.

Daily Expenditure (in `) 100–150 150–200 200–250 250–300 300–350

No. of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method. [CBSE 2019, (30/1/3)]

28. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details
are tabulated as given below:

Mileage (km/l) 10–12 12–14 14–16 16–18

Number of Cars 7 12 18 13

Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km/l.

Do you agree with this claim?

29. The following are the ages of 300 patients getting medical treatment in a hospital on a particular
day:

Age (in years) 10–20 20–30 30–40 40–50 50–60 60–70

No. of patients 60 42 55 70 53 20

Form:
(i) Less than type cumulative frequency distribution.

(ii) More than type cumulative frequency distribution.

30. The following data gives the information on the observed lifetimes (in hours) of 225 electrical
components:

412 Xam idea Mathematics–X

Lifetimes (in hours) 0–20 20–40 40–60 60–80 80–100 100–120

No. of components 10 35 52 61 38 29

Determine the modal lifetimes of the components.

QQ Long Answer Questions: [5 marks each]

31. The mean of the following distribution is 18. Find the frequency f of the class 19 – 21.

Classes 11–13 13–15 15–17 17–19 19–21 21–23 23–25

Frequency 3 6 9 13 f 5 4

[CBSE 2018, (30/1)]

32. The following distribution gives the daily income of 50 workers of a factory:

Daily Income (in ™) 100–120 120–140 140–160 160–180 180–200

Number of Workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution and draw

its ogive. [CBSE 2018, (30/1)]

33. Find the mean and mode for the following data:

Classes 10–20 20–30 30–40 40–50 50–60 60–70 70–80

Frequency 4 8 10 12 10 4 2

[CBSE 2018, (C) (30/1)]

34. If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.

Class 0–10 10–20 20–30 30–40 40–50 50–60 60–70 Total

Frequency f1 5 9 12 f2 3 2 40

[CBSE 2019, (30/1/1)]

35. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in `) 200–220 220–240 240–260 260–280 280–300

Number of workers 12 14 8 6 10

Convert the distribution above to a ‘less than type’ cumulative frequency distribution and draw

its ogive. [CBSE 2019, (30/1/3)]

36. Change the following data into ‘less than type’ distribution and draw its ogive:

Class Interval 30–40 40–50 50–60 60–70 70–80 80–90 90–100

Frequency 7 5 8 10 6 6 8

[CBSE 2019, (30/2/1)]

37. Change the following distribution to a ‘more than type’ distribution. Hence draw the ‘more than
type’ ogive for this distribution.

Class Interval 20–30 30–40 40–50 50–60 60–70 70–80 80–90
6 25 15
Frequency 10 8 12 24
[CBSE 2019, (30/3/1)]


38. Calculate the mean of the following frequency distribution:

Class 10–30 30–50 50–70 70–90 90–110 110–130

Frequency 5 8 12 20 3 2
[CBSE 2019, (30/4/2)]

Statistics 413

39. The following table gives production yield in kg per hectare of wheat of 100 farms of a village:

Production yield 40–45 45–50 50–55 55–60 60–65 65–70
(kg/hectare)

Number of farms 4 6 16 20 30 24

Change the distribution to a ‘more than type’ distribution, and draw its ogive.
[CBSE 2019, (30/4/2)]

40. If the mean of the following frequency distribution is 62.8, then find the missing frequency x:

Class 0–20 20–40 40–60 60–80 80–100 100–120

Frequency 5 8 x 12 7 8

[CBSE 2019, (C) (30/1/1)]

41. The annual rainfall record of a city of 66 days is given in the following table:

Rainfall (in cm) 0–10 10–20 20–30 30–40 40–50 50–60
Number of days 22 10 8 15 5 6

Calculate the median rainfall using ogive (of more than type and of less than type).

42. Draw ‘less than’ ogive and ‘more than’ ogive for the following distribution and hence find its
median.

Classes 20–30 30–40 40–50 50–60 60–70 70–80 80–90
Frequency 10 8 12 24 6 25 15

43. The following is the frequency distribution of duration for 100 calls made on a mobile phone:

Duration (in seconds) 95–125 125–155 155–185 185–215 215–245
Number of calls 14 22 28 21 15

Calculate the average duration (in sec.) of a call and also find the median from a cumulative

frequency curve.

44. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50.
Compute the missing frequency f1 and f2.

Classes 0–20 20–40 40–60 60–80 80–100 100–120

Frequency 5 f1 10 f2 7 8

45. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years) 5–15 15–25 25–35 35–45 45–55 55–65

No. of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures
of central tendency.

Answers

1. (i) (b) (ii) (b) (iii) (b) (iv) (d) (v) (c)

2. (i) median (ii) n (n + 1) (iii) frequency of the class succeeding the modal class
2

(iv) most frequent (v) Mean

414 Xam idea Mathematics–X

3. 30–40 4. 17.5 5. 30–40 6. xi – a 7. 82 8. 19
9. 15 10. Ogive h 13. 8

11. 3 median = mode + 2 mean 12. Median

14. 38.33 15. 30–40 16. False, it depends on the data 17. 25, 30 18. k = 4

19. ™ 13.42 thousand or ™ 13420 (approx) 21. Mode=36

22. Mean = 14.1 23. Mode = 201.7 24. p = 1 25. 36.36

26. 31 years 27. ™211 28. 14.48 km/L, No

29.

(i) Less than Type (ii) More than Type

Age (in years) Number of students Age (in years) Number of students

Less than 20 60 More than or equal to 10 300

Less than 30 102 More than or equal to 20 240
Less than 40 157 More than or equal to 30 198

Less than 50 227 More than or equal to 40 143
Less than 60
280 More than or equal to 50 73

Less than 70 300 More than or equal to 60 20

30. 65.63 hours 31. f = 8 33. Mean = 42.2, Mode = 45

34. f1 = 3, f2 = 6

38. 65.6 40. x = 10 41. 21.25 cm 42. 58.33

43. Average = 170.3 sec, Median = 170 sec. 44. f1 = 8, f2 = 12
45. Mode = 36.8 year, Mean = 35.38 years

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) Which of the following is not a measure of central tendency?

(a) class mark (b) mean (c) median (d) mode

(ii) An ogive is useful in determining the

(a) mean (b) median (c) mode (d) all of the above

(iii) In the formula x = a + / fi di for finding the mean of a grouped data, di's are deviations
/ fi
from a of

(a) lower limits of the classes (b) upper limits of the classes

(c) mid-points of the classes (d) frequencies of the class marks

(iv) If xi's are the mid-points of the class intervals of a grouped data fil s are the corresponding

∑frequencies and x is the mean, then fi (xi − x) is equal to

(a) 0 (b) – 1 (c) 1 (d) 2

Statistics 415

2. Fill in the blanks. [3 × 1 = 3]

(i) ________________ is the positional mid value of the observations in a data.

(ii) Number of times a particular observation occurs is called ________________ . n
2
(iii) A class interval, in which cumulative frequency is greater than and nearest to is called
________________ .

3. Solve the following questions. (3 × 1 = 3)

(i) The mean of 11 observations is 50. If the mean of first 6 observations is 49 and that of the
last six observations is 52, what is the value of 6th observation?

(ii) What is the mean of first five prime numbers?

(iii) Consider the following distribution.

Marks obtained Number of observations
More than or equal to 0 68
More than or equal to 10 53
More than or equal to 20 50
More than or equal to 30 45
More than or equal to 40 38
More than or equal to 50 25

Find the number of students having marks more than 29 but less than 40.

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. Consider the following frequency distribution of the heights of 60 students of a class.

Height (in cm) 150–155 155–160 160–165 165–170 170–175 175–180

Number of students 15 13 10 8 9 5

What is the sum of the lower limit of the modal class and upper limit of the median class?
5. If the mean of the following distribution is 27, find the value of p.

Classes 0–10 10–20 20–30 30–40 40–50

Frequency 8 p 12 13 10

6. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in
the following table:

Number of seats 100–104 104–108 108–112 112–116 116–120
Frequency 15 20 32 18 15

Determine the mean number of seats occupied over the flights.

416 Xam idea Mathematics–X

QQ Solve the following questions. (3 × 3 = 9)
7. Calculate the median from the following data:
Number of tenants
Rent (in `) 8
1500–2500 10
2500–3500 15
3500–4500 25
4500–5500 40
5500–6500 20
6500–7500 15
7500–8500 7
8500–9500

8. Calculate the missing frequency from the following distribution, it being given that the median
of the distribution is 24.

Age ( in years) 0–10 10–20 20–30 30–40 40–50

Number of persons 5 25 f 18 7

9. Draw ‘less than’ ogive and ‘more than’ ogive for the following distribution and hence find its median.

Classes 20–30 30–40 40–50 50–60 60–70 70–80 80–90
Frequency 10 8 12 24 6 25 15

QQ Solve the following questions. (3 × 5 = 15)
10. The following distribution gives the daily income of 50 workers of a factory:

Daily income (in `) 100–120 120–140 140–160 160–180 180–200
Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution and draw
its ogive. Find the median from this ogive.

11. The annual profits earned by 30 shops of a shopping complex in locality give rise to the following

distribution:

Profit Number of shops
(in lakhs in `) (frequency)

More than or equal to 5 30

More than or equal to 10 28

More than or equal to 15 16

More than or equal to 20 14

More than or equal to 25 10

More than or equal to 30 7

More than or equal to 35 3

Draw both ogives for the above data and hence obtain the median.

Statistics 417

12. The marks obtained by 100 students of a class in an examination are given below.

Marks No. of students
0–5 2
5 –10 5
6
10–15 8
15–20 10
20–25 25
25–30 20
30–35 18
35–40 4
40–45 2
45–50

Draw ‘a less than’ type cumulative frequency curves (ogive). Hence find median.
[CBSE 2019(30/1/2)]

Answers

1. (i) (a) (ii) (b) (iii) (c) (iv) (a)

2. (i) median (ii) frequency (iii) median class

3. (i) 56 (ii) 5.6 (iii) 7

4. 315 5. p = 7 6. 110(approx) 7. ™5,800 8. f = 25

9. 58.33(approx)

10. ™138 11. Median = ™17.5 lakh 12. 29.5

zzz

418 Xam idea Mathematics–X

Probability 15

BASIC CONCEPTS – A FLOW CHART

Probability 419

MORE POINTS TO REMEMBER



 Equally likely outcomes: The outcomes of a random experiment are said to be equally
likely outcomes, when each outcome is as likely to occur as the other.

For example, when a coin is tossed, both outcomes H and T are equally likely to appear. Thus
H and T are equally likely outcomes. Similarly, when we throw a die then the outcomes, 1,
2, 3, 4, 5 and 6 are equally likely.

 Favourable outcomes: The outcomes which ensure the occurrence of an event are called
favourable outcome of that event.

For example, suppose experiment is throwing a die and event is getting even number, then
favourable outcomes are 2, 4 and 6.

 Odds in favour or against of occurrence of event: If E and Er are complementary events,

then the ratio P(E) : P( Er ) is called as odds in favour of occurrence of event E while the ratio

P( Er ) : P(E) is called as odds against the occurrence of event E.

For example, suppose experiment is throwing a pair of die and event E is getting doublet.
6
P (E) 36 6
Then, odds in favour of occurrence of E= P (Er ) = 30 = 30 = 1
5
36
30
P (Er ) 36 30
Odds against occurrence of E= P (E) = 6 = 6 = 5

36

 Playing cards: The details of playing card having 52 cards are as:

10.

Fig. 15.1

Multiple Choice Questions [1 mark]

C hoose and write the correct option in the following questions.

1. If an event cannot occur, then its probability is [NCERT Exemplar]

(a) 1 (b) 3 (c) 1 (d) 0
4 2

2. Which of the following cannot be the probability of an event? [NCERT Exemplar]

(a) 1 (b) 0.1 (c) 3% (d) 17
3 16

420 Xam idea Mathematics–X

3. An event is very unlikely to happen. Its probability is closest to

(a) 0.0001 (b) 0.001 (c) 0.01 (d) 0.1

4. Which of the following cannot be the probability of an event?

(a) 1 (b) 0 (c) – 1 (d) 0.8
4 2

5. The probability expressed as a percentage of a particular occurrence can never be

(a) less than 100 (b) less than 0
(c) greater than 1 (d) anything but a whole number

6. If P(A) denotes the probability of an event A, then [NCERT Exemplar]

(a) P(A) < 0 (b) P(A) > 1 (c) 0 ≤ P(A) ≤ 1 (d) –1 ≤ P(A) ≤ 1

7. The probability that a non leap year selected at random will contain 53 sunday's is

[NCERT Exemplar]

(a) 1 (b) 2 (c) 3 (d) 5
7 7 7 7

8. When a die is thrown once, the probability of getting an odd number less than 3 is

(a) 1 (b) 1 (c) 1 (d) 0
6 3 2

9. A die is thrown once, the probability of getting a prime number is

(a) 2 (b) 1 (c) 1 (d) 1
3 3 2 6

10. The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot
is

(a) 7 (b) 14 (c) 21 (d) 28

11. The probability that a number selected at random from the numbers 1, 2, 3 ... 15 is a multiple

of 4 is

(a) 4 (b) 2 (c) 1 (d) 1
15 15 5 3

12. A coin is tossed 1000 times and 640 times a ‘head’ occurs. The empirical probability of

occurrence of a head in this case is

(a) 0.6 (b) 0.64 (c) 0.36 (d) 0.064

13. A card is selected from a deck of 52 cards. The probability of its being a face card is

(a) 3 (b) 1 (c) 3 (d) 2
26 2 13 13

14. Two coins are tossed simultaneously. The probability of getting atmost one head is

(a) 1 (b) 1 (c) 3 (d) 1
4 2 4

15. If the probability of an event is p, the probability of its complementary event will be

[NCERT Exemplar]
(a) p – 1
(b) 1 – p (c) 1– 1 (d) p
p

16. A card is selected from a deck of 52 cards. The probability of being a red face card is

[NCERT Exemplar]

(a) 3 (b) 3 (c) 2 (d) 1
26 13 13 2

Probability 421

17. A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The

number of outcomes favourable to E is [NCERT Exemplar]

(a) 4 (b) 13 (c) 48 (d) 51

18. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6,000

tickets are sold, how many tickets has she bought? [NCERT Exemplar]

(a) 40 (b) 240 (c) 480 (d) 750

19. A bag contains three green marbles, four blue marbles and two orange marbles. If a marble is
picked at random, then the probability that it is not an orange marble is

(a) 7 (b) 2 (c) 4 (d) none of these
9 9 9

20. If a number x is chosen from the numbers 1, 2, 3 and a number y is selected from the numbers
1, 4, 9. Then P(xy < 9) is

(a) 3 (b) 4 (c) 1 (d) 5
9 9 9 9

Answers

1. (d) 2. (d) 3. (a) 4. (c) 5. (b) 6. (c)
7. (a) 8. (a) 9. (c) 10. (b) 11. (c) 12. (b)
13. (c) 14. (c) 15. (b) 16. (a) 17. (d) 18. (c)
19. (a) 20. (d)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. Probability of an event E + Probability of the event ‘not E’= ______________ .
2. The probability of an event that cannot happen is ______________ . Such an event is called

______________ .
3. The probability of an event that is certain to happen is ______________ . Such an event is called

______________ .
4. The sum of the probabilities of all the elementary events of an experiment is ______________.
5. The probability of an event is greater than or equal to ______________ and less than or equal to

______________ .
6. The events which have equal chances to occur or no one is preferred over the other are called

______________.
7. When sum of probability of two events is 1, the events are called ______________ .
8. On a single roll of a die, the probability of getting a number 8 is ______________.
9. On a single roll of a die, the probability of getting a number less than 7 is______________ .
10. Someone is asked to mark a number from 1 to 100. The probability that it is a prime is

______________ .
11. A single letter is selected random from the word Probability, the probability it is a vowel

is______________ .
12. If I toss a coin 3 times and get head each time then the probability of getting head in next toss is

______________ .

13. Total number of red face card in a pack of cards is ______________ .

422 Xam idea Mathematics–X

14. The outcomes which ensure the occurrence of an event are called ______________ outcomes.
15. Probability of an event cannot be ______________ .

Answers

1. 1 2. 0, impossible event 3. 1, sure or certain event 4 1
5. 0, 1 8. zero
9. one 6. equally likely events 7. complementary events
1 4 1 14. favourable
15. negative 10. 4 11. 11 12. 2 13. 6

Very Short Answer Questions [1 mark]

1. Two coins are tossed simultaneously. Find the probability of getting exactly one head.

Sol. Possible outcomes are {HH, HT, TH, TT}

P (exactly one head) = 2 = 1
4 2

2. From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting
a black queen.

Sol. Number of black queens in a pack of cards = 2

\ P (black queen) = 2 = 1
52 26

3. If P (E) = 0.05, what is the probability of ‘not E’ ? [NCERT]

Sol. As we know that,

P (E) + P (not E) = 1

P (not E) = 1 – P (E) = 1 – 0.05 = 0.95

4. What is the probability of getting no head when two coins are tossed simultaneously?

Sol. Favourable outcome is TT;

\ P (no head) = 1
4

5. In a single throw of a pair of dice, what is the probability of getting the sum a perfect square?

Sol. Total outcomes = 36

Favourable outcomes are {(1, 3), (3, 1), (2, 2), (3, 6), (6, 3), (4, 5), (5, 4)}

\ Required probability = 7
36

6. Someone is asked to choose a number from 1 to 100. What is the probability of it being a prime
number?

Sol. Total prime numbers between 1 to 100 = 25

\ P (prime number) = 25 = 1
100 4

7. Cards marked with number 3, 4, 5, ....., 50 are placed in a box and mixed thoroughly. A card

is drawn at random from the box. Find the probability that the selected card bears a perfect

square number. [CBSE Delhi 2016]

Sol. Possible outcomes are 4, 9, 16, 25, 36, 49, i.e., 6.

\ P (perfect square number) = 6 or 1
48 8

Probability 423

8. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability

of getting neither a red card nor a queen. [CBSE (AI) 2016]

Sol. Number of possible outcomes = 52

Number of red cards and queens = 28
Number of favourable outcomes = 52 – 28 = 24

P (getting neither a red card nor a queen) = 24 = 6
52 13

9. 20 tickets, on which numbers 1 to 20 are written, are mixed thoroughly and then a ticket is

drawn at random out of them. Find the probability that the number on the drawn ticket is a

multiple of 3 or 7. [CBSE (F) 2016]

Sol. n(S) = 20, multiples of 3 or 7, A: {3, 6, 9, 12, 15, 18, 7, 14}, n(A) = 8

\ Required probability = 8 or 2
20 5

10. A number is chosen at random from the numbers −3,−2,−1,0,1,2,3. What will be the

probability that square of this number is less than or equal to 1 ? [CBSE Delhi 2017]

Sol. Favourable outcomes are –1, 0, 1 = 3

Total outcomes = 7 3
\ Required probability = 7

Short Answer Questions-I [2 marks]

1. Two dice are thrown at the same time and the product of numbers appearing on them is noted.

Find the probability that the product is a prime number. [NCERT Exemplar]

Sol. Product of the number on the dice is prime number, i.e., 2, 3, 5.

The possible ways are (1, 2), (2, 1), (1, 3), (3, 1), (5, 1), (1, 5).

So, number of possible ways = 66 = 1
\ Required probability = 36 6

2. Find the probability that a number selected from the numbers 1 to 25 is not a prime number
when each of the given numbers is equally likely to be selected.

Sol. Total prime numbers from 1 to 25 = 9.

\ Non-prime numbers from 1 to 25 = 25–9 = 16

⇒ P (non-prime number) = 16
25

3. One card is drawn at random from a pack of 52 cards. Find the probability that the card drawn

is an ace and black.

Sol. Number of black aces in a pack of cards = 2
2
\ P (an ace and black card) = 52 = 1
26
4. 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

(i) an orange flavoured candy? (ii) a lemon flavoured candy? [NCERT]

  Sol. (i) As the bag contains only lemon flavoured candies. So, the event related to the experiment
of taking out an orange flavoured candy is an impossible event. So, its probability is 0.

(ii) As the bag contains only lemon flavoured candies. So, the event related to the experiment

of taking out lemon flavoured candies is sure event. So, its probability is 1.

5. 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. [NCERT]

424 Xam idea Mathematics–X

Sol. Here, total number of pens = 132 + 12 = 144

\ Total number of elementary outcomes = 144

Now, favourable number of elementary events =132 132 = 11 .
\ Probability that a pen taken out is good one = 144 12

6. A child has a die whose six faces show the letters as given below:

A B C A A B

The die is thrown once. What is the probability of getting (i) A? (ii) B? [CBSE 2019 (30/5/1)]

Sol. The total number of elementary events associated with random experiment of throwing a die is 6.

(i) Let E be the event of getting a letter A.

\ Favourable number of elementary events = 3

\ P(E) = 3 = 1
6 2
(ii) Let E be the event of getting a letter B.

\ Favourable number of elementary events = 2
\
P(E) = 2 = 1
6 3

7. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card
drawn is neither a red card nor a black king.

Sol. Let E be the event ‘card drawn is neither a red card nor a black king’

The number of outcomes favourable to the event E = 24 ( 26 red cards and 2 black kings are not

there, so 52 – 28 = 24)

\ P(E) = 24 = 6
52 13

8. Rahim tosses two different coins simultaneously. Find the probability of getting at least one

tail. [CBSE Delhi 2014]

Sol. The sample space is {HH, HT, TH, TT}

Total number of outcomes = 4

Outcomes for getting at least one tail is {HT, TH, TT}

Number of favourable outcomes =3

\ Probability of getting at least one tail = Number of favourableoutcomes = 3
Total number ofoutcomes 4

9. Subhash enjoys walking every morning. As it is very cold outside, he usually take his sweaters
when he goes out for walking. He has 3 red sweaters, 7 black sweaters and 10 white sweaters.
One day, while he was about to take his sweater from the cupboard, the power goes off.

(i) If he picks one sweater at random, how likely is it that he will get a white sweater?

(ii) If he picks one sweater at random, how likely is it that he will get either red or black
sweater?

Sol. We have,

Total possible outcomes = 3 + 7 + 10 = 20 sweaters

(i) Total number of white sweater = 10 10 1
20 2
∴ Probability that he will get a white sweater = =

(ii) Total number of favourable outcomes i.e., either red or black s12w00e=ate12r = 3 + 7 = 10
∴ Probability that he will get either a red or black sweater =

Probability 425

Short Answer Questions-II [3 marks]

1. Harpreet tosses two different coins simultaneously (say, one is of ™1 and other of ™2). What is
the probability that she gets at least one head?

Sol. 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).

So, the number of outcomes favourable to E is 3.

Therefore, P(E) = 3 3
4 4
i.e., the probability that Harpreet gets at least one head is .

2. A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh

wins the game if all the tosses give the same result (i.e. three heads or three tails) and loses

otherwise. Find the probability of Ramesh losing the game. [CBSE (F) 2016, Delhi 2017 (C)]

OR
A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same

result in all the tosses is a success, find the probability of losing the game.

[CBSE 2019 (30/1/2)]

Sol. The outcomes associated with this experiment are given by HHH, HHT, HTH, THH, TTH,
THT, HTT, TTT.

\  Total number of possible outcomes = 8

Now, Ramesh will lose the game if he gets HHT, HTH, THH, TTH, THT, HTT.


\ Favourable number of events = 6

\ Probability that he lose the game = 6 = 3
8 4

OR

Solution is similar as above and answer remains same.

3. Three unbiased coins are tossed together. Find the probability of getting: [CBSE (AI) 2016]

(i) all heads. (ii) exactly two heads. (iii) exactly one head.

(iv) at least two heads. (v) at least two tails.

Sol. Elementary events associated to random experiment of tossing three coins are HHH, HHT,
HTH, THH, HTT, THT, TTH, TTT.

\ Total number of elementary events =8

(i) The event ‘‘getting all heads’’ is said to occur, if the elementary event HHH occurs,

i.e.favourable number of elementary events = 1
1
Hence, required probability = 8

(ii) The event ‘‘getting two heads’’ will occur, if one of the elementary events HHT, THH, HTH

occurs i.e.; favourable number of elementary events = 3

Hence, required probability = 3
8

(iii) The event of “getting one head”, when three coins are tossed together, occurs if one of
the elementary events HTT, THT, TTH, occurs i.e.; favourable number of elementary
events = 3

426 Xam idea Mathematics–X

Hence, required probability = 3 .
8

(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say

that the event ‘‘getting at least two heads’’ occurs i.e.; favourable number of elementary

events = 4 4 1
8 2
Hence, required probability = = .

(v) Similar as (iv) P (getting at least two tails) = 4 = 1
8 2

4. A die is thrown once. Find the probability of getting:

(i) a prime number. (ii) a number lying between 2 and 6.

(iii) an odd number. [NCERT, CBSE 2019 (30/1/2)]

Sol. We have, the total number of possible outcomes associated with the random experiment of
throwing a die is 6 (i.e., 1, 2, 3, 4, 5, 6).

(i) Let E denotes the event of getting a prime number.

So, favourable number of outcomes = 3 (i.e., 2, 3, 5)

\ P(E) = 3 = 1
6 2
(ii) Let E be the event of getting a number lying between 2 and 6.

\ Favourable number of elementary events (outcomes) = 3 (i.e., 3, 4, 5)

\ P(E) = 3 = 1
6 2
(iii) Let E be the event of getting an odd number.

\ Favourable number of elementary events = 3 (i.e., 1, 3, 5)
\
P(E) = 3 = 1
6 2

5. 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?

Sol. (i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes

are 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) = 6 = 3

(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.
4 2
Therefore, P(F) = 6 = 3

6. 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 ?

(ii) 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? [NCERT]

Sol. Here, the total number of possible outcomes = 5.

(i) Since, there is only one queen

\ Favourable number of elementary events = 1
\ 1
Probability of getting the card of queen = 5

Probability 427

(ii) Now, the total number of possible outcomes = 4.

(a) Since, there is only one ace

\ Favourable number of elementary events = 1
1
(b)  \ Probability of getting an ace card = 4

Since, there is no queen (as queen is put aside)

\ Favourable number of elementary events = 0

\ Probability of getting a queen = 0 = 0.
4
7. 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) not

green?

Sol. Here, total number of marbles = 17

\ Total number of possible outcomes = 17

(i) Since, there are 5 red marbles in the box.

\ Favourable number of elementary events = 5

\ Probability of getting red marble = 5
17

(ii) Since, there are 5+8=13 marbles which are not green in the box.

\ Favourable number of elementary events = 13

\ Probability of not getting a green marble = 13
17

8. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find
the probability that the drawn ball is:

(i) red or white. (ii) not black. (iii) n either white nor black.

Sol. Total number of balls = 5 + 8 + 7 = 20

(i) P (red or white ball) = 5+8 = 13
20 20
7 13
(ii) P (not black ball) = 1 – P (black ball) = 1 – 20 = 20

(iii) P (neither white nor black ball) = P (red ball) = 5 = 1 .
20 4

9. 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?

[NCERT]

Sol. Let E be the event of having the same birthday.

Therefore, Er is the event of not having the same birthday.

i.e., P ( Er ) = 0.992 (Given)

Now, we have

P(E) + P( Er ) = 1 ⇒ P(E) = 1 – P( Er ) = 1 – 0.992 = 0.008.

10. 1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Saket has purchased
one lottery ticket, what is the probability of winning a prize ?

Sol. Out of 1000 lottery tickets, one ticket can be chosen in 1000 ways.

\ Total number of elementary events = 1000

It is given that there are 5 prizes on these 1000 tickets.

Therefore, number of ways of selecting a prize ticket = 5

428 Xam idea Mathematics–X

Hence, P (winning a prize) = 5 = 1 .
1000 200

11. In a single throw of a pair of different dice, what is the probability of getting (i) a prime

number on each dice ? (ii) a total of 9 or 11 ? [CBSE Delhi 2016]

Sol. Total outcomes = 36

(i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3,3) (3, 5) (5, 2) (5, 3) (5, 5) i.e., 9.

P (a prime number on each die) = 9 or 1
36 4

(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e., 6 outcomes

P (a total of 9 or 11) = 6 or 1
36 6

12. Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than 7 (ii) have a product less than 16 (iii) is a doublet of
odd numbers [CBSE Delhi 2017]

Sol. Total number of outcomes = 36

(i) Favourable outcomes are (1, 1,) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 2) (2, 3) ( 2, 4) (3, 1) (3, 2)

(3, 3) (4, 1) (4, 2) (5, 1) i.e., 15.

∴ P (sum less than 7) = 15 or 5
36 12

(ii) Favourable outcomes are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5)

(2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (4, 1) (4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) i.e., 25.

∴ P (product less than 16) = 25
36
(iii) Favourable outcomes are

(1, 1) (3, 3) (5, 5) i.e., 3. 3 1
36 12
∴ P (doublet of odd number) = or

Long Answer Questions [5 marks]

1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

(i) a king of red colour. (ii) a face card. (iii) a red face card.

(iv) the jack of hearts. (v) a spade. (vi) the queen of diamonds.

[NCERT]

Sol. Here, total number of possible outcomes = 52

(i) As we know that there are two suits of red card, i.e., diamond and heart and each suit
contains one king.

\ Favourable number of outcomes = 2

\ Probability of getting a king of red colour = 2 = 1
52 26
(ii) As we know that kings, queens and jacks are called face cards i.e., there are 12 face cards.

\ Favourable number of elementary events = 12

\ Probability of getting a face card = 12 = 3
52 13
(iii) As we know there are two suits of red cards, i.e., diamond and heart and each suit contains

3 face cards.

\ Favourable number of elementary events = 2 × 3 = 6
\
Probability of getting red face card = 6 = 3
52 26

Probability 429

(iv) Since, there is only one jack of hearts.

\ Favourable number of elementary events = 1

\ Probability of getting the jack of heart = 1 .
52
(v) Since, there are 13 cards of spade.

\ Favourable number of elementary events = 13

\ Probability of getting a spade = 13 = 1 .
52 4

(vi) Since, there is only one queen of diamonds.

\ Favourable number of outcomes (elementary events) = 1
\
Probability of getting a queen of diamond = 1 .
52

2. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be
drawn. Find the probability that the card drawn is:

(i) an ace. (ii) red. (iii) either red or king. (iv) red and a king.

(v) a face card. (vi) a red face card. (vii) ‘2’ of spades. (viii) ‘10’ of a black suit.

Sol. Out of 52 cards, one card can be drawn in 52 ways.

So, total number of elementary events = 52

(i) There are four ace cards in a pack of 52 cards.

\ Favourable number of elementary events = 4

Hence, required probability = 4 = 1 .
52 13

(ii) There are 26 red cards in a pack of 52 cards. Therefore favourable number of elementary

events = 26

Hence, required probability = 26 = 1 .
52 2

(iii) There are 26 red cards, including two red kings, in a pack of 52 playing cards. Also, there
are 4 kings, two red and two black. Therefore, card drawn will be a red card or a king if it
is any one of 28 cards (26 red cards and 2 black kings).

\ Favourable number of elementary events = 28

Hence, required probability = 28 = 7 .
52 13

(iv) A card drawn will be red as well as king, if it is a red king. There are 2 red kings in a pack
of 52 playing cards.

\ Favourable number of elementary events = 2

Hence, required probability = 2 = 1 .
52 26

(v) In a deck of 52 cards kings, queens, and jacks are called face cards. Thus, there are 12 face

cards. Favourable number of elementary events = 12

Hence, required probability = 12 = 3 .
52 13

(vi) There are 6 red face cards 3 each from diamonds and hearts. Favourable number of

elementary events = 6.

Hence, required probability = 6 = 3 .
52 26

430 Xam idea Mathematics–X

(vii) There is only one ‘2’ of spades.

\ Favourable number of elementary events = 1

Hence, required probability = 1 .
52

(viii) There are two suits of black cards viz. spades and clubs. Each suit contains one card bearing
number 10.

\ Favourable number of elementary events = 2

Hence, required probability = 2 = 1 .
52 26

3. A game of chance consists of spinning an arrow which comes to rest pointing
at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.2), and these are

equally likely outcomes. What is the probability that it will point at:

(i) 8 ? (ii) an odd number? (iii) a number greater than 2?

(iv) a number less than 9? [NCERT]

Sol. Here, total number of elementary events (possible outcomes) = 8

(i) We have only one ‘8’ on the spining plant. Fig. 15.2

\ Favourable number of outcomes = 1 1
8
Hence, the probability that arrow points at 8 = .

(ii) We have four odd points (i.e.,1, 3, 5 and 7)

\ Favourable number of outcomes = 4 4 1
8 2
Hence, probability that arrow points at an odd number = = .

(iii) We have 6 numbers greater than 2, i.e., 3, 4, 5, 6, 7 and 8.

Therefore, favourable number of outcomes = 6 6 = 3 .
Probability that arrow points at a number greater than 2 = 8 4

(iv) We have 8 numbers less than 9, i.e, 1, 2, 3, ... 8.

\ Favourable number of outcomes = 8 8
8
Hence, probability that arrow points at a number less than 9 = = 1.

4. 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? (ii) 13? (iii) less than or equal to 12?

Sol. 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’, ‘5’ 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.

So, the number of possible outcomes
= 6 × 6 = 36.

(i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted by
E, are :

Probability 431

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see figure) i.e.; 5

Hence, P(E) = 5
36
(ii) As you can see from figure, there is no outcome favourable to the event F, ‘the sum of two

numbers is 13’.

So, P(F) = 0 =0
36
(iii) As you can see from figure, all the outcomes are favourable to the event G, ‘sum of two

numbers ≤ 12’.

So, P(G) = 36 = 1.
36

5. A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after

mixing the cards throughly. Find the probability that the number on the drawn card is:

[CBSE Delhi 2014]

(i) an odd number. (ii) a multiple of 5. (iii) a perfect square. (iv) an even prime number.

OR

Cards marked with numbers 5 to 50 (one number on one card) are placed in a box and mixed
throughly. One card is drawn at random from the box. Find the probability that the number
on the card taken out is

(i) a prime number less than 10 (ii) a number which is a perfect square.

[CBSE 2019 (30/5/1)]

Sol. Total number of cards = 49

\ Total number of outcomes = 49
(i) Odd number

Favourable outcomes : 1 , 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 2 9, 31, 33, 35, 37, 39,
41, 43, 45, 47, 49 i.e., 25

Probability (E) = NToo.taolf favourable outcomes = 25
number of outcomes 49

(ii) A multiple of 5

Favourable outcomes : 5, 10, 15, 20, 25, 30, 35, 40, 45 i.e., 9

PArpoebrafbeiclittysq(uEa)r e = NToo.taolf favourable outcomes = 9
(iii) number of outcomes 49

Favourable outcomes : 1, 4, 9, 16, 25, 36, 49 i.e.; 7

Probability (E) = No. of favourable outcomes = 7 = 1
Total number of outcomes 49 7
(iv) An even prime number

Favourable outcome : 2 i.e., 1

Probability (E) = NToo.taolf favourable outcomes = 1
number of outcomes 49

OR

Similar solution as above only values are changed.

Ans. (i) 1 (ii) 5
23 46

6. All the black face cards are removed from a pack of 52 playing cards. The remaining cards are
well shuffled and then a card is drawn at random. Find the probability of getting a:

(i) face card. (ii) red card. (iii) black card. (iv) king. [CBSE Delhi 2014]

432 Xam idea Mathematics–X

Sol. Cards remaining after removing black face cards = red cards + black cards excluding face cards

Total number of possible outcomes = 26 + 20 = 46

(i) Face card

Favourable outcomes : 6 red face cards (king, queen and jack of d iamond and heart suits)

Probability (E) = No. of favourable outcomes = 6 = 3
Total number of outcomes 46 23

(ii) Red card

No. of favourable outcomes : 26 (13–13 cards of heart and diamond suits)

Probability (E) = Number of favourable outcomes = 26 = 13
Total number of outcomes 46 23

(iii) Black card

No. of favourable outcomes : 20 (10–10 cards of club and spade suits)

Probability (E) = Number of favourable outcomes = 20 = 10
(iv) King Total number of outcomes 46 23

No. of favourable outcomes : 2 (king of heart and diamond suits)

Probability (E) = Number of favourable outcomes = 2 = 1
Total number of outcomes 46 23

7. Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box,

find the probability that the number on the drawn card is: [CBSE Delhi 2014]

(i) an odd number. (ii) a perfect square number. (iii) divisible by 5.

(iv) a prime number less than 20.

Sol. No. of possible outcomes = 60 – 11 + 1 = 50.

(i) An odd number

Favourable outcomes : 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45,

47, 49, 51, 53, 55, 57, 59 i.e., 25

Probability (E) = Number of favourable outcomes = 25 = 1
Total number of outcomes 50 2

(ii) A perfect square number

Favourable outcomes : 16, 25, 36, 49 i.e., 4

Probability (E) = Number of favourable outcomes = 4 = 2
Total number of outcomes 50 25

(iii) Divisible by 5

Favourable outcomes : 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 i.e., 10

Probability (E) = Number of favourable outcomes = 10 = 1
Total number of outcomes 50 5

(iv) A prime number less than 20

Favourable outcomes : 11, 13, 17, 19 i.e., 4

Probability (E) = Number of favourable outcomes = 4 = 2
Total number of outcomes 50 25

8. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected

at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less

than 16. [CBSE (AI) 2016]

Sol. x can be any one of 1, 2, 3 or 4 and y can be any one of 1, 4, 9 or 16

Probability 433

Total number of cases of product of x and y = 16

Product less than 16 = (1×1, 1×4, 1×9, 2×1, 2×4, 3×1, 3×4, 4×1)

Number of cases, where product is less than 16 = 8
8 1
\ Required probability = 16 or 2

9. In Fig. 15.3, shown a disc on which a player spins an arrow twice. The function a is formed,
b
where ‘a’ is the number of sector on which arrow stops on the first spin and ‘b’ is the number

of the sector in which the arrow stops on second spin. On each spin, each sector has equal
a
chance of selection by the arrow. Find the probability that the fraction b > 1.

[CBSE (F) 2016]

Sol. For a/b > 1, when a = 1, b can not take any value,

a = 2, b can take 1 value,

a = 3, b can take 2 values,

a = 4, b can take 3 values,

a = 5, b can take 4 values, Fig. 15.3
a = 6, b can take 5 values.

Total possible outcomes = 36

\ P( a/b > 1) = 1+2+3+ 4+5 = 15 or 5
36 36 12

10. Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum (ii) even product. [CBSE (AI) 2017]

Sol. Total number of outcomes = 36 [(1, 1), (1, 2) ... (6, 6)]

Number of outcomes when sum is even = 18 [(1, 1), (1, 3) ...(6, 6)]

Number of outcomes when product is even = 27 [(1, 2), (1, 4) ... (6, 6)]

(i) P (even sum) = 18 = 1
36 2
27 3
(ii) P (even product) = 36 = 4

11. Reshma and Ritam are two friends studying in same class 10 of different school in Delhi. In

order to complete their holidays project work on historical monuments they decide to visit

Redfort. For this they decided to toss coin. If heads appear they decide to visit otherwise not.

For Redfort they toss coin separately and result of both was head. Redfort remains close on

Monday. So none of them visited Redfort. Now to decide day of visit from remaining days of

week, each of them throw a die. On the basis of outcome they went to Redfort.

(i) What is the probability of visiting Redfort on Monday?

(ii) What is the probability of not visiting Redfort on Monday?

(iii) What is the probability of visiting Redfort on same day of week?

(iv) What is the probability of visiting Redfort on different day of week?

(v) What is the probability of visiting Redfort on consecutive day of week?

Sol. Since the Redfort remains close on Monday, therefore only six days in a week to visit the Redfort.

(i) ∴ Probability of visiting Redfort on Monday = 0

(Since on Monday Redfort is closed)

(ii) Probability of not visiting Redfort on Monday = 1

(Since on Mondey Redfort is closed) 6 1
36 6
(iii) Probability of visiting Redfort on same day = =

434 Xam idea Mathematics–X

(iv) Probability of visiting Redfort on different day = 1– 1 = 5
6 6
(v) Favourable outcomes for consecutive day except Monday in a week are

{(Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}

and each pair has two choices i.e., if on first day Reshma goes and second day Ritam goes
or if on first day Ritam goes then second day Reshma goes.

Hence, total number of favourable outcomes = 10

and total number of possible outcomes = 36
10
∴ Required probability = 36

HOTS [Higher Order Thinking Skills]

1. A die is thrown twice. What is the probability that

(i) 5 will not come up either time? (ii) 5 will come up at least once?

Sol. (i) Favourable outcomes are

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)] i.e.; 25

Total outcomes = 36

\ Required probability = 25
36
(ii) Probability that 5 will come atleast once = 1 – P (5 will not come up either time)

   = 1– 25 = 11
36 36

2. Find the probability that in a leap year there will be 53 Tuesdays. [CBSE (F) 2017]

Sol. Leap year = 366 days = (52×7+2) days = 52 weeks and 2 days.

Thus, a leap year always has 52 Tuesdays.

The remaining 2 days can be:

(i) Sunday and Monday (ii) Monday and Tuesday (iii) Tuesday and Wednesday

(iv) Wednesday and Thursday (v) Thursday and Friday (vi) Friday and Saturday

(vii) Saturday and Sunday

Out of these 7 cases, we have Tuesdays in two cases.

\ P(53 Tuesdays) = 2 .
7

3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from

the bag is thrice that of a red ball, find the number of blue balls in the bag.

Sol. Let there be x blue balls in the bag.

\ Total number of balls in the bag = (5 + x) x
5+x
Now, P1 = Probability of drawing a blue ball =

P2 = Probability of drawing a red ball = 5
5+x

We are given that,

P1 = 3 × P2 ⇒ 5 x x = 3× 5 5 x
+ +

⇒ x(5 + x) = 15(5 + x) ⇒ x = 15

Hence, there are 15 blue balls in the bag.

Probability 435

4. Apoorv throws two dice once and computes the product of the numbers appearing on the dice.

Peehu throws one die and squares the number that appears on it. Who has the better chance

of getting the number 36? Why? [NCERT Exemplar]

Sol. Apoorv throws two dice once.

So, total number of outcomes, n(S) = 36.

Number of outcomes for getting product 36,

n(E1) = 1 [(6 × 6)] n (E1) = 1
\ Probability for Apoorv getting the number 36 = n (S) 36

Also, Peehu throw one die.

So, total number of outcomes n(S) = 6

Number of outcomes for getting square of a number as 36.

n(E2) = 1 ( 62 = 36) n (E2)
n (S)
\ Probability for Peehu getting the number 36 = = 1 = 6
6 36

Hence, Peehu has better chance of getting the number 36.

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) One card is drawn from a well shuffled deck of 52 cards. The probability that it is black
queen is

(a) 1 (b) 1 (c) 1 (d) 2
26 13 52 13

(ii) Which of the following can be probability of an event?

(a) 2 (b) –1 (c) 0.3 (d) 1.12

(iii) A die is thrown once. The probability of getting an even number is

(a) 1 (b) 1 (c) 1 (d) 1
3 6 4 2

(iv) The probability of throwing a number greater than 2 with a fair die is

(a) 1 (b) 2 (c) 1 (d) 3
3 3 4 5

(v) The probability of getting exactly one head in tossing a pair of coins is

(a) 0 (b) 1 (c) 1 (d) 1
2. Fill in the blanks. 3 2

(i) The probability of getting a spade card from a well shuffled deck of 52 cards is ________ .
(ii) Total number of events in throwing ten coins simultaneously is ___________ .
(iii) Three coins are tossed simultaneously the probability of getting all heads is ___________.
(iv) The letters of the word "SOCIETY" are placed at random. The probability of getting a

vowel is ___________.
(v) Probability of a sure event is ______________.

436 Xam idea Mathematics–X

QQ Very Short Answer Questions: [1 mark each]

3. A card is drawn from a deck of 52 cards. The event E is defined as ‘card is not a face card’. What
will be the number of outcomes favourable to E?

4. In a single throw of a die, what is the probability of getting a prime number?

5. A number is chosen from the numbers 1 to 50. What is the probability that the selected number
is multiple of 5?

6. The probability of getting a bad egg in a lot of 400 is 0.035. What is the number of bad eggs in
the lot?

7. A number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3. What is the probability
that |x|< 2?

8. A bag contains 3 black marbles, 5 red marbles and 6 white marbles. If a marble is picked at
random, then what is the probability that it is not a white marble?

9. Avni and Arushi draws one ball each from a bag containing 2 red and 3 green balls. Avni draws a
red ball first which is not put back. What is the probability that Arushi who draws next also gets
a red ball?

10. A letter of English alphabet is chosen at random. What is the probability that it is a letter of the
word ‘MATHEMATICS’?

11. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is

the number of rotten apples in the heap? [CBSE (AI) 2017]

12. A bag contains 3 red and 5 black balls. A ball is drawn at random from the bag. What is the

probability that the drawn ball is not red? [CBSE Delhi 2017 (C)]

QQ Short Answer Questions–I: [2 marks each]

13. Two different dice are tossed together. Find the probability: [CBSE 2018 (30/1)]
(i) of getting a doublet

(ii) of getting a sum 10, of the numbers on the two dice.

14. An integer is chosen at random between 1 and 100. Find the probability that it is:

(i) divisible by 8. (ii) not divisible by 8. [CBSE 2018 (30/1)]

15. The probability of selecting a blue marble at random from a jar that contains only blue, black
1
and green marbles is 5 . The probability of selecting a black marble at random from the same

jar is 1 . If the jar contains 11 green marbles, find the total number of marbles in the jar.
4

[CBSE 2019 (30/2/1)]

16. A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime

number. [CBSE 2019 (30/3/1)]

17. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the

probability that Poonam selects a card which is a multiple of 7? [CBSE 2019 (30/3/1)]

18. A bag contains 15 balls, out of which some are white and the others are black. If the probability
2
of drawing a black ball at random from the bag is 3 , then find how many white balls are there

in the bag. [CBSE 2019 (30/4/2)]

19. A card is drawn at random from a pack of 52 playing cards. Find the probability of drawing a

card which is neither a spade nor a king. [CBSE 2019 (30/4/2)]

20. Three different coins are tossed simultaneously. Find the probability of getting exactly one

head. [CBSE 2019 (30/4/3)]

Probability 437

21. A die is thrown once. Find the probability of getting (a) a prime number (b) an odd number.
[CBSE 2019 (30/4/3)]

22. A pair of dice is thrown once. Find the probability of getting (i) even number on each die (ii) a

total of 9. [CBSE 2019 (C) (30/1/1)]

23. A bag contains some balls of which x are white, 2x are black and 3x are red. A ball is selected at

random. What is the probability that it is (i) not red (ii) white [CBSE 2019 (C) (30/1/1)]

24. A bag contains 6 white balls numbered 1 to 6 and 4 red balls numbered 7 to 10. Find the
probability of getting a:

(a) red ball with even number on it.

(b) an odd number ball.
25. A card is drawn at random from a well-shuffled pack of 52 playing cards. Find the probability of

getting a red face card.
26. Two coins are tossed simultaneously. What is the probability of getting at least one head?
27. What is the probability of getting atmost one tail when two coins are tossed simultaneously?
28. A letter of English alphabet is chosen at random. Determine the probability that the letter is a

consonant.
29. If P(E) = 0.75, what is the probability of ‘not E’?
30. If probability of success is 63%, what is the probability of failure?
31. There are 30 cards of same size in a bag on which the numbers 1 to 30 are written. One card is

taken out of the bag at random. Find the probability that the number on the selected card is not
divisible by 3.

In a simultaneous throw of a pair of dice, find the probability of getting:

32. a doublet of even numbers.

33. an even number on one and a multiple of 3 on the other.

34. neither 9 nor 11 as the sum of the numbers on the faces.

35. A number is selected at random from first 50 natural numbers. Find the probability that it is a
multiple of 3 and 4.

36. Two different dice are rolled simultaneously. Find the probability that the sum of numbers

appearing on the two dice is 10. [CBSE (F) 2014]

37. Two different dice are tossed together. Find the probability: [CBSE (AI) 2014]
(i) that the number on each die is even.
(ii) that the sum of numbers appearing on the two dice is 5.

QQ Short Answer Questions–II: [3 marks each]

38. Two dice are thrown simultaneously. Find the probability of getting the sum [NCERT Exemplar]

(i) 9.

(ii) 1.

(iii) a prime number.

39. Two dice are thrown at the same time. Determine the probability that the difference of the

numbers on the two dice is: [NCERT Exemplar]

(i) 0.

(ii) 2.

40. Two different dice are thrown together. Find the probability that the product of the numbers

appeared is less than 18. [CBSE (F) 2017]

438 Xam idea Mathematics–X

41. A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score

is recorded. [NCERT Exemplar]

(i) How many different scores are possible?

(ii) What is the probability of getting a total of 7?

42. A bag contains white, black and red balls only. A ball is drawn at random from the bag. The

probability of getting a white ball is 3 and that of a black ball is 2 . Find the probability of
10 5
getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.

43. A die is thrown twice. What is the probability that:
(i) 3 will not come up either time?
(ii) 3 will come up at least once?

44. A bag contains 8 red, 7 orange and 9 green balls. A ball is drawn at random from the bag. Find
the probability that the drawn ball is:

(i) orange or green.
(ii) not orange.
(iii) neither green nor red.

45. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card
drawn is neither a black card nor a king.

46. All the jacks, queens and kings are removed from a deck of 52 playing cards and then well
shuffled. Then one card is drawn at random. If an ace is given a value 1, find the probability that
the card has a value:

(i) 5.
(ii) less than 5.
(iii) greater than 5.

47. A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag.

Find the probability that it bears:

(i) a two digit number.

(ii) a number which is a perfect square.

48. A carton of 24 bulbs contain 6 defective bulbs. One bulb is drawn at random. What is the

probability that the bulb is not defective? If the bulb selected is defective and it is not replaced

and a second bulb is selected at random from the rest, what is the probability that the second

bulb is defective? [NCERT Exemplar]

49. A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are

blue and rest are red. One piece is lost at random. Find the probability that it is a: [NCERT Exemplar]

(i) triangle.

(ii) square.
(iii) square of blue colour.

(iv) triangle of red colour.

50. Box a contains 25 slips of which 19 are marked ™1 and other are marked ™5 each. Box b contains

50 slips of which 45 are marked ™1 each and others are marked ™13 each. Slips of both the boxes

are poured into a third box and reshuffled. A slip is drawn at random. What is the probability

that it is marked other than ™1? [NCERT Exemplar]

Probability 439

51. A lot of 60 bulbs contain 12 defective ones. One bulb is drawn at random from the lot. What is the
probability that this bulb is defective? Suppose the bulb drawn in first attempt is 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?

52. Rajani and Kanika are friends. What is the probability that both will have:

(i) different birthdays?

(ii) the same birthday? (ignoring a leap year)

QQ Long Answer Questions: [5 marks each]

53. The King, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining

cards are well shuffled. A card is selected from the remaining cards. Find the probability of

getting a card [CBSE 2018 (C) 30/1]

(i) of spade

(ii) of black king

(iii) of club

(iv) of jacks

54. The probability of selecting a green marble at random from a jar that contains only green, white

and yellow marbles is 1 . The probability of selecting a white marble at random from the same
4

jar is 1 . If this jar contains 10 yellow marbles, what is the total number of marbles in the jar?
3

55. Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at

random from the remaining cards, after reshuffling them. Find the probability that the drawn

card is: [CBSE (AI) 2014]

(i) a king.

(ii) of red colour.

(iii) a face card.

(iv) a queen.

56. Cards numbered 1 to 30 are put in a bag. A card is drawn at random from this bag. Find the

probability that the number on the drawn card is:

(i) not divisible by 3.

(ii) a prime number greater than 7.

(iii) not a perfect square number. [CBSE (F) 2014]

57. A dice is rolled twice. Find the probability that: [CBSE (F) 2014]
(i) 5 will not come up either time.
(ii) 5 will come up exactly one time.

58. 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 number divisible by 5.
[CBSE (F) 2017]

440 Xam idea Mathematics–X

Answers

1. (i) (a) (ii) (c) (iii) (d) (iv) (b) (v) (d)

2. (i) 1 (ii) 1024 (iii) 1 (iv) 3 (v) 1
4 8 7

3. 40 4. 1 5. 1 6. 14 7. 3 8. 4
2 5 7 7

9. 1 10. 4 11. 162 12. 5 13. (i) 1 (ii) 1
4 13 8 6 12

14. (i) 6 (ii) 43 15. 20 marbles 16. (i) 1 (ii) 1 17. 5
49 49 3 2 34

18. Number of white balls=5 19. 9 20. 3 21. (i) 1 (ii) 1
13 8 2 2

22. (i) 1 (ii) 1 23. (i) 1 (ii) 1 24. (a) 1 (b) 1
4 9 2 6 5 2

25. 3 26. 3 27. 3 28. 21 29. 0.25 30. 37%
26 4 4 26

31. 2 32. 1 33. 11 34. 5 35. 2 36. 1
3 12 36 6 25 12

37. (i) 1 (ii) 1 38. (i) 1 (ii) 0 (iii) 5
4 9 9 12

39. (i) 1 (ii) 2 40. 13 41. (i) 6 (ii) 1 42. 3 , 50 balls
6 9 18 3 10

43. (i) 25 , (ii) 11 44. (i) 2 (ii) 17 (iii) 7 45. 7
36 36 3 24 24 13

46. (i) 1 (ii) 2 (iii) 1 47. (i) 81 (ii) 8
10 5 2 89 89

48. P (not defective) = 3 , P (2nd bulb defective) = 5 49. (i) 94   (ii) 5  (iii) 1   (iv) 5
4 23 9 3 18

50. 11 51. 1 , 48 52. (i) 364 (ii) 1 53. (i) 3 (ii) 1 (iii) 10 (iv) 3
75 5 59 365 365 49 49 49 49

54. 24 55. (i) 1  (ii) 1  (iii) 1  (iv) 1 56. (i) 2   (ii) 1  (iii) 5
12 2 6 24 3 5 6

57. (i) 25 (ii) 5 58. (i) 9 (ii) 1
36 18 10 5

Probability 441

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) Which of the following can be the probability of an event?

(a) – 0.04 (b) 1.004 (c) 18 (d) 8
23 7

(ii) A card is selected at random from a well shuffled deck of 52 playing cards. The probability
of its being a face card is

(a) 3 (b) 4 (c) 6 (d) 9
13 13 13 13

(iii) A bag contains 3 red balls, 5 white balls and 7 black balls. What is the probability that a ball

drawn from the bag at random will be neither red nor black?

(a) 1 (b) 1 (c) 7 (d) 8
5 3 15 15
(iv) When a die is thrown, the probability of getting an even number less than 4 is

(a) 1 (b) 0 (c) 1 (d) 1
4 2 6

2. Fill in the blanks. (3 × 1 = 3)

(i) The probability of an event that is certain to happen is _______________ and such an event
is called _______________ .

(ii) If the probability of an event is P, the probability of its complementary event will be
_______________ .

(iii) The probability of an impossible event is _______________ .

3. Solve the following questions. (3 × 1 = 3)

(i) In a game, a number is chosen at random from the set 1, 2, 3, ...28, 29, 30. What is the
probability that the number chosen is a product of exactly two different prime numbers?

(ii) What is the probability that an ordinary year has 53 Sundays?

(iii) The probability of getting a bad egg from a lot of 400 eggs is 0.035. What is the number of
bad eggs?

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. Two different dice are rolled simultaneously. Find the probability that the sum of numbers
appearing on the two dice is 10.

5. Two dice are thrown at the same time. Find the probability of getting:

(i) same number on both dice.

(ii) different numbers on both dice.

6. A coin is tossed 2 times. List the possible outcomes. Find the probability of getting:

(i) at least one head. (ii) atmost one head.

442 Xam idea Mathematics–X