NCERT Exemplar Solutions Class 10 Maths Chapter 13 – Free PDF Download
NCERT Exemplar Class 10 Maths Chapter 13 Statistics and Probability, provided here for students to prepare for CBSE exams. These problems and solutions are prepared by our subject experts in accordance with CBSE latest syllabus (20212022), to help students score better. Students can use the NCERT Exemplar for Class 10 Maths as a productive tool for studying as well as practising sums. The exemplar also contains solved questions relevant to the exercise problems present in the textbook. Moreover, it will allow students to have a thorough revision of the entire chapter and be prepared to face the exams.
Chapter 13 basically has problems based on the different statistical measures like mean, mode, and median. Students will learn about how to solve these problems and also about the concept of cumulative frequency distribution, cumulative frequency curves, and more. Additionally, students will also develop an experimental approach to probability. Students will explore concepts like multiplication rule of probability, the Bayes’ theorem, and independence of events. These Exemplars problems and solutions cover the following topics of Statistics and Probability given below:
 Determining the mean of grouped data by a direct method, assumed mean method, stepdeviation method
 Finding the mode of the given data
 To find the median of the grouped data
 Representation of Cumulative Frequency Distribution graphically
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Exercise 13.1 Page No: 157
Choose the correct answer from the given four options:
1. In the formula
For finding the mean of grouped data di’s are deviations 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
Solution:
(C) Mid points of the classes
Explanation:
We know,
d_{i} = x_{i} – a
Where,
x_{i} are data and ‘a’ is the assumed mean
So, d_{i} are the deviations from of mid – points of the classes.
Hence, the option (C) is correct
2. While computing mean of grouped data, we assume that the frequencies are
(A) Evenly distributed over all the classes
(B) Centred at the class marks of the classes
(C) Centred at the upper limits of the classes
(D) Centred at the lower limits of the classes
Solution:
(B) Centered at the class marks of the classes
Explanation:
In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.
Hence, the option (B) is correct
3. If xi’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and x is the mean, then (f_{i}x_{i} – \(\overline{x}\) ) is equal to
(A)0 (B) –1 (C) 1 (D) 2
Solution:
(A) 0
Explanation:
Mean (x) = Sum of all the observations/ Number of observations
x = (f_{1}x_{1} + f_{2}x_{2} + …..+ f_{n}x_{n}) / f_{1} + f_{2} +……+ f_{n}
x = Σfixi / Σfi, Σfi = n
x = Σfixi / n
n x = Σfixi …………… (1)
Σ (fixi – x) = (f_{1}x_{1} – x) + (f_{2}x_{2} – x)+ …..+ (fnxn – x)
Σ (fixi – x) = (f_{1}x_{1} + f_{2}x_{2} + …..+ fnxn) – (x +x +….n times)
Σ (fixi – x) = Σfixi –nx
Σ(fixi – x) = nx – nx ( From eq1)
Σ(fixi – x) = 0
Hence, option (A) is correct
4. In the formula x = a + h(f_{i}u_{i}/f_{i}), for finding the mean of grouped frequency distribution, u_{i} =
(A) (x_{i}+a)/h
(B) h (xi – a)
(C) (x_{i }–a)/h
(D) (a – x_{i})/h
Solution:
(C) (x_{i }–a)/h
Explanation:
According to the question,
x = a + h(f_{i}u_{i}/f_{i}),
Above formula is a step deviation formula.
In the above formula,
x_{i} is data values,
a is assumed mean,
h is class size,
When class size is same we simplify the calculations of the mean by computing the coded mean of u_{1},u_{2},u_{3}…..,
Where u_{i} = (x_{i} – a)/h
Hence, option (C) is correct
5. 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
(A) mean (B) median
(C) mode (D) all the three above
Solution:
(B) Median
Explanation:
Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa, 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
Hence, option (B) is correct
6. For the following distribution :
Class 005 510 1015 1520 2025
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
Solution:
(B) 25
Explanation:
Class  Frequency  Cumulative Frequency 
05  10  10 
510  15  25 
1015  12  37 
1520  20  57 
2025  9  66 
From the table, N/2 = 66/2 = 33, which lies in the interval 10 – 15.
Hence, lower limit of the median class is 10.
The highest frequency is 20, which lies in between the interval 15 – 20.
Hence, lower limit of modal class is 15.
Therefore, required sum is 10 + 15 = 25.
Hence, option (B) is correct
7. Consider the following frequency distribution:
Class 005 611 1217 1823 2429
Frequency 13 10 15 8 11
The upper limit of the median class is
(A)17 (B) 17.5 (C) 18 (D) 18.5
Solution:
(B) 17.5
Explanation:
According to the question,
Classes are not continuous, hence, we make the data continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.
Class  Frequency  Cumulative Frequency 
0.55.5  13  13 
6.511.5  10  23 
11.517.5  15  38 
17.523.5  8  46 
23.529.5  11  57 
According to the question,
N/2 = 57/2 = 28.5
28.5 lies in between the interval 11.5 – 17.5.
Therefore, the upper limit is 17.5.
Hence, option (B) is correct
8. For the following distribution:
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
(A)1020 (B) 2030 (C) 3040 (D) 5060
Solution:
(C) 3040
Explanation:
Marks  Number of students  Cumulative Frequency 
Below 10  3=3  3 
1020  (12 – 3) = 9  12 
2030  (27 – 12) = 15  27 
3040  (57 – 27) = 30  57 
4050  (75 – 57) = 18  75 
5060  (80 – 75) = 5  80 
Here, we see that the highest frequency is 30, which lies in the interval 30 – 40.
Hence, option (C) is correct
9. Consider the data :
Class  6585  85105  105125  125145  145165  165185  185205 
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
A0 (B) 19 (C) 20 (D) 38
Solution:
(C) 20
Explanation:
Class  Frequency  Cumulative Frequency 
6585  4  4 
85105  5  9 
105125  13  22 
125145  20  42 
145165  14  56 
165185  7  63 
185205  4  67 
Here, N/2 = 67/2 = 33.5 which lies in the interval 125 – 145.
Hence, upper limit of median class is 145.
Here, we see that the highest frequency is 20 which lies in 125 – 145.
Hence, the lower limit of modal class is 125.
∴ Required difference = Upper limit of median class – Lower limit of modal class
= 145 – 125 = 20
Hence, option (C) is correct
10. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below
Class  13.814  1414.2  14.214.4  14.414.6  14.614.8  14.815 
Frequency  2  4  5  71  48  20 
The number of athletes who completed the race in less than 14.6 seconds is :
A11 (B) 71 (C) 82 (D) 130
Solution:
(C) 82
Explanation:
The number of athletes who completed the race in less than 14.6 second= 2 + 4 + 5 + 71 = 82
Hence, option (C) is correct
11. Consider the following distribution :
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 3040 is
(A) 3 (B) 4 (C) 48 (D) 51
Solution:
(A) 3
Explanation:
Marks Obtained  Number of students  Cumulative Frequency 
010  (63 – 58) = 5  5 
1020  (58 – 55) = 3  3 
2030  (55 – 51) = 4  4 
3040  (51 – 48) = 3  3 
4050  (48 – 42) = 6  6 
50<  42 = 42  42 
Hence, frequency in the class interval 30 – 40 is 3.
Hence, option (A) is correct
12. If an event cannot occur, then its probability is
(A)1 (B) ¾ (C) ½ (D) 0
Solution:
(D) 0
Explanation:
The event which cannot occur is said to be impossible event.
The probability of impossible event = zero.
Hence, option (D) is correct
13. Which of the following cannot be the probability of an event?
(A)1/3 (B) 0.1 (C) 3% (D)17/16
Solution:
(D)17/16
Explanation:
Probability of an event always lies between 0 and 1.
Probability of any event cannot be more than 1 or negative as (17/16) > 1
Hence, option (D) is correct
Exercise 13.2 Page No: 161
1. The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.
Solution:
In order to calculate the median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed or equally spaced. Hence, we cannot say that the statement “the median of an ungrouped data and the median calculated when the same data is grouped are always the same” is always correct.
2. In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
where a is the assumed mean. a must be one of the midpoints of the classes. Is the last statement correct? Justify your answer.
Solution:
No, the statement is not correct. It is not necessary that assumed mean should be the mid – point of the class interval. a can be considered as any value which is easy to simplify it.
3. Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.
Solution:
No, the values of mean, mode and median of grouped data can be the same as well, it depends on the type of data given.
4. Will the median class and modal class of grouped data always be different? Justify your answer.
Solution:
The median class and modal class of grouped data is not always different, it depends on the data given.
5. In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is ¼. Is this correct? Justify your answer.
Solution:
No it is not correct that in a family having three children, there may be no girl, one girl, two girls or three girls, the probability of each is ¼. .
Let boys be B and girls be G
Outcomes can be BBB , GGG , BBG , BGB , GBB, GGB, GBG , BGG
Then Probability of 3 girls = 1/8
Probability of 0 girls = 1/8
Probability of 2 girls = 3/8
Probability of 1 girl = 3/8
6. A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. 13.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.
Solution:
Total no. of outcome = 360
p(1)= 90/360 =1/4
p(2) = 90/ 360 = 1/4
p(3) = 180/360 =1/2
Hence, it is clear that the outcome are not equal
7. 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?
Solution:
Apoorv throw two dice at once.
Hence, the total number of outcomes = 36
Number of outcomes for getting product 36 = 1(6×6)
∴ Probability for Apoorv = 1/36
Peehu throws one die,
Hence, the total number of outcomes = 6
Number of outcomes for getting square = 36
∴ Probability for Peehu = 6/36 = 1/6
Therefore, Peehu has a better chance of getting the number 36.
Exercise 13.3 Page No: 166
1. Find the mean of the distribution :
Class 13 35 57 710
Frequency 9 22 27 17
Solution:
We first, find the class mark x_{i} of each class and then proceed as follows.
Class  Class Marks (x_{i})  Frequency (f_{i})  f_{i}x_{i} 
13  2  9  18 
35  4  22  88 
57  6  27  162 
710  8.5  17  144.5 
Σf_{i} = 75  Σf_{i}x_{i} = 412.5 
Mean,
Therefore, mean of the given distribution = 5.5.
2. Calculate the mean of the scores of 20 students in a mathematics test :
Marks 1020 2030 3040 4050 5060
Number of students 2 4 7 6 1
Solution:
We first, find the class mark x_{i} of each class and then proceed as follows
Class  Class Marks (x_{i})  Frequency (f_{i})  f_{i}x_{i} 
1020  15  2  30 
2030  25  4  100 
3040  35  7  245 
4050  45  6  270 
5060  55  1  55 
Σf_{i} = 20  Σf_{i}x_{i} = 700 
Mean,
Therefore, mean of scores of 20 students in mathematics test = 35.
3. Calculate the mean of the following data :
Class 4 – 7 8 –11 12– 15 16 –19
Frequency 5 4 9 10
Solution:
The given data is not continuous.
So, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Class  Class Marks (x_{i})  Frequency (f_{i})  f_{i}x_{i} 
3.5 – 7.5  5.5  5  27.5 
7.5 – 11.5  9.5  4  38 
11.5 – 15.5  13.5  9  121.5 
15.5 – 19.5  17.5  10  175 
Σf_{i} = 28  Σf_{i}x_{i} = 362 
Mean,
Therefore, mean of the given data = 12.93.
4. The following table gives the number of pages written by Sarika for completing her own book for 30 days :
Number of pages written per day 1618 1921 2224 2527 2830
Number of days 1 3 4 9 13
Find the mean number of pages written per day.
Solution:
Class Marks  Mid – Value (x_{i})  Number of days (f_{i})  f_{i}x_{i} 
15.5 – 18.5  17  1  17 
18.5 – 21.5  20  3  60 
21.5 – 24.5  23  4  92 
24.5 – 27.5  26  9  234 
27.5 – 30.5  29  13  377 
Σf_{i} = 30  Σf_{i}x_{i} = 780 
The given data is not continuous.
Hence, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Therefore, the mean of pages written per day = 26.
5. The daily income of a sample of 50 employees are tabulated as follows :
Income (in Rs) 1200 201400 401600 601800
Number of employees 14 15 14 7
Find the mean daily income of employees.
Solution:
C.I  x_{i}  d_{i} = (x_{i} – a)  F_{i}  f_{i}d_{i} 
1 – 200  100.5  – 200  14  – 2800 
201 – 400  300.5  0  15  0 
401 – 600  500.5  200  15  2800 
601 – 800  700.5  400  7  2800 
Σf_{i} = 50  Σf_{i}x_{i} = 2800 
∴ Assumed mean, a = 300.5 and d_{i} = (x_{i} – a)
= 300.5 + 2800/50
= 356.5
Hence, the average daily income of employees = Rs.356.5
6. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :
Number of seats 100104 104108 108112 112116 116120
Frequency 15 20 32 18 15
Determine the mean number of seats occupied over the flights.
Solution:
Class Interval  Class Marks (x_{i})  Frequency (f_{i})  Deviation (d_{i} = x_{i} – a)  f_{i}d_{i} 
100 – 104  102  15  – 8  – 120 
104 – 108  106  20  – 4  – 80 
108 – 112  110  32  0  0 
112 – 116  114  18  4  72 
116 – 120  118  15  8  120 
N = Σf_{i} = 100  Σf_{i}d_{i} = – 8 
∴ Assumed mean, a = 110
Class width, h = 4
And total observations, N = 100
Hence, finding mean,
= 110 + (8/100)
= 110 – 0.08
= 109.92
But we know that the seats cannot be in decimal.
Therefore, the number of seats = 109.
7. The weights (in kg) of 50 wrestlers are recorded in the following table :
Weight (in kg)  100110  110120  120130  130140  140150 
Number of wrestlers  4  14  21  8  3 
Find the mean weight of the wrestlers.
Solution:
Weight (in kg)  Number of Wrestlers (f_{i})  Class Marks (x_{i})  Deviation
(d_{i} = x_{i} – a) 
f_{i}d_{i} 
100 – 110  4  105  – 20  – 80 
110 – 120  14  115  – 10  – 140 
120 – 130  21  125  0  0 
130 – 140  8  135  10  80 
140 – 150  3  145  20  60 
N = Σf_{i} = 50  Σf_{i}d_{i} = – 80 
∴ Assumed mean, (a) = 125
Class width, (h) = 10
and total observations, (N) = 50
By step deviation method,
= 125 – 16
= 123.4kg
Hence, mean weight of wrestlers = 123.4kg
8. 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) 1012 1214 1416 1618
Number of cars 7 12 18 13
Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?
Solution:
Mileage (km L^{1})  Class –Marks (x_{i})  Number of cars (f_{i})  f_{i}x_{i} 
10 – 12  11  7  77 
12 – 14  13  12  156 
14 – 16  15  18  270 
16 – 18  17  13  221 
Total  Σf_{i} = 50  Σf_{i}x_{i} = 724 
Here, Σf_{i} = 50
Σf_{i}x_{i} = 724
= 724/50 = 14.48
Hence, mean mileage = 14.48 km/h
No, I don’t agree with the claim because the manufacturer is claiming mileage 1.52 km/h more than average mileage.
9. The following is the distribution of weights (in kg) of 40 persons :
Weight (in kg) 4045 4550 5055 5560 6065 6570 7075 7580
Number of persons 4 4 13 5 6 5 2 1
Construct a cumulative frequency distribution (of the less than type) table for the data above.
Solution:
Weight (in kg)  Cumulative frequency 
Less than 45  4 
Less than 50  4 + 4 = 8 
Less than 55  8 + 13 = 21 
Less than 60  21 + 5 = 26 
Less than 65  26 + 6 = 32 
Less than 70  32 + 5 = 37 
Less than 75  37 + 2 = 39 
Less than 80  39 + 1 = 40 
10. The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
Marks Number of students
Below 10 10
Below 20 50
Below 30 130
Below 40 270
Below 50 440
Below 60 570
Below 70 670
Below 80 740
Below 90 780
Below 100 800
Construct a frequency distribution table for the data above.
Solution:
The frequency distribution table for the given data is:
Class Interval  Number of students 
010  10 
1020  50 – 10 = 40 
2030  130 – 50 = 80 
3040  270 – 130 = 140 
4050  440 – 270 = 170 
5060  570 – 440 = 130 
6070  670 – 570 = 100 
7080  740 – 670 = 70 
8090  780 – 740 = 40 
90100 

11. Form the frequency distribution table from the following data :
Marks (out of 90) Number of candidates
More than or equal to 80 4
More than or equal to 70 6
More than or equal to 60 11
More than or equal to 50 17
More than or equal to 40 23
More than or equal to 30 27
More than or equal to 20 30
More than or equal to 10 32
More than or equal to 0 34
Solution:
The frequency distribution table for the given data is:
Class Interval  Number of students 
010  34 – 32 = 2 
1020  32 – 30 = 2 
2030  30 – 27 = 3 
3040  27 – 23 = 4 
4050  23 – 17 = 6 
5060  17 – 11 = 6 
6070  11 – 6 = 5 
7080  6 – 4 = 2 
8090 
12. Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class:
Height Frequency Cumulative frequency
(in cm)
150155 12 a
155160 b 25
160165 10 c
165170 d 43
170175 e 48
175180 2 f
Total 50
Solution:
Height (in cm)  Frequency  Cumulative frequency given  Cumulative frequency 
150 – 155  12  a  12 
155 – 160  b  25  12 + b 
160 – 165  10  c  22 + b 
165 – 170  d  43  22 + b + d 
170 – 175  e  48  22 + b + d + e 
175 – 180  2  f  24 + b + d + e 
Total  50 
On comparing last two tables, we get
a = 12
∴ 12 + b = 25
⇒ b = 25 – 12 = 13
22 + b = c
⇒ c = 22 + 13 = 35
22 + b + d = 43
⇒ 22 + 13 + d = 43
⇒ d = 43 – 35 = 8
22 + b + d + e = 48
⇒ 22 + 13 + 8 + e = 48
⇒ e = 48 – 43 = 5
24 + b + d + e = f
⇒ f = 24 + 13 + 8 + 5 = 50
13. The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years) 1020 2030 3040 4050 5060 6070
Number of patients 60 42 55 70 53 20
Form:
ALess than type cumulative frequency distribution.
 More than type cumulative frequency distribution.Solution:(i)Less than type cumulative frequency distribution of the data is given below.
(i) Less than type Age (in year) Number of patients Less than 10 0 Less than 20 60 + 0 = 60 Less than 30 60 + 42 = 102 Less than 40 102 + 55 = 157 Less than 50 157 + 70 = 227 Less than 60 227 + 53 = 280 Less than 70 280 + 20 =300 (ii)
More than type cumulative frequency distribution of the data is given below.
(i) More than type Age (in year) Number of patients More than or equals 10 60 + 42 + 55 + 70 + 53 + 20 = 300 More than or equals 20 42 + 55 + 70 + 53 + 20 = 240 More than or equals 30 55 + 70 + 53 + 20 = 198 More than or equals 40 70 + 53 + 20 = 143 More than or equals 50 53 + 20 = 73 More than or equals 60 20 More than or equals 70 0 14. Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class:
Marks Below 20 Below 40 Below 60 Below 80 Below 100
Number of students 17 22 29 37 50
Form the frequency distribution table for the data.
Solution:
The frequency distribution table for given data.
Marks Number of students 0 – 20 12 20 – 40 22 – 17 = 5 40 – 60 29 – 22 = 7 60 – 80 37 – 29 = 8 80 – 100 50 – 37 = 13 15. Weekly income of 600 families is tabulated below :
Weekly income Number of families
(in Rs)
01000 250
10002000 190
20003000 100
30004000 40
40005000 15
50006000 5
Total 600
Compute the median income.
Solution:
Weekly Income Number of families (f_{i}) Cumulative frequency (cf) 01000 250 250 10002000 190 250 + 190 = 400 20003000 100 440 + 100 = 540 30004000 40 540 + 40 = 580 40005000 15 580 + 15 = 595 50006000 5 595 + 5 = 600 According to the question,
n = 600
∴ n/2 = 600/2 = 300
Cumulative frequency 440 lies in the interval 1000 – 2000.
Hence, lower median class, l = 1000
f = 190,
c_{f} = 250,
Class width, h = 1000
And total observation n = 600
= 1000 + 5000/19
= 1000 + 263.15 = 1263.15
Hence, the median income is Rs.1263.15.
16. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows:
Speed (km/h) 85100 100115 115130 130145
Number of players 11 9 8 5
Calculate the median bowling speed.
Solution:
First we construct the cumulative frequency table
Speed ( in km/h) Number of players Cumulative frequency 85 – 100 11 11 100 – 115 9 11 + 9 = 20 115 – 130 8 20 + 8 = 28 130 – 145 5 28 + 5 = 33 It is given that, n = 33
∴ n/2 = 33/2 = 16.5
Hence, the median class is 100 – 115.
Where, lower limit(l) = 100
Frequency (f) = 9
Cumulative frequency (cf) = 11
And class width(h) = 15
= 100 + 82.5/9
= 100 + 9.17
= 109.17
Hence, the median bowling speed is 109.17 km/h.
17. The monthly income of 100 families are given as below :
Income (in Rs) Number of families
05000 8
500010000 26
1000015000 41
1500020000 16
2000025000 3
2500030000 3
3000035000 2
3500040000 1
Calculate the modal income.
Solution:
According to the data given,
The highest frequency = 41,
41 lies in the interval 10000 – 15000.
Here, l = 10000, f_{m} = 41,f_{1} = 26,f_{2} = 16 and h = 5000
= 10000 + 15×125
= 10000 + 1875
= 11875
Hence, the modal income = Rs.11875 per month.
18. The weight of coffee in 70 packets are shown in the following table :
Weight (in g) Number of packets
200201 12
201202 26
202203 20
203204 9
204205 2
205206 1
Determine the modal weight.
Solution:
In the given data, the highest frequency is 26, which lies in the interval 201 – 202
Here, l = 201,f_{m} = 26,f_{1} = 12,f_{2} = 20 and (class width) h = 1
Hence, the modal weight = 201.7 g.
19. Two dice are thrown at the same time. Find the probability of getting
ASame number on both dice.
Different numbers on both dice.Solution:
Two dice are thrown at the same time.
So, total number of possible outcomes = 36
(i) Same number on both dice.
Possible outcomes = (1,1), (2,2), (3, 3), (4, 4), (5, 5), (6, 6).
Hence, number of possible outcomes = 6
Therefore, the probability of getting same number on both dice = 6/36 = 1/6
(ii) Different number on both dice.
Hence, number of possible outcomes
= 36 – Number of possible outcomes for same number on both dice
= 36 – 6 = 30
Therefore, the probability of getting different number on both dice = 30/36 = 5/6
20. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
(i) 7? (ii) a prime number? (iii) 1?
Solution:
According to the question,
Two dice are thrown simultaneously.
So, that number of possible outcomes = 36
(i) Sum of the numbers appearing on the dice is 7.
So, the possible outcomes = (1, 6), (2,5), (3, 4), (4, 3), (5, 2), (6, 1).
Hence, number of possible outcomes = 6
∴ the probability that the sum of the numbers appearing on the dice is 7 = 6/36 = 1/6
(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.
So, the possible outcomes are (1, 1), (1,2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) = (6, 5).
Hence, number of possible outcomes = 15
∴ the probability that the sum of the numbers appearing on the dice is a prime number = 15/36 = 5/12
(iii) Sum of the numbers appearing on the dice is 1.
It is not possible, so its probability is zero.
∴ the probability that the sum of the numbers appearing on the dice is 1 = 0
21. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i) 6 (ii) 12 (iii) 7
Solution:
Number of total outcomes = 36
(i) When product of the numbers on the top of the dice = 6.
The possible outcomes = (1, 6), (2,3), (3, 2), (6, 1).
Hence, number of possible ways = 4
∴ Probability that the product of the numbers on the top of the dice is 6= 4/36 = 1/9
(ii) When product of the numbers on the top of the dice = 12.
The possible ways are (2, 6), (3,4), (4, 3), (6, 2).
Hence, number of possible ways = 4
∴ Probability that the product of the numbers on the top of the dice is 12= 4/36 = 1/9
(iii) Product of the numbers on the top of the dice cannot be 7.
Hence, the probability is zero.
∴ Probability that the product of the numbers on the top of the dice is 7 = 0
To facilitate easy learning and help students understand the chapter in full detail, free NCERT Exemplar for Class 10 Maths Chapter 13 is provided here which can be further downloaded in the form of a PDF. Students can also download the other source of learning from BYJU’S such as notes, exemplar books, NCERT Maths Solutions for Class 10 and question papers to prepare for their exams and score good marks. Solving sample papers and previous year question papers also gives an idea of question types asked in the board exam from chapter Chapter 13 Statistics and Probability.
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Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Maths Chapter 13
List out the topics and sub topics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths.
The topics and sub topics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths are listed below.
1. Determining the mean of grouped data by a direct method, assumed mean method, stepdeviation method
2. Finding the mode of the given data
3. To find the median of the grouped data
4. Representation of Cumulative Frequency Distribution graphicallyExplain the concept of mean in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths.
Mean is nothing but the average of the given set of values. It denotes the equal distribution of values for a given data set. Central tendency is the statistical measure that recognizes a single value as representative of the entire distribution. It strives to provide an exact description of the whole data. It is the unique value that represents collected data. The mean, median and mode are the three commonly used measures of central tendency.Why is NCERT Exemplar Solutions for Class 10 Maths Chapter 13 beneficial for the students to score well in the board exams?
NCERT Exemplar Solutions for Class 10 Maths Chapter 13 is provided by the experts at BYJU’S based on the understanding abilities of students. The solution module PDF can be downloaded from BYJU’S based on the requirements of students. All the important concepts from the exam point of view are explained in a simple language to help students ace the exam without fear. The solutions for all the problems present in the NCERT textbook helps students to cross check their answers and analyse their areas of weakness.