Green C Math Class 10

EXERCISE 13.1

A. Very Short Questions

1. Find the mode of the following.

a. a, b, a, b, a, c, c, d, c, c, a, c b. 4, 5, 4, 4, 5, 6, 4, 6, 4

c. 3, 4, 4, 5, 5, 6, 5, 6, 7 d. 15, 16, 16, 16, 17, 18, 18, 19, 19, 18, 16, 18, 20

2. Find the mean (x) of the data:

a. 50, 60, 70, 80, 90 b. 10, 20, 12, 30, 18, 60
c. 8, 12, 6, 4, 14, 16 d. 20, 30, 40, 50, 60, 70

3. Write the tally marks for the following:

Number 8 12 20 7 3
Tally marks |||| |||

B. Short Questions
4. Find the mean (x) of the given distribution table:

a. Marks 10 20 30 40 50
No. 2 1354

b. Wage 220 320 420 520 620
No. of workers 10 9 12 11 7

5. Find the mean (x) of the data given:

a. Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Frequency 2 1 3 5 4

b. Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students 2 5 12 8 9

c. Wage (Rs.) 500 - 700 700 - 900 900 - 1100 1100 -1300 1300 - 1500

No. of workers 10 12 9 3 17

6. a. If Σfx = 600 and Σf = 30, find the mean (x).

b. If mean (x) = 12 and Σf = 30, find the value of Σfx.
c. If a = assumed mean = 20, Σfd = 40 and N = 10, find mean (x).
d. The mean of a, b, c and d is 12. What will be the mean of a, b, c, d and e if e = 50?

GREEN Mathematics Book-10 245

e. The mean weight of some boys is 62 kg and the sum of their weights is 744kg.
Find the number of boys.

f. If the value of Σfm = 170 + 10a, N = 34 + 2a, find the mean (x).

7. a. In a grouped data, if Σfm = 240 + k, mean (x) = 25 and Σf = 12, find the value of k.
b. Find the mean of prime numbers less than 20.
c. Find the mean of all the odd numbers between 90 and 100.
d. If Σfx = 60 + 45 p and Σf = 12 + 9p, find (x).
e. Find the value of x if the mean of 61, 67, x, 77, 79, 81 and 82 is 74.
f. If the mean of 6 observations x, x + 3, x + 6, x + 9, x + 12 and x + 15 is 18, find the
mean (x) of the first four observations.
g. If mean (x) = 16 and Σfm = 320, find the number of terms.
h. If mean (x) = 10 and Σfx = 700 + 5m and Σf = N = 40 + 3m, find the value of m.
i. If Σfx = 240 + 15p, N = 17 + p and mean (x) = 14.25, find the value of p.
j. The mean (x) of 3 different numbers is 5 and the mean (x) of 4 different numbers
is 12. Find the mean of 7 numbers.

D. Long Questions

8. a. Construct a frequency table of class interval 10 from the given data and find the
mean (x).

23, 5, 17, 28, 39, 52, 16, 21, 29, 25, 41, 33, 9, 19, 34, 59, 7, 11, 51, 31, 22, 41, 32, 55, 18

b. Find the mean of the following data making suitable class interval.
20, 22, 30, 10, 5, 9, 29, 32, 40, 47, 50, 52, 59

9. a. If the mean (x) of the data given below is 39, find the value of m.

Class 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
f 3 8 15 m 4

b. If mean (x) = 39 and m is unknown, find the value of m from the data given.

Class 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65
f 4 6 12 m 3

c. In the given distribution table, N = 20, and arithmetic mean (x)= 33.5. Find the
values of p and q.

x 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

f 4 p63q

246 GREEN Mathematics Book-10

13.2 Quartiles-Introduction

In the given figure, Q2 or median divides the 100%
whole part into two halves. They are called
lower half and upper half. Q1 is known as 25% 25% 25% 25%
first quartile as well as lower quartile. Q2 is Q1 Q2 Q3
known as second quartile as well as median.
Similarly, Q3 is known as third quartile as
well as upper quartile. As we observe the
figure we find 25% means Q1 50% means Q2
and 75% means Q3.

Let's learn how to find quartiles.

Worked Out EXAMPLES

EXAMPLE 1 Data are 10, 30, 40, 20, 60, 50. Find Q1, Q2, Q3.
Solution : We have data in random form.

At first we need to arrange the data in ascending order,

10, 20, 30, 40, 50, 60

∴ Number of data (N) = 6

To find first quartile:

∴ The position of Q1 = o N + 1 th
4
p item

= o 6 + 1 pthitem
4

= o 7 pthitem
4

= 1.75th item.

Hence, The first quartile (Q1) = 1st item + 0.75 (2nd item – 1st item)
= 10 + 0.75 (20 – 10)

= 10 + 0.75 × 10

= 10 + 7.5

= 17.5

To find second quartile (median):

∴ The position of Q2 = o N + 1 th
2
p item

GREEN Mathematics Book-10 247

= o 6 + 1 pthitem
2

= o 7 pthitem
2

= 3.5th item.

Hence, median (Q2) = 3rd item + 0.5 (4th item – 3rd item)

= 30 + 0.5 (40 – 30)

= 130 + 0.5 × 10

= 30 + 5

= 35

To find third quartile:
∴ The position of Q3 = o3(N4+ 1)pthitem
= o3 (6 + 1)pthitem

4
= o3 × 7pthitem

4
= 5.25th item.
Hence, The third quartile (Q3) = 5th item + 0.25 (6th item – 5th item)
= 50 + 0.25 (60 – 50)
= 50 + 0.25 × 10
= 50 + 2.5
= 52.5

EXAMPLE 2 Find Q1, Q2, Q3 from the data given.

Marks 20 30 40 50 60 70

No. of students 5 9 6 7 8 10

Solution : Marks (x) f c.f.
20 5 5

30 Add 14

9

40 6 20

50 7 27

60 8 35

70 10 45
N = 45

248 GREEN Mathematics Book-10

Here, N = 45

To find Q1 :

∴ The position of Q1 = o N + 1 th
4
p item

= o45 + 1pthitem
4

= o 46 pthitem
4

= 11.5th item

While finding the item, we need to search exact or nearest larger value
in c.f. column.
Hence, our item is 14.
Therefore, according to rule,
The first quartile (Q1) = 30

Now, to find Q2 : = o N + 1 th ∵Which is the corresponding value of
The position of Q2 14th term in marks.
2 p item

= o45 + 1pthitem
2

= o 46 th 23th item.

2 p item

As we did Q1, we do Q2 also,
∴ Our required item is 27.

Median, Q2 = 50

Similarly, to find Q3:

The position of Q3 = o 3(N + 1) th
4
p item

= o3(45 + 1)pthitem
4

= o 3 × 46 th

4 p item

= o 138 th 34.5th item.

4 p item

As we did Q1 and Q2, we do Q3 also,
∴ Our required item is 35.
Hence, Q3 = 60.

GREEN Mathematics Book-10 249

EXAMPLE 3 Find the Q1, Q2, Q3 from the following frequency distribution table.

Marks in maths 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

No. of students(f) 5 20 10 15 7

Solution : Marks (x) f c.f.
0 – 10 5 5

10 – 20 20 25

20 – 30 10 35

30 – 40 15 50

40 – 50 7 57

N = 57

Here, we have

N = 57

To find Q1 :

∴ The position of Q1 = o N th
4
p item

= o 57 th

4 p item

= 14.25th item
As we know, the nearest item of 14.25 in c.f. column is 25.

∴ Q1 class is 10 - 20

∴ Q1 lower limit (L) = 10 and N = 57

∴ Differences of class interval (i) = upper limit – lower limit

= 20 – 10

= 10
∴ Frequency (f) = 20
∵ Corresponding frequency to the class.

∴ Cumulative frequency (c.f.) = 5 ∵Above the item 25 in c.f. column

o N – c.fp
4
Q1 =L+q f × ir

Putting the value of L, c.f., f and i in the formula, we get,

14.25 – 5
The first quartile (Q1) = 10 + 20 × 10
= 10 + o 14.25 – 5 p

2

= 10 + 9.25 ∵Final result must not be greater than
2 upper limit.
= 14.62

250 GREEN Mathematics Book-10

To Find Q2 :

The position of Q2 = oNpthitem
2

= o 57 th

2 p item

= 28.5th item

∴ Required item is 35

Hence, Q2 lies in the class 20 – 30

∴ L = 20

i = 10

f = 10

c.f. = 25

Now, o N – c.fp

∴ Median (Q2) = L + q 2 × ir
f

= L + q 28. 5– 25 × 10r
10

= 20 + 3.5

= 23.5

Again, to find Q3 :

The position of Q3 = o 3N th

4 p item

= o 3 × 57 th

4 p item

= (3 × 14.25)th item

= 42.75th item

∴ Just greater than 42 . 75th item is 50.

∴ Q3 lies in the class 30 – 40
∴ L = 30

i = 10

f = 15 and c.f. = 35.

GREEN Mathematics Book-10 251

Now,

o 3N – c.fp
4
Q3 =L+q f × ir

42.75 – 35
= L + q 15 × 10r

= 30 + 7.75 × 10
15

= 30 + 77.5
15

= 30 + 5.16
= 35.16

EXAMPLE 4 If the lower quartile of the data given below is 16, find the value of x.
14, x, 18, 20, 22

Solution : Here, we have

Lower quartile Q1 = 16

No. of data N = 5

We know that,

The position of Q1 = o N + 1 th
4
p item

= o 5 + 1 th

4 p item

= o 6 th

4 p item

= 1.5th item.

Now,
∴ Lower quartile (Q1) = 1st item + 0.5 (2nd item – 1st item)
or, 16 = 14 + 0.5 (x – 14)
or, 2 = 0.5 (x – 14)
2 = 0 . 5x – 7
or, 0.5x = 9
∴ x = 18

252 GREEN Mathematics Book-10

EXAMPLE 5 Find the value of p in the given table, if Q2 (median) of the data is 30
and corresponding c.f. is 7.

Marks in maths 10 20 30 40 50

No. of students(f) 1 p 4 3 1

Solution : Marks (x) f c.f.
10 1 1
1+P
20 P 5+P
8+P
30 4 9+P

40 3

50 1
N=9+P

Here,

Q2 = 30
∴C orresponding

c.f. = 5 + p

i.e. 5 + p = 7

or, p = 7 – 5

∴ p = 2

EXAMPLE 6 Find the value of p in the table; if Q3 = 35.5.
Marks 0 – 10 10 – 20 20 – 30 30 – 40
f 6 p 12 10 40 – 50 50 – 60
4 2

Solution : Marks (x) f c.f.
0 – 10 6 6

10 – 20 p 6+p

20 – 30 12 18 + p

30 – 40 10 28 + p

40 – 50 4 32 + p

50 – 60 2 34 + p

N = 34 + P

GREEN Mathematics Book-10 253

Here, we have

The third quartile Q3 = 35.5, No. of item (N) = 34 + p
class is 30 – 40 since 35.5 lies between 30 - 40

∴ L = 30

i = 10

f = 10

c.f. = p

o 3N – c.fp
4
q Q3 = L + f × ir

o 3 (34 + p) – (18 + p) p

35.5 = 30 + q 4 ×10r

10

or, 5 . 5 = 3(34 + p) – 4 (18 + p)

4

or, 22 = 102 + 3p – 72 – 4p

or, 22 = 30 – p

∴ p = 8

EXAMPLE 7 Find the frequencies m and n from the following distribution if N =
100 and the median of the distribution is 30.

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

No. of students 10 m 25 30 n 10

Solution : Marks (x) f c.f.
0 – 10 10 10
10 + m
10 – 20 m 35 + m
65 + m
20 – 30 25 65 + m + n
75 + m + n
30 – 40 30

40 – 50 n

50 – 60 10

N = 75 + m + n

254 GREEN Mathematics Book-10

Here, we have
N = 100
And from the table,
N = 75 + m + n
i.e. 75 + m + n = 100
or, m + n = 25 ............................ (i)
As we know,
Q2 lies in the class 30 – 40
Q2 = Median = 30, so its corresponding class 30 - 40
∴ L = 30
i = 10

c.f. = 35 + m

Now, o N – c.fp

q Q2 = L + 2 × ir
f

o100 – (35 + m)p
30 = 30 + q 2 × 10r
30

or, 0 = (50 – 35 – m) 10
or, 0 = (15 – m)
∴ m = 15
Again, putting the value of m in
m + n = 25
or, 15 + n = 25
∴ n = 10
Hence, m = 15
n = 10

GREEN Mathematics Book-10 255

EXAMPLE 8 Find the lower quartile from the given data.

Class < 10 < 20 < 30 < 40 < 50
Class 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50
f
2 5 7 12 18

Solution : Marks (x) f c.f.
0 – 10 2 2
5
10 – 20 5 – 2 = 3 7
12
20 – 30 7 – 5 = 2 18

30 – 40 12 – 7 = 5

40 – 50 18 – 12 = 6
N = 18

The Position of Q1 = oN4 th

p item

= o18pthitem
4

= 4.5th item

∴J ust greater than c.f. of 4.5 is 5

∴Q 1 lies in class 10 – 20
∴ L = 10

i = 10

f = 3 and

c.f. = 2

Now, o N – c.fp

qT he first quartile (Q1) = L + 4 × ir
f

o 18 – 2 p

= 10 + q 4 1 × 10r

3
= 10 + 8.33
= 18.33

256 GREEN Mathematics Book-10

EXAMPLE 9 Find the median from the data given.

Class 0 – 10 10 – 30 30 – 40 40 – 70 70 – 80
f 28 35 1

Solution : Arranging the class with frequency.

Marks (x) f c.f.
0 – 10 2 2

10 – 30 8 10

30 – 40 3 13

40 – 70 5 18

70 – 80 1 19

N = 19 `

Now,

The position Median = o N th

2 p item

= o19pthitem Remember :
2
When class interval is different,

= 9.5th item we can calculate adjustment fre-

∴ Greater than c.f. is 10 quency to calculate median.

∴ Q2 lies in class 20 – 30 Median value can also be calculat-
ed by using upper limit.
L = 20
Let's take example 9
i = 20
Where,
f = 5
c.f = 5 Md = U – q o c.f – Np × ir
2
f
Now, o N – c.fp
[Here, cf and f lies in median class]

q Q2 = L + 2 r× i Md = 30 – 10–9.5 × 10
f 5

= 10 + 9.5–2 × 20 = 30 – 0.5 × 10
8 5

= 10 + 7.5×20 = 30 – 5
8 5

= 29

= 10 + 37.5
2

= 10 + 18.75

= 28.75

GREEN Mathematics Book-10 257

EXERCISE 13.2

A. Very Short Questions
1. Arrange in ascending order.

a. 5, 10, 12, 18, 20, 30, 22

b. x + 1, x + 2, x + 5, x + 7, x + 3, x + 4

2. Write the formula to find following values for the continuous series.

a. Q1 b. Q2 c. Q3

3. Solve the following:

a. Find the lower quartile of the data 10, 20, 12, 8, 22, 6

b. Find the value of x in data given, if median of the data is 12.

x, x + 1, x + 2, x + 3, x + 4

c. If Q3 = 15, find the value of x.
1, 5, 7, 2x – 4, x + 7, 2x + 1 and 3x + 2 are the given data.

d. 2x + 1, 3x + 1, 3x + 5, 5x – 7, 63 and 70 are in the ascending order and Q1 = 20, find
the value of x .

e. Find 25% of quartile from the data 100, 108, 102, 110, 106, 104.

4. a. If Q1 class is 15 – 20, cf of the preceding class is 7, corresponding frequency is 6
and Q1 = 16.46, find the total number of items.

b. If Q3 class is 20 – 25, cf of the preceding class is 25, corresponding frequency is
10 and Q3 = 22.5, find the total number of items.

c. Construct a frequency distribution table from the data given .

5, 10, 5, 10, 20, 22, 22, 24, 10, 2, 10, 25

d. What is the mid value of class interval of 20 – 40?

e. How many frequencies are there altogether if the class 0 – 10 considered as ex-
cluding type?

f. In the given table, find the marks having maximum frequencies.

Marks 10 20 30 40 50 60
f 2 1 3 10 4 2

g. Out of 25%, 50%, 75% and 100%, which one is equivalent to median? 700
h. Find the cumulative frequencies of the table. 2

Wage 100 200 300 400 500 600

Workers 1 3 2 1 5 3

258 GREEN Mathematics Book-10

i. Find the class of Q3.

Marks 0-10 10-20 20-30 30-40 40-50 50-60
f 10 4 2
213

B. Short Questions

5. a. Find the median from the data given.

Marks 20 30 40 50 60 70
f 213254

b. Arrange the classes with frequencies.

Class More More More More More More
f than 10 than 20 than 30 than 40 than 50 than 60

20 15 12 7 4 2

c. Arrange the classes with frequencies.

Class Less than 10 Less than 20 Less than 30 Less than 40 Less than 50
f 2 5 9 12 14

d. Find the value of x if N = 18 from the data given.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
f 1 x 2 4 3 2

6. a. 9, 11, 12, 14, 18, x + 2 and 30 are in ascending order. If third quartile of this data
is 22, find the lower quartile.

b. Find the median class of the given data :

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
f 5 5 20 4 2

c. Find the class for upper quartile from the data given.

Class 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
f 2 3 1 4 2

d. Construct frequency distribution table with a class 0 – 5 from the data given

1, 0, 5, 2, 10, 6, 7, 9, 11, 17, 22, 29, 33, 42

e. Construct cf of cumulative frequency table for the data given below:

10 20 30 40 50 60 70 80

25132412

GREEN Mathematics Book-10 259

C. Long Questions

7. a. Find the lower quartile from the data:

Class 0 – 15 15 – 30 30 – 45 45 – 60 60 – 75
f 1 2 5 3 1

b. Find the median of the data using upper limit formula :

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
f 2 5 3 7 6

c. Find the lower quartile from the table given below:

x 50 – 70 70 – 90 90 – 110 110 – 130 130 – 150

f 5 10 20 4 2

d. If the Q3 of the data is 60, find the value of m.

Class 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
f 4 5m8 7 4 5

e. In the following distribution table median is 32.5. Find the missing frequency

x 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65
20 6
f 36 ? 46

8. a. Find the upper quartile from the data given:

x 0–4 4–8 8 – 12 12 – 16 16 – 20
20 22
f 3 8 17

b. The median of the data is 24. Find the value of x.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
f 4 12 x 9 5

c. The median of the data is 32. Find p.

Class 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65
f 5 8 P 9 7 1

d. Find the missing frequencies from the data, if N = 60 and Q2 = 28.5.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
f 5 x 20 15 y 5

e. Find the minimum marks obtained by above 75% students from the data given.

x 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100

f 12 20 25 18 5

260 GREEN Mathematics Book-10

f. If Q1 = 11372, find ‘m’ from the data.

Marks 6–8 8 – 10 10 – 12 12 – 14
m 100
f 85 65

g. Find lower quartile from the data given below:

Class 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50
f 2 5 9 11 13

h. If Q3 = 60, find the value of w.

x 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
f 4 5w8 766

i. From the marks distribution of 30 students, 2 frequencies are missing. However
median marks is known as 45. Find the values of missing frequencies.

Marks 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
f 3 ? 10 ? 4

13.3 Cumulative frequency curves or ogives

Ogives or cumulative frequency curves help to determine as well as portray the number
or proportion of cases above or below the given value. Medians, first quartile and third
quartile can be estimated when the cumulative frequency of a grouped data is given.

There are two method of constructing an ogive.

They are:

a. The “Less than” method

b. The “More than” method

a. Less than method:

In this method, we start with the upper limits of the classes and go on adding the
frequencies. When these frequencies are plotted on the graph we get ogive.

For Example:

Marks class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

f 21341

GREEN Mathematics Book-10 261

f cf Before constructing ogive, we need to make
Marks class 2 coordinates with upper limit of the class and its
2 corresponding c.f.
Less than 10

Less than 20 1 3

Less than 30 3 6 So, the coordinates are (10, 2), (20, 3), (30, 6),
(40, 10) and (50, 11)

Less than 40 4 10
Now, plotting the coordinates on the graph as
Less than 50 1
11 shown below:
Y

16 The corresponding value of the last point of ogive is the
14
12 total number of frequency. From the graph N = 11. Here,
10
Frequency 8 Q1 = oNpth = o 11 th = 2.75th items
6 Q2 4
4 X Q3 4 p
2
= oNpth = o 11 th = 5.5th items
O 2
10 20 30 40 50 60 70 2 p
(Marks)
= o 3N th = o 33 th = 8.25th items

4 p 4 p

b. The “More than” method

In the "more than" method we start with the lower limits of the classes and from the total
frequency we subtract the frequency of each class. When these frequencies are ploted we
get a declining curve instead of rising curve.

Q1 = o 3N th

Q2 4 p

Q3 = o N th
For Example:
2 p

= oNpth
4

Marks class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
f 4 6 20 10 7 3

Marks class f cf Before constructing ogive, we need to make coordi-
Less than 10 4 nates with lower limit of the class and its correspond-
50
ing c.f. Y

Less than 20 6 50 – 4 = 46 Frequency 60
So, the coordinates are (0, 50
40
Less than 30 20 46 – 6 = 40 50), (10, 46), (20, 40), (30, 30
Less than 40 10 40 – 20 = 20 20), (40, 10) and (50, 3)
20

Less than 50 7 20 – 10 = 10 Now, Plotting the coor- 10

dinates on the graph as O X
shown alongside: 10 20 30 40 50 60 70
Less than 60 3 10 – 7 = 3
(Marks)

262 GREEN Mathematics Book-10

Combination of less than and more than method

We learnt the construction of less than and more than ogives. Now the combination of
these both ogives is known as combined ogives. Individually as the proper rule of less
than and more than method, we need to draw both the ogives. In this ogive we need
to find only the median (Q2) and related class. The intersection of both the ogives is the
vital-point. When we draw the vertical line through this vital point, it cuts x – axis. The
corresponding value is median and where does it lie. The class is the proper class for
median.

For example:

Marks class 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
f 10 30 60 40 10

c.f. Marks c.f. Now, we need to construct both the
Marks ogives as shown below,
Less than 20 10 0 or more 150
Y

Less than 40 40 20 or more 140 Cumulative Frequency 160
140
Less than 60 100 40 or more 110 120
100
Less than 80 140 60 or more 50
80
Less than 100 150 80 or more 10 60
40
Here median = 60, its class is 60 – 70. 20 X

O

20 40 60 80 100
(Marks)

Worked Out EXAMPLES

EXAMPLE 1 Find the first quartile and class for lower quartile from the given ogive.
Solution :
As we know, corresponding value of last point of ogive in y-axis is total
number of frequency.

∴ The position of Q1 = o N th = o 20 th = 5th item.
4 4
p item, p item

Now, passing a horizontal line through Y

5 of y–axis and parallel to the x-axis as Frequency 20
shown on graph, we get the next point 16
on ogive. Again a vertical line is drawn 12 X
through the point on x-axis, which lies in 8
class 10 – 20. 4

Hence, class for Q1 = 10 – 20 O
10 20 30 40 50
(Marks)

GREEN Mathematics Book-10 263

EXAMPLE 2 Find the class for third quartile from the given ogive.
Solution :
Y

N = 20 20
18
The position of Q3 = o N th in more than method,
4 16
p item 14
12
= o 20 th Frequency 10
8
4 p item 6
4
= 5th item. 2

∴ Q3 class = 40 – 50 O X
10 20 30 40 50
(Wages Rs.)

EXAMPLE 3 Draw the less than ogive from the following data and answer the
Solution : questions.

Wages class 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
f 5 8 10 9 3

i. Find the number of workers having wages more than 40.

ii. Find the number of workers having wages less than 80.

iii. Find the number of workers having wages between 80 – 100.

Wages f cf Frequency Y
0 – 20 55
20 – 40 8 13 35
40 – 60 10 23 30
60 – 80 9 32 25
80 – 100 3 35 20
15
Here, 10
5
We have the coordinates to draw less
than ogive: (20, 5), (40, 13), (60, 23), OX
(80, 32) and (100, 35) 20 40 60 80 100 120
(Wages Rs.)

i. No of workers having wages more than 40 = 22
ii. No. of workers having wages less than 80 = 32
iii. No. of workers having wages between 80 and 100 = 3

264 GREEN Mathematics Book-10

EXERCISE 13.3

A. Very Short Questions
1. a. Construct less than ogive from the data given.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
10 4
f 213

b. Construct more than ogive form the data given:

Wage (Rs.) 0 – 20 20 – 40 40 – 60 60 – 80

No.s 3521

c. Construct less than and more than ogive individually from the data given.

Marks 20 – 40 40 – 60 60 – 80 80 – 100
f 5 3 7 6

2. a. Draw the less than ogive of the data and answer the following questions.

Wage 0 – 50 50 – 100 100 – 150 150 – 200 200 – 250 250 – 300
6
f 12 18 27 20 17

i. Find the total number of workers.
ii. Find the number of workers who got less than Rs. 250.
iii. Find the median of the wages.
iv. Find the number of workers who got more than Rs. 100.

b. The wages of 260 workers in a factory are given below:

Wages 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
f 15 30 40 60 75 28 12

i. Construct less than and more than ogive separately.
ii. Find out the median, Q1 and Q3 from both the ogives.

B. Short Questions

3. Find the class for Q1, Q2 and Q3 from the given ogives.

Y Y

a. b.

Frequency12 60
Frequency1050
8 40
6 X 30
4 20
2 10

O OX
10 20 30 40 50 20 40 60 80 100
(Marks) (Wages Rs.)

GREEN Mathematics Book-10 265

c. Y Y

Frequency 28 Frequencyd.
24
20 X 16 X
16
12 14
8 12
4 10
8
O 6
10 20 30 40 50 60 4
(Wages Rs.) 2

O
40 80 100 120140 180
(Wages Rs.)

4. Find the median and class for median from the given curve.

Y Y

a. b.

28 35
24
Frequency 20 Frequency 30
16
12 25
8
4 20

OX 15
20 40 60 80 100
(Marks) 10

5

O X
50 100 150 200250300

(Wages Rs.)

Project Work:

Visit 15 families of your friends and ask their parents about income and expenditure.
Make data chart of the expenses and change it to the class interval.
1. Calculate the mean, median and mode of the data.
2. Draw the pie-chart, bar-diagram, less than curve and more than curve.
3. Present your work in the class.

266 GREEN Mathematics Book-10

14

Probability

Estimated Teaching Periods : 11

The pascal is the SI derived unit of pressure used to quantify internal
pressure, stress, Young's modulus and ultimate tensile strength. It is
defined as one newton per square meter. It is named after the French
polymath Blaise Pascal.

Contents

8.1 Review/Introduction
8.2 Tree diagram

Objectives

At the end of this unit, students will be able to:
introduce probability, know the types of probability
define probability scale
state the law of probability
solve the problems related to probability

Materials

Coin, dice, playing cards, spinner, balls of different colours, bag, thumb pins, number
cards, etc.

GREEN Mathematics Book-10 267

14.1 Review/Introduction

'Probability' and 'Chance' are very commonly used words in our day - to day conversation.
These words are used in the same sense. For example, we say that there is probability that
I may not be able to come to the party. It is likely to rain today. Similarly, we say that
they have little chance of winning the football match. All these events are not certain.
Probability theory is a concept which numerically measures the degree of uncertainty.
Probability is the measure of the chance that a particular event will happen.

There are two possibilities – either an event may take place or it may not take place. The
probability of two possibilities never exceed one. If it is sure that an event can never take
place the probability is 0 and if it is sure to happen, its probability is 1. Chance between
absolute certainty and impossibility lies between 0 and 1.

0 1

Impossibility Absolute certainity
(events never take place) (events always take place)

Some basic terms used in probability

Experiment:

An activity or operation which produces result is called an experiment. For example,
when we toss a coin either a 'Head' or a 'Tail' appears on the top.

Random experiment:

An experiment whose outcome cannot be predicted in advance is called a random
experiment. For example, the event of occuring head or tail cannot be predicted
before tossing a coin.

Trial and outcome (event) :

Performing a random experiment is a trial and result of it is an event or outcome.
Rolling a dice is a trial and getting 1, 2, 3, 4, 5, or 6 is an event.

Equally likely events:

Events of an random experiment which have equal chance of occurrence are called
equally likely events. For example : When a die is rolled, the outcomes 1, 2, 3, 4, 5 and
6 are equally likely to appear.

Sample space:

The set of all possible outcomes of a random experiment is called sample space. The
sample space is generally denoted by 'S'.

268 GREEN Mathematics Book-10

For example:

a. When a unbiased coin is tossed once, sample space (S) = {H, T}

b. When a coin is tossed two times in succession, S = {HH, HT, TH, TT}

Exhaustive event:

The total number of all possible outcomes of a random experiment is called the
exhaustive event.

Example : In a deck of playing cards, the total possible outcomes= 52.

Favourable events:

The events (outcomes) of a random experiment which are expected are called
favourable events.

For example : While rolling a dice, S = {1, 2, 3, 4, 5, 6} and the favourable events
getting prime number are 2, 3, and 5.

Mutually exclusive events :

Two or more events of a random experiment are said to be mutually exclusive

events if only one of them can occur in a single trial.

For example, in rolling a dice once, the occurrence of 1 A B

excludes the other events 1, 2, 3, 4, 5 and 6. So in rolling

a dice once, the events 1, 2, 3, 4, 5, and 6 are mutually

exclusive events.

Thus A and B are two events of an experiment and
(A∩B) = φ then A and B are mutually exclusive events.

Addition Theorem of Probability for mutually exclusive events

If A and B are two mutually exclusive events, the probability of the occurrence of at least

one event A or B. (i.e A or B) is the sum of the individual probabilities of the occurrence

of A and B. AB

Symbolically, P(A∪B) = P(A or B)

= P(A) + P(B)

Similarly, if A, B and C are the mutually exclusive events, then P(A∪B) = P(A) + P(B)
P(A∪B∪C) = P (A) + P (B) + P (C) .

Example:

Probability of getting a king or a queen from a pack of 52 cards = P(K) + P(Q)

= 542 + 4
52

= 8
52

=2
13

GREEN Mathematics Book-10 269

Non-mutually exclusive events

Two events A and For example:
B of an experiment
are said to be non Let A be the event 'the number is SA B
mutually exclusive prime number' and B be the event
events if A∩B ≠ 0. 'the number is odd number' in 3
rolling a dice. 251
46
Then A = {2, 3, 5}, B = {1, 3, 5} and
(A∩B) = {3,5} A∩ B ≠ 0

∴ A and B are non-mutually exclusive events.

Here, Probability of A or B AB
= P(A∪B)
= P(A) + P(B) – P (A∩B)

Independent Events P(A∪B) = P(A) + P(B) – P(A∩B)

If the occurrence or non- For example:
occurrence of one event does When two coins are tossed together, the
not affect the occurrence occurrence of Head or Tail on the first coin does
or non - occurrence of the not affect the occurrence of Head or Tail on the
other, then the events are second coin. So tossing two coins together are
independent events. independent events.

Multiplication Law of probability for independent events 2nd H T
1st
If two coins are tossed together, we get the following outcomes of
H HH HT
these two independent events. {HH, HT, TH, TT}
T TH TT
Here, P (H) = 1 , P (T) = 1
22

and P(HH) = 1 = 1 × 1 = P(H) × P(H)
422

P(HT) = 1 = 1 × 1 = P(H) × P(T)
422

P(TH) = 1 = 1 × 1 = P(T) × P(H)
422

P(TT) = 1 = 1 × 1 = P(T) × P(T)
422

∴ If A and B are two independent events, then P(A∩B) = P(A and B) = P(A) × P(B)

270 GREEN Mathematics Book-10

8.2Dependent events

If the occurrence For example:
of one event affects
the other or is If second ball is drawn out of a bag containing 3 red and 5
affected by other, green balls of same size with the replacement of the first ball
then the events are drawn, it is the case of independent events. If the second
dependent events. ball is drawn without the replacement of the first ball, it is
the case of dependent events.

Die

A die is a small cubical solid object used in gambling or playing game. Its six surfaces are
marked with 1, 2, 3, 4, 5, and 6.

Playing cards (52 cards)

Red cards (26) Black cards (26)

♥ Heart (13) ♦Diamond (13) ♠ Spade (13) ♣ Club (13)

Face cards (Picture cards) : 12

Ace : 4
Jack : 4
Queen : 4
King : 4

Remember :
Normal year ➟ 365 days = 52 weeks + 1 day
Leap year ➟ 366 days = 52 weeks + 2 days

GREEN Mathematics Book-10 271

Worked Out EXAMPLES
EXAMPLE 1 When a card is drawn at random from a pack of well-shuffled 52

cards, find the probability of getting (i) a king (ii) not getting king.
Solution : Sample spaces = total cards n(S) = 52

Number of favourable events = no of king , n (k) = 4

Now, P (K) = n(K)
n(S)

P (K ) = 4
52

=1
13

= 1 – P(K)

=1– 1
13

= 12
13

EXAMPLE 2 What is the probability of giving a child birth on Tuesday by a

pregnant woman?

Solution : Here,
Sample space, S = {Sun, Mon, Tue, Wed, Thus, Fri, Sat}
n(s) = 7
Favourable event, = {Tuesday}

∴ n (T) = 1

Now, P (T) = n(T)
n(S)

= 1
7

∴P robability of giving a child birth on Tuesday is 1 .
7

272 GREEN Mathematics Book-10

EXAMPLE 3 From a pack of 52 cards, a card is drawn at random. What is the

probability of getting a king or a queen.

Solution : Here,

Total events = total cards, n(S) = 52

Number of king, n(K) = 4

Number of queen n(Q) = 4

Now,

P (K or Q) = P(K) + P (Q) ∵ King and Queen are mutually

= 4 + 4 exclusive events.

52 52

= 8
52

= 123
∴ Probability of getting king or queen = 2

13

EXAMPLE 4 There are 4 red, 3 green and 2 white identical balls in a bag. If a ball

Solution : is drawn randomly from the bag, find the probability that it may be

either red or green.

Here, exIctluissitvheeecvaesnetsof mutually
The number of red balls n (R) = 4
T he number of green balls, n(G) = 3
The number of white balls, n(W) = 2

The number of total balls, n(S) = 4 + 3 + 2

= 9

Now, P(R or G) = P(R) + P(G)

= 4 + 3
9 9

= 7
9

GREEN Mathematics Book-10 273

EXAMPLE 5 From the number cards, numbered from 7 to 27 a card is drawn at

random. Find the probability of getting a card having a prime num-
ber or an even number.

Solution : Here,

Total number cards, n(S) = 21 (27 – 6)

Set of prime number (P) = {7, 11, ,13, 17, 19, 23}

∴ n(P) = 6

Set of even number,(E) = {8, 10, 12, 14, 16, 18, 20, 22, 24, 26}

∴ n(E) = 10
Now,
P(P or E) = 6 + 10
21 21

= 16
21

EXAMPLE 6 From the number cards numbered from 1 to 30, a card is drawn at

random. Find the probability that the number is either divisible by
4 or by 5.

Solution : Here, The total card, n(S) = 30

Set of numbers divisible by 4, (D4) = {4, 8, 12, 16, 20, 24, 28}
n(D4) = 7

Set of numbers divisible by 5, (D5) = {5, 10, 15, 20, 25, 30}

n(D5) = 6

Iat lilsytehxeccluasseivoe fenvoenn-tms.utu- n((DD44 ∩ D5) = {20}
∩ D5) = 1

Now, n(D4 ∪ D5) = P(D4) + P (D5) – P(D4 ∩ D5)

= 7 + 6 – 1
30 30 30

= 12
30

= 2
5

274 GREEN Mathematics Book-10

EXAMPLE 7 Two dice are thrown together. What is the probability that the sum

of the numbers on the two faces is either less than 6 or divisible
by 4?

Solution : Here, Total events = 36

Let A = {set of pair of faces whose sum is less than 6}

and B = {set of pair of faces which is divisible by 4}

Thus,

A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), 3,2)}

∴ n(A) = 9
B = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6) }

∴ n(B) = 9
(A∩B) = {(1, 3), (3, 1)}

n(A∩B) = 2

2nd 1 23 4 56
1st

1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Total outcomes when two dice are thrown together

Now,

P(A or B) = P(A) + P (B) – P (A∩B)

= 9 + 9 – 2
36 36 36

= 16
36

= 4
9

∴ The probability that the sum of the numbers on the two faces is

either less than 6 or divisible by 4 = 4
9

GREEN Mathematics Book-10 275

EXAMPLE 8 The probability that A gets scholarship is 80% and B will get is
90%. What is the probability that at least one of them gets the
Solution : scholarship?

Here,

P(A) = 80% = 80 = 0.8
100

P(B) = 90% = 90 = 0.9
100

P(A∩B) = P(A) × P(B)
Now,
= 0.8 × 0.9

= 0.72

P(A∪B) = P(A) + P(B) – P(A∩B)

= 0.8 + 0.9 – 0.72

= 1.7 – 0.72
= 0.98

= 98%

∴ The probability that at least one of them gets the scholarship = 98%


EXERCISE 14.1

A. Very Short Questions
1. a. Define probability.

b. Write the probability of certain and impossible events.

c. What is the probability of an event other than sure and impossible event?

d. If the probability of an event is 2 , find the probability of not happening of the
4
events.

e. Write the sample space when a die is tossed twice.

f. Define independent and dependent event with examples.

2. a. In the given Venn-diagram 'S' represents sample S Q
space and 'P' and 'Q' are events of S. What types of P B
events are P and Q? Also write the probability of P or
Q.

b. From the given figure find the probability of A or B if A S
A
and B are the events of sample space S.

276 GREEN Mathematics Book-10

3. a. If A, B and C are the mutually exclusive events, write the probability of the
occurrence of either A or B or C.

b. If X and Y are non-mutually exclusive events, what is the probability of the
occurrence of either X or Y?

c. If P(A) = 2 , P(B) = 1 and A and B are mutually exclusive events, find (i) P(A∪B)
99

and (ii) P (A∪B).

d. A and B are mutually exclusive events. If P(A∪B) = 70% and P(A) = 40%, find
P(B).

e. M and N are non-mutually exclusive event. If n(M) = 4, n(N) = 3, n(M∩N) = 1
and n(S) = 15, find P(M∪N).

4. a. A bag contains some balls of red and white colour. A ball is drawn from the bag
and it is replaced to draw another ball.

i. Are the probabilities of outcomes independent?

ii. Find the probability of getting red balls in both draw if number of red balls
is 3 and white balls is 5.

b. If A, B and C are independent events, find P(A∩B∩C)

B. Short Questions

5. a. A and B are two mutually exclusive events. If P(A) = 50% and P(B) = 30%, find
P(A∪B) and P (A∪B).

b. A and B are two mutually exclusive events. If 3P(A) = 2P(B) = 0.6, find P(A∪B)
and P (A∪B).

6. a. A and B are two non-mutually exclusive events, if P(A) = 0.5, P(B) = 0.4, P(A∩B)

= 0.2, find P(A∪B) and P (A∪B).

b. If P(X) = 0.6, P(Y) = 0.5 and X and Y are non-mutually exclusive events find
P(X∩Y).

7. There are 4 green and 6 white balls in a bag. Two balls are drawn randomly one
after another with the replacement of the first ball. Find the probabilities of the
following events :

a. both are green b. both are white c. balls of same colour
e. none of them are green.
d. the first is green and the second is white.

8. There are two bags. One contains 5 red 4 white and another contains 6 red and 3
white. If one ball is taken out from each bag find the probability of getting :

a. both red b. both white c. the same colour.

GREEN Mathematics Book-10 277

9. Two cards are taken out one after another from a well-shuffled pack of 52 cards. If
the first card drawn is replaced, find the probability of the following.

a. both are king b. both of them are not king

c. one is an ace and another is a queen

d. one is a king and another is not queen

10. Two cards are drawn randomly one after another. If the first card is not replaced,
find the probability of getting king in both drawing.

11. A coin is tossed two times in succession. Find the probability of the following
events;

a. both are head b. head and tail in order c. at least one head
d. at least one tail e. both are not head

12. Two dice are rolled two times. Find the probability of getting

a. 1 in first and 6 in second.
b. not 1 in first and not prime number in second.
c. not getting 5 in both drawing.

13. A fair coin is tossed and an unbiased dice is rolled. Find the probabilities of the
following events.

a. 1 and the head b. 1 and head or tail

14. a. There are 5 red, 4 green and 3 white balls in a bag. A ball is drawn at random
from the bag. Find the probabilities of the following events.

i. the ball is either red or green. ii. the ball is neither red nor green.

iii. the ball is red or white. iv. the ball is not green

b. A card is drawn from the set of number cards from 8 to 27. Find the probability
that the card drawn is exactly divisible by 6 or 4.

c. Find the probability of getting a card having a prime number or a square number
when the card is drawn randomly from the number cards, numbered from 3 to
32.

d. From the number cards numbered from 1 to 30, a card is drawn at random. Find
the probabilities of the following events.

i. The number is either exactly divisible by 5 or 9.
ii. The number is either a multiple of 7 or a multiple of 8.
iii. The number is a cube number or a square number.

278 GREEN Mathematics Book-10

15. A card is drawn at random from a well-shuffled pack of 52 cards. Find the
probabilities of the following events.

a. It is a king or queen.
b. It is either a king or black queen.
c. It is either a black king or a red king.
d. It is either a spade or a heart.
e. It is either a faced card or an ace.
f. It is neither an ace nor a queen.

16. A card is drawn out from a pack of 52 cards. Find the probabilities of the following
events.

a. It is a king or a red card. b. It is either a heart or a jack.

c. It is either a faced card or a red card. d. It is not a king nor a black card.

17. Find the probabilities of the following when a fair die is rolled.

a. Getting 1 or 3 or 5.
b. Getting a number less than 3 or 6.
c. Getting either 1 or a prime number.
d. Getting a prime number or a squared number.
e. Not getting 1 or 6.

18. a. A die is rolled once. Find the probability of getting a multiple of 2 or a multiple
of 3.

b. a prime number or an even number.

19. Two unbiased dice are rolled together. Find the probability that:

a. The faces have the same numbers.
b. The sum of the numbers on the two faces is either exactly divisible by 5 or by 6.

20. A problem of Mathematics is given to two students A and B. Their chance of
solving the problem are 1 and 1 respectively. What is the probability that the
23
problem will be solved?

21. A speaks truth in 75% cases and B in 80% cases. In what percentage of cases they
are likely to contradict each other in stating the same fact.

GREEN Mathematics Book-10 279

14.2 Tree diagram

The tree diagram is the diagrammatical representation of all the possible outcomes of
series and their respective probabilities of a random experiment. It is used to find the
combined occurrence of two or more simultaneous events. Every branch of tree diagram
shows the probability of respective event of the random experiment.

The probability of independent and dependent events can be calculated by using their
tree diagram. Let's study the following examples:

Worked Out EXAMPLES

EXAMPLE 1 If a coin is tossed two times, show the probabilities of all events in
a tree diagram.
Solution :
Let, H and T represent head and tail of a coin.

n(H) = 1 [H = head]

n(T) = 1 [T = Tail]

n(S) = 1 + 1 = 2

Tree diagram: Sample space Probabilities

= 1 2 H HH 1 × 1 = 1
T HT 2 2 4
independent P(H ) H TH
H TT
T 2

= 1 2 H

Iteivsetnhtes. case of P(H ) P(T2) 1 × 1 = 1
2 2 4
1

= 1
2

P(H2) = 1 1 1 1
2 2 2 4
P(T
1 ) = 1 T × =

2 P(T
2
) = 1 T

2 1 1 1
2 2 4
First toss Second toss × =

EXAMPLE 2 Two cards are drawn randomly one after another without replace-
ment of the first card. Find the probability of getting king in both
Solution : draw by drawing a tree diagram.

Let K and K represent king and non-king cards respectively.
Total cards, n(S) = 52
Number of kings cards, n(K) = 4
Number of other cards, n( K) = 52 – 4 = 48

Tree diagram: (Showing all the events and their respective probabilities)

280 GREEN Mathematics Book-10

3 K
K
Iteivsetnhte. case of dependent P(K ) = 51 K

2

4

P(K ) = 52 K
K
1 48
51
K P( K 2) =
K
4

P(K2) = 51

P( K ) 48
1
=
52 P(
K
) = 47 K
2

51

First toss Second toss

Now, P(both king) = P(K1) × P(K2) = 4×3= 1
52 51 221

EXAMPLE 3 There is one red, one green and one yellow ball in a bag. Two balls

are taken out randomly one after another without the replacement
of first ball. Find the probability of getting a red and a yellow in
order by drawing a tree digram.

Solution : Here, let R, G and Y represent red, green and yellow balls respectively.

Number of red ball, n(R) = 1

Number of green ball, n(G) = 1

Number of yellow ball, n(Y) = 1

Total balls, n(S) = 3

Tree diagram: (Showing all possible events and their respective
probabilities)

1 G RG

) = 2

P(G 2

evIetnitss.the case of dependent 1R P(Y2) = 1
2
3 Y RY
) = GR
GY
P(R 3 YR

R 1 P(R 2) = 1 R YG
3 2

G G P(Y ) = 1
Y 2 2

Y

P(Y ) = P(R 2) = 1 R P(RY) = P(R1) × P(Y2)
1 2
1
3 Y

P(G = 1 × 1
3 2
2 ) = 1

2 G =1
6

GREEN Mathematics Book-10 281

EXAMPLE 4 Two children were given birth by the married couple. Illustrate the
probability of son or daughter by drawing a tree diagram and cal-
Solution : culate the probability so that both are sons.

Let 'S' and 'D' denote son and daughter respectively.

P(S) = 1 , P(D) = 1
2 2

Tree diagram: (Showing all possible events and their respective
probabilities)

= 1 2 S

P(S )

2

evIetntiss. the case of independent = 1 2 S

P(S ) P(D2) D
S
1 D

S = 1
2

D P(D P(S2) = 1
1 2

) = 1 D

2 P(D
2
) = 1

2

First birth Second birth

Sample space = {SS, SD, DS, DD}

Now, P(both son) = P(SS) 1×1= 1
22 4
= P(S1) × P(S2) =

EXAMPLE 5 There are 20 boys and 30 girls in a class. If two students are selected
at random, show the probability of selecting a boy and a girl by
Solution : drawing a tree diagram.

Let 'B' and 'G' denote boy and girl student respectively.

Number of boys n(B) = 20, Number of girls n(G) = 30

Total number of ball n(S) = 50

P(B) = 20 P(G) = 30
50 50

Tree diagram: (Showing all possible events and their respective
probabilities)

P(B ) = 1949 B

2

evIetntiss. the case of independent P(B ) = 20 B
50
1 G

B P(G2) = 30
49
20
G P(G P(B2) = 49 B
1
) = 30 G

50 P(G
2
) = 29 G

49

First selection Second selection

282 GREEN Mathematics Book-10

Sample space = {BB, BG, GB, GG}

Now, P(selecting a boy and a girl) = P(B1G2 or G1B2)

= P(B1G2) + (G1B2)

= P(B1) × P(G2) + P(G1) × P(B2)

= 20 × 30 + 30 × 20
50 49 50 49

= 600 + 600
2450 2450

= 1200 = 24
2450 49

EXAMPLE 6 There are 10 Nepali books and 6 English books in a bag. If 3 books
are drawn randomly one after another without replacement, find
Solution : the probability that all books are Nepali by drawing tree diagram.

Let 'N' and 'E' denote Nepali and English books respectively.

Number of Nepali books n(N) = 10, Number of English books n(E) = 6

Total number of books n(S) = 16

P(N) = 10 P(E) = 6
16 16

Tree diagram: (Showing all possible events and their respective
probabilities)

8 14 N

) =

P(N 3

9 15 N EP(E3) =

evIetntiss. the case of dependent P(N ) = 6
14
2 NP(N

10 = 9 14

) = 16 )

1 3

P(N N E E
N
N P(E2) = 6 P(E3) = 5
15 14
9
E P(N2) = 10 P(N3) = 14
15
P(E ) N
1 = E
6 P(E
16 3 ) = 5
E
P(E 14
2 )
= 5 E NP(N3) = 10
14
First draw 15

P(E ) = 4
3

Second draw 14 E

Third draw

Sample space = {NNN, NNE, NEN, NEE, ENN, ENE, EEN, EEE}

Now, P(all books are Nepali) = P(N1N2N3)

= P(N1) × P(N2) × P(N3)

= 10 × 9 × 8 = 3
16 15 14 14

GREEN Mathematics Book-10 283

EXERCISE 14.2

A. Very Short Questions 4

P(A ) = 8 A

2

1. From the given tree diagram answer the following questions. 5

a. i. What is the value of n(A) + n(B)? P(A ) = 9 A B

1 P(B2) 4
8
A =

ii. What is the probability of P (A A) ? B P(A2) = 5 A
iii. What is the probability of P(A B)? 8
P(B
1 ) = 4 B

iv. Is the second event dependent on the first? 9 P(B
2
) = 3 B

8

First outcome Second outcome

b. The tree diagram shows all the events when a coin is tossed twice. Answer

the following questions from the tree = 1 2 H
diagram.
P(H )

2

i. Are the first and second toss = 1 H
independent?
P(H ) 2 T
ii. Find the sample space. H
1 P(T2) T
iii. Find the probability of getting head
in both toss. H = 1
2
iv. Find the probability of getting head
and tail in order. T P(H2) = 1
2
P(H
1 ) = 1 T

2 P(T
2
) = 1

2

First outcome Second outcome

c. The tree diagram shows all the events when two balls are taken out one after

another from a bag containing red and 7 12 R
green balls of same size.
P(R ) =
Complete the diagram and answer the
following questions. 2

i. Find the number of green balls. = 7 12 R

ii. Find the total number of balls. P(R ) 5 G
12 R
iii. Is the first ball taken out replaced? 1 G

iv. Are the first and the second draw R P(G2) =
independent?
G P(G P(R2) = 7
1 12

) = 5 G

12 P(G
2
) = 5

12

First outcome Second outcome

284 GREEN Mathematics Book-10

B. Short Questions

2. A coin is tossed twice. Draw a tree diagram by showing the probabilities of all the
events. Also find the:

a. sample space b. probabilities of getting head in both toss.

3. A coin is tossed three times in succession. By drawing a tree diagram find the
probability of

a. getting at least one head b. getting no head
c. all three heads

4. A bag contains 3 red and 4 white balls. Two balls are taken out randomly one after
another. If the first ball is replaced, drawing a tree diagram, find the probability of
getting:

a. red balls in both draw. b. white balls in both draw
c. balls of same colour.

5. A bag contains 7 black and 5 white balls. Two balls are drawn randomly one by
one without replacing the first ball.

a. Show the probabilities of all events in a tree diagram.
b. Find the probability that the first is black and second is white
c. Find the probability that the first is white and second is black.
d. Find the probability that both of them are of different colour.

6. Two children are born in a family. By drawing a tree diagram, find the probability
of

a. having at least one son. b. having no daughter.

c. a son and a daughter.

7. A class contains 20 boys and 30 girls. If two students are selected at random, show
the probability of selecting a boy and a girl on a tree diagram.

8. There are one red, one green and one blue sweet in a pot. A sweet is taken out
randomly and not replaced, then another sweet is drawn. Write a sample space
using a tree diagram.

9. From a well-shuffled pack of 52 cards two cards are drawn at random without
replacing the first one. Find the probability that both of them are faced cards.

10. A die is rolled two times. By drawing a tree diagram find the probability of getting
3 in both roll.

11. Two cards are drawn from a well-shuffled deck of 52 cards. Find the probability
that both cards are king (without replacement) by drawing a tree diagram.

GREEN Mathematics Book-10 285

12. Two children were given birth by a couple at the interval of 5 years. Find the
probablity of son or daughter by drawing a tree diagram, and calculate the
probability so that both are daughter.

13. A bag contains 10 blue balls and 14 green balls. If two balls are drawn one after
another without replacement, find the probability of getting both balls of different
colours.

14. A coin is tossed three times. Find the probablity that all the three are heads by
drawing a tree diagram.

15. A bag contains 5 blue balls, 3 yellow balls and 2 white balls. If two balls are drawn
at random with replacement, show the above experiment in a tree diagram.

16. A and B appear in an interview for two different posts. The probability of selecting

A is 1 and selecting B is 4 . What is the probability that :
7 5
i. Both of them will be selected? ii. Only one of them will be slelected?

iii. None of them will be selected?

Project Work:
Write the uses of probability. Discuss its advantages and disadvantages and list them.

286 GREEN Mathematics Book-10

Revision Problems for Examination

Mensuration

1 Mark Questions

1. What is the formula for lateral surface area of square base pyramid?
2. Find the volume of sphere having diameter 14cm.
3. Write down the formula for area of rhombus having diameter d1 and d2.

A

4. Calculate the area of ABC. 70cm

B C
24cm

5. Calculate the base area of prism with volume 250 3 cm³ and height 10cm .

6. What is the formulae for total surface area of hemisphere?

7. Calculate the radius of base of cone having real height 12cm and slanted height
13cm.

8. Which solid object has same TSA and CSA?

2 Marks Questions : P

Area of Triangle and Triangular Prism

1. Find the area of the DPQR from the adjoining figure. R
Q

2. Find the area of an equilateral triangle whose perimeter is 30m.

3. The ratio of the sides of a triangle is 3:4:5 and its perimeter 12cm. Find its area.

A
F

8cm

4. Find the volume of the given triangular prism. B E

6cm D

A C A2' 0cm

5cm B'

5. What height of the prism shown in B

the figure makes its volume 48cm³. 4cm C' X
C
A
6. In the given triangular prism, the lateral surface area is 8cm C

480 cm². If XY = 8cm, YZ = 6cm and ∠XYZ = 90°, find the Y B
length of the prism. 6cm ?

Z

GREEN Mathematics Book-10 287

Cylinder

1. The height of cylindrical tank is 21cm and diameter of the base is 14cm. Find the
volume.

2. The sum of radius and height of a cylinder is 10cm and circumference of base is
308cm. Find the total surface area of that cylinder.

3. Find the radius of the base of a solid cylinder whose height is 7cm, base area is 50cm²
and total surface area 188 square cm.

21cm
4. A right circular cylinder has height 21cm and the base radius
5cm. Find the curved surface area of the cylinder.

5cm

5. The area of the curved surface of a right circular cylinder is 4400 cm² and the
circumference of its base is 110cm. Find the height of the cylinder.

Sphere and Hemisphere

1. The volume of a sphere is 3611 cu. cm. Find its diameter.
2. If the circle circumference of the greatest circle of sphere is pcm, find its volume.
3. Find the circumference of the base of a hemi-shpere whose volume is 486p cu. cm.

4. Find the total surface area of the given solid hemisphere.
42cm

5. Three spheres of diameters 6cm, 8cm and 10cm are melted and formed a single

sphere. Find the diameter of the sphere.

Cylinder 14

1. Find the volume of: r

Find the radius of the base of the cylinder .

28cm

2. A cylinder whose height is 7cm and total surface area is 276 4 cm².
7

3. A cylinder water tank contains 462000 liters of water and radius is 3.5m. Then find
the height of the tank.

4. The volume of a cylindrical tank is 936m³. If its height is 6m, determine the base area
of that tank.

5. If the cylinder given in the figure is filled with ?
2200cm³ of water, how much is the height of 10cm
the water surface?

288 GREEN Mathematics Book-10

Cone A
1. Find the volume of
51cm10cm

OB B
O
2. In the given cone, its radius and height are equal. If the volume of A
the cone is 9702cm³, find the height of the cone.

O 9cm A

3. Find the curve surface area of the given cone. 12cm

14cm

4. Find the total surface area of : 24cm B

5. The circumference of the base of a solid cone is 42πcm. Qr
P l
If the sum of its slant height and the radius of its base is
49cm, what will be the slant height of the cone? Find it. O

24cm

4 Marks Questions 84cm
Combined solids formed by cylinder, hemisphere and cone
1. Find the total surface area of : 14cm

2. Find the volume of : 28cm

28cm 80cm

3. In the given solid the radius of the hemisphere is
equal to the radius of the cylinder. The height of the
cylinder is 80cm and the volume of the hemisphere
is 144πcm³. Find the total surface of that solid object.

4. Find the total surface area of the given solid: 13cm

10cm

5. The total surface area and the height of solid are 770 sq cm and 14cm
14cm respectively. Find the height of the cylinder.

GREEN Mathematics Book-10 289

Pyramid O 34cm
1. Find the volume of the square based pyramid. A B

DC 24cm
6cm
E
12cm
15cm 17cm
D
2. Find the total surface area of the square based pyramid. A

O
BC

E

3. In the given figure the total surface area of the given square D C
based pyramid is 96cm² and the side of the square base is 6cm.
Find the slant height of the pyramid.

A B
6cm

4. The total surface area of the square based pyramid is 144cm² and the ratio between
the side and vertical height is 8:3. Find the slant height of the pyramid.

E

5. The volume of the given pyramid is 4032cm³. Find its slant height. C
D

5 Marks Questions O
Combined solid
AB
24cm

1. A right circular cone is 8cm high and the radius of its base is 2 cm. The cone is melted
and recast into a sphere. Determine the diameter of the sphere and also find its total
surface.

2. Calculate the change in height of cylinder when the
given combined solid figure is melted and made
the cylinder of same base and same total surface 80cm 10cm
area.

14cm

4. If the water filled in the cylinder is poured into
the cone, to what height will the surface of the
water each?

14cm

290 GREEN Mathematics Book-10

5. Find the total surface area of the given figure.

15cm
12cm

Pyramid D

1. The lateral edge of a triangular pyramid is 10cm. A C
If the area of equilateral triangular base is 36 3 E
cm², find the total surface area. E

F

2. A pyramid with square base of vertical height D
16cm and of slant height 20cm is shown in the
figure. Calculate lateral surface area and volume 20cm
of the pyramid. 16cm
AD
B
F
F

C

Geometry

Area of Triangle and Quadrilateral: 1 Mark Questions A
D
1. If area of ∆DBE is 70cm², then find the area of DADE.
E
B

2. At which condition, the area of triangle and parallelogram are equal when they are

constructed under same parallel line? A B

3. Find the area of following parallelogram. 5cm

A B
DC

7cm
5cm

4. In the figure, area of //gm ABCD is 96cm² then calculate the

height of //gm. D C

5. What is the formula of area of trapezium? 10cm E
2 Marks Questions F
A
M 8cm

1. From the given figure, find the area of DACM. D

B
6cm C

A B
DA
2. In the adjoining figure, area of a parallelogram

ABCD is 69sq. cm find the area of DADC and O

DCOD. C B
D
3. In the given figure, ED = EC, EF⊥BC, BE = 10cm and CD = 16cm.

Find the area of parallelogram ABCD. A B C

4. In the adjoining figure, CQ = 6cm, Q

EP = 7cm and AD = 14cm, then find the D F

measure of BF. CE

GREEN Mathematics Book-10 291

5. In the given figure, PS//QR, PQ//ST and M is the mid point P S

of QT. If the area of PQRS is 60 sq cm, find the area of

DPMT. QR
4 Marks Questions M
T

1. Parallelogram PQRS and QRTU are standing on the same base QR and between

the same parallel lines PT and QR. Prove that the area of parallelogram PQRS and

parallelogram QRTU are equal.

2. Prove that parallelogram ABCD and rectangle ABXY on the same base AB and
between the same parallel line AB and MN are equal in area.

3. `If a rectangle ABXY and a triangle ABC stand on the same base AB and between the

same parallel lines AB and MN, prove that the area of rectangle ABXY is double of

the area of triangle ABC. DP

4. The line drawn through the vertex C of the quad. ABCD

parallel to the diagonal DB is CP. Prove that the quad A

ABCD = quad ABPD in area. B

P C
S

5. In the adjoining figure, PQRS and CQMN are two L O N
parallelograms equal in area. Prove that LR//SN. M
E
Q
R A
M
F
E
5 Marks Questions G D
I
A
H
1. In the figure, AB//DC. AE//BC. BF//CF and CG//

EF, then prove that: ABCD = GCEF.
N

2. In DMON, E and H are BC

the mid point of MN and H
PN respectively MP is
R

trisected by points O and M OW P
W. Prove that : 2 DONE =

DTWO. TB D

3. In the adjoining figure, D and M are the mid points O

of AB and AC respectively. Then prove that: Area

2DABCD = Area ( CEMO) S

4. In DABC: D, E, F and G are the mid-points C

BC, AD, BE and CF respectively. Prove F E
that DABC = 8DEFG.
G
B D C

H

5. In the adjoining figure, ABCD is a

parallelogram. If DE = CE and AG//HF, then G
DE
prove that FAGN = ABCD. F C

AB

292 GREEN Mathematics Book-10

Circle

1 Marks Questions A
D
Aa C
C
1. Calculate the value of a. 2. Find ∠BAD. 105°

B O 2a B
3. Find ∠AOC: A 30° O B

4. What is the value of ∠RPQ? P ? Q
75° S
CT

R

O

5. Calculate the value of a. a B
4a
A
C

2 Marks Questions

R

P O R

1. Find ∠PRQ. O 2. Find ∠POQ. 30°

Q P
QA

PS R
C
3. Find ∠QPS. O 4. Find angle ∠BDF.

Q P B 52°
R S

D

5. If ∠QTP = 90° and ∠QTS = 45°, find ∠PST. F
T

Q

4 Marks Questions

1. Prove that the angle QOR at the center of circle PQR is twice the angle QPR at the
circumference standing on the same arc QR.

2. Prove that the inscribed ∠QPR and ∠AQSR standing on the same arc QR or a circle
are equal.

3. ABCD is a cyclic quadrilateral. Prove that ∠BAD + ∠BCD = 180° and ∠ABC + ∠ADC
= 180°.

4. In the figure, O is the center of the circle. If chords O

KL and MN intersect at the point p, prove that K N

∠NOL + ∠KOM = 2 ∠KPM. P

P QM L

5. If PC is the bisector of ∠APB, then X Y
B
prove that XY//AB. A

C

GREEN Mathematics Book-10 293

5 Marks Questions AD
P

1. In the given figure if BC = DC, prove that PQ//DB. C

EQ E
A
B
D
2. In the given figure, AB = AC and BD = CE. Then prove that : AB = AE.B
C
P
M

3. In the given figure, O is the center of the circle. If O Q
LM = MQ, then prove that N

B

(i) PN = NQ (ii) ON//PQ. D

4. a. Two chords AB and CD intersect at point P at right A M

angle. M is mid-point of BD. MP is produced to meet P

AC at N. Prove that PN⊥AC. N B

C

b. In the given diagram, O is the center of the O

circle. If AB = CD , then prove that: OP = OQ.A P Q D

EBC
AB

c. In the given figure O is the centre O
of circle. If AB||CD, prove that
A
∠AOC = 2 ∠BED. CD C

5. a. In the figure, AB||CD and ∠ABC = ∠EBC. Prove B E

that BE = CD. C
O
D

b. In the given figure, O is the centre of the P R
circle. Q and S are two points on the chord
O

AB. If AQ = BS and AP = BR , prove that

∠A QP= ∠BSR. AQ SB

D

c. In the given figure, OA is the radius of the circle ABC A
and diameter of the circle AOD. Prove that AD = DC.

4 Marks Questions B

1. Verify experimentally that the angle at the circumference ACB is half of the angle
APB at the centre P of a circle standing on the APB at the centre P of a circle standing
on the same arc AB. (Two circles with radii at least 3cm are necessary. )

2. Verify experimentally that ∠PQR and ∠PSR are supplementary if the angle at the
circumference of semi circle PQR is a right angle.

3. Verify experimentally that the angle at the circumference of semi circle PQR is a right
angle.

294 GREEN Mathematics Book-10