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NCERT Solutions Class 10 Maths Chapter 14 Probability

Probability explains how likely an event is when all possible outcomes are equally likely.
These NCERT Solutions help students solve Chapter 14 questions on coins, dice, cards, marbles and complementary events.

Chapter 14 Probability first moves students from experimental probability to theoretical probability. The chapter uses simple cases such as tossing a coin, throwing a die, drawing cards and selecting marbles. NCERT Solutions Class 10 Maths Chapter 14 cover all 25 questions from Exercise 14.1 in textbook order. Students practise favourable outcomes, total outcomes, impossible events, sure events and complementary events. These solutions are useful for 2026-27 CBSE exam practice because each answer shows the counting step before the final probability.

Key Takeaways

Theoretical probability: Probability is favourable outcomes divided by total possible outcomes.
Range of probability: Probability always lies between 0 and 1.
Complementary events: P(not E) = 1 - P(E).
Equally likely outcomes: Coin, die, card and marble questions use this idea.

NCERT Solutions Class 10 Maths Chapter 14 Structure 2026-27

Exercise	Topic	Question Count
Exercise 14.1	Theoretical probability	25
Main Formula	Favourable outcomes / Total outcomes	Used throughout
Key Applications	Coins, dice, cards, marbles and complements	Full exercise

NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1

Exercise 14.1 has 25 questions on theoretical probability. These Probability Class 10 questions and answers follow the 2026-27 NCERT textbook order.

Q1. Complete the following statements.

(i) Probability of an event E + Probability of the event ‘not E’ = ___.

Answer: Probability of an event E + Probability of the event ‘not E’ = 1.

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

(ii) The probability of an event that cannot happen is _. Such an event is called _.

Answer: The probability is 0.

Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is _. Such an event is called _.

Answer: The probability is 1.

Such an event is called a sure event.

(iv) The sum of probabilities of all elementary events is ___.

Answer: The sum is 1.

All elementary events together cover every possible outcome.

(v) The probability of an event is greater than or equal to _ and less than or equal to _.

Answer: The probability is greater than or equal to 0 and less than or equal to 1.

So, 0 ≤ P(E) ≤ 1.

Q2. Which experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Answer: The outcomes are not equally likely.

A car may be more likely to start if it is in good condition.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer: The outcomes are not equally likely.

The result depends on the player’s skill and position.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Answer: The outcomes are equally likely, if the answer is guessed.

There are only two possible choices.

(iv) A baby is born. It is a boy or a girl.

Answer: The outcomes can be treated as equally likely at this level.

The possible outcomes are boy or girl.

Q3. Why is tossing a coin considered fair for deciding which team gets the ball?

Answer: Tossing a coin is fair because both outcomes are equally likely.

A fair coin has two outcomes: head and tail.

Each team has the same chance of winning the toss.

Probability of head = 1 / 2

Probability of tail = 1 / 2

Q4. Which of the following cannot be the probability of an event?

Options:
(A) 2 / 3
(B) -1.5
(C) 15%
(D) 0.7

Answer: -1.5 cannot be the probability of an event.

Probability cannot be less than 0.

So, option (B) is correct.

Q5. If P(E) = 0.05, what is the probability of ‘not E’?

Given:
P(E) = 0.05

Formula:
P(not E) = 1 - P(E)

Step 1: Substitute the value
P(not E) = 1 - 0.05

Step 2: Solve
P(not E) = 0.95

Final Answer:
The probability of ‘not E’ is 0.95.

Q6. A bag contains lemon flavoured candies only. Find the probability of taking out an orange candy and a lemon candy.

(i) Orange flavoured candy

Answer: The probability is 0.

The bag has only lemon flavoured candies.

So, taking out an orange candy is impossible.

(ii) Lemon flavoured candy

Answer: The probability is 1.

Every candy in the bag is lemon flavoured.

So, taking out a lemon candy is certain.

Q7. Probability of 2 students not having the same birthday is 0.992. Find the probability that they have the same birthday.

Given:
P(not same birthday) = 0.992

Formula:
P(same birthday) = 1 - P(not same birthday)

Step 1: Substitute the value
P(same birthday) = 1 - 0.992

Step 2: Solve
P(same birthday) = 0.008

Final Answer:
The probability is 0.008.

Q8. A bag contains 3 red balls and 5 black balls. Find the probability that the ball drawn is red and not red.

Given:
Red balls = 3
Black balls = 5
Total balls = 8

(i) Probability of red ball

Formula:
Probability = Favourable outcomes / Total outcomes

Step 1: Substitute the values
P(red) = 3 / 8

Final Answer:
Probability of drawing a red ball is 3 / 8.

(ii) Probability of not red

Not red means black.

P(not red) = 5 / 8

Final Answer:
Probability of drawing a ball that is not red is 5 / 8.

Q9. A box contains 5 red, 8 white and 4 green marbles. Find the probabilities.

Given:
Red marbles = 5
White marbles = 8
Green marbles = 4
Total marbles = 17

(i) Probability of red

P(red) = 5 / 17

Final Answer:
Probability of red marble is 5 / 17.

(ii) Probability of white

P(white) = 8 / 17

Final Answer:
Probability of white marble is 8 / 17.

(iii) Probability of not green

Not green marbles = Red + White

Not green marbles = 5 + 8

Not green marbles = 13

P(not green) = 13 / 17

Final Answer:
Probability of not green is 13 / 17.

Q10. A piggy bank contains 100 fifty-paise coins, 50 one-rupee coins, 20 two-rupee coins and 10 five-rupee coins.

Given:
50p coins = 100
Rs 1 coins = 50
Rs 2 coins = 20
Rs 5 coins = 10
Total coins = 180

(i) Probability of a 50p coin

P(50p coin) = 100 / 180

P(50p coin) = 5 / 9

Final Answer:
Probability of getting a 50p coin is 5 / 9.

(ii) Probability of not getting a Rs 5 coin

Coins that are not Rs 5 = 100 + 50 + 20

Coins that are not Rs 5 = 170

P(not Rs 5 coin) = 170 / 180

P(not Rs 5 coin) = 17 / 18

Final Answer:
Probability of not getting a Rs 5 coin is 17 / 18.

Q11. A tank contains 5 male fish and 8 female fish. Find the probability of taking out a male fish.

Given:
Male fish = 5
Female fish = 8
Total fish = 13

Formula:
Probability = Favourable outcomes / Total outcomes

Step 1: Substitute the values
P(male fish) = 5 / 13

Final Answer:
The probability of taking out a male fish is 5 / 13.

Q12. A spinner has numbers 1 to 8. Find the required probabilities.

Given:
Possible outcomes = 1, 2, 3, 4, 5, 6, 7, 8
Total outcomes = 8

(i) Probability of 8

Favourable outcome = 8

P(8) = 1 / 8

Final Answer:
Probability is 1 / 8.

(ii) Probability of an odd number

Odd numbers = 1, 3, 5, 7

Favourable outcomes = 4

P(odd number) = 4 / 8

P(odd number) = 1 / 2

Final Answer:
Probability is 1 / 2.

(iii) Probability of a number greater than 2

Numbers greater than 2 = 3, 4, 5, 6, 7, 8

Favourable outcomes = 6

P(number greater than 2) = 6 / 8

P(number greater than 2) = 3 / 4

Final Answer:
Probability is 3 / 4.

(iv) Probability of a number less than 9

All numbers are less than 9.

Favourable outcomes = 8

P(number less than 9) = 8 / 8

P(number less than 9) = 1

Final Answer:
Probability is 1.

Q13. A die is thrown once. Find the probabilities.

Given:
Possible outcomes = 1, 2, 3, 4, 5, 6
Total outcomes = 6

(i) Probability of a prime number

Prime numbers = 2, 3, 5

Favourable outcomes = 3

P(prime number) = 3 / 6

P(prime number) = 1 / 2

Final Answer:
Probability is 1 / 2.

(ii) Probability of a number lying between 2 and 6

Numbers between 2 and 6 = 3, 4, 5

Favourable outcomes = 3

P(number between 2 and 6) = 3 / 6

P(number between 2 and 6) = 1 / 2

Final Answer:
Probability is 1 / 2.

(iii) Probability of an odd number

Odd numbers = 1, 3, 5

Favourable outcomes = 3

P(odd number) = 3 / 6

P(odd number) = 1 / 2

Final Answer:
Probability is 1 / 2.

Q14. One card is drawn from a well-shuffled deck of 52 cards.

Given:
Total cards = 52

(i) Probability of a king of red colour

Red kings = 2

P(red king) = 2 / 52

P(red king) = 1 / 26

Final Answer:
Probability is 1 / 26.

(ii) Probability of a face card

Face cards = 12

P(face card) = 12 / 52

P(face card) = 3 / 13

Final Answer:
Probability is 3 / 13.

(iii) Probability of a red face card

Red face cards = 6

P(red face card) = 6 / 52

P(red face card) = 3 / 26

Final Answer:
Probability is 3 / 26.

(iv) Probability of the jack of hearts

Jack of hearts = 1

P(jack of hearts) = 1 / 52

Final Answer:
Probability is 1 / 52.

(v) Probability of a spade

Spades = 13

P(spade) = 13 / 52

P(spade) = 1 / 4

Final Answer:
Probability is 1 / 4.

(vi) Probability of the queen of diamonds

Queen of diamonds = 1

P(queen of diamonds) = 1 / 52

Final Answer:
Probability is 1 / 52.

Q15. Five diamond cards are shuffled. Find the probabilities.

Cards are ten, jack, queen, king and ace of diamonds.

(i) Probability that the card is queen

Given:
Total cards = 5
Queen = 1

P(queen) = 1 / 5

Final Answer:
Probability is 1 / 5.

(ii) Queen is drawn and put aside. Find second card probabilities.

Now, total cards left = 4.

(a) Probability that second card is ace

Ace left = 1

P(ace) = 1 / 4

Final Answer:
Probability is 1 / 4.

(b) Probability that second card is queen

Queen is already put aside.

Queen left = 0

P(queen) = 0 / 4

P(queen) = 0

Final Answer:
Probability is 0.

Q16. 12 defective pens are mixed with 132 good pens. Find the probability of drawing a good pen.

Given:
Defective pens = 12
Good pens = 132
Total pens = 144

Formula:
Probability = Favourable outcomes / Total outcomes

Step 1: Substitute the values
P(good pen) = 132 / 144

Step 2: Simplify
P(good pen) = 11 / 12

Final Answer:
Probability of drawing a good pen is 11 / 12.

Q17. A lot of 20 bulbs contains 4 defective bulbs.

(i) Probability that one bulb drawn is defective

Given:
Total bulbs = 20
Defective bulbs = 4

P(defective bulb) = 4 / 20

P(defective bulb) = 1 / 5

Final Answer:
Probability is 1 / 5.

(ii) First bulb drawn is not defective and not replaced. Find probability that second bulb is not defective.

After one good bulb is removed:

Total bulbs left = 19

Good bulbs left = 15

P(second bulb is not defective) = 15 / 19

Final Answer:
Probability is 15 / 19.

Q18. A box contains 90 discs numbered 1 to 90.

Given:
Total discs = 90

(i) Probability of a two-digit number

Two-digit numbers from 1 to 90 are 10 to 90.

Favourable outcomes = 81

P(two-digit number) = 81 / 90

P(two-digit number) = 9 / 10

Final Answer:
Probability is 9 / 10.

(ii) Probability of a perfect square number

Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Favourable outcomes = 9

P(perfect square) = 9 / 90

P(perfect square) = 1 / 10

Final Answer:
Probability is 1 / 10.

(iii) Probability of a number divisible by 5

Numbers divisible by 5 from 1 to 90 are 5, 10, 15, ..., 90.

Favourable outcomes = 18

P(number divisible by 5) = 18 / 90

P(number divisible by 5) = 1 / 5

Final Answer:
Probability is 1 / 5.

Q19. A die has faces A, B, C, D, E, A. Find the probabilities.

Given:
Total outcomes = 6
A appears twice.
D appears once.

(i) Probability of A

P(A) = 2 / 6

P(A) = 1 / 3

Final Answer:
Probability is 1 / 3.

(ii) Probability of D

P(D) = 1 / 6

Final Answer:
Probability is 1 / 6.

Q20. A die is dropped on a rectangle of 3 m by 2 m. Find probability of landing inside a circle of diameter 1 m.

Given:
Rectangle length = 3 m
Rectangle breadth = 2 m
Circle diameter = 1 m
Circle radius = 1 / 2 m

Formula:
Probability = Favourable area / Total area

Step 1: Find rectangle area
Area of rectangle = 3 × 2

Area of rectangle = 6 square metres

Step 2: Find circle area
Area of circle = πr²

Area of circle = π × (1 / 2)²

Area of circle = π / 4

Step 3: Find probability
P(inside circle) = (π / 4) / 6

P(inside circle) = π / 24

Using π = 22 / 7:

P(inside circle) = 22 / 168

P(inside circle) = 11 / 84

Final Answer:
Probability is π / 24 or 11 / 84 using π = 22 / 7.

Q21. A lot has 144 ball pens, of which 20 are defective. Find the probabilities.

Given:
Total pens = 144
Defective pens = 20
Good pens = 124

(i) Probability that Nuri will buy it

Nuri will buy a good pen.

P(good pen) = 124 / 144

P(good pen) = 31 / 36

Final Answer:
Probability that she will buy it is 31 / 36.

(ii) Probability that Nuri will not buy it

Nuri will not buy a defective pen.

P(defective pen) = 20 / 144

P(defective pen) = 5 / 36

Final Answer:
Probability that she will not buy it is 5 / 36.

Q22. Complete the probability table for the sum on two dice.

When two dice are thrown, total outcomes = 36.

Sum	Favourable Outcomes	Probability
2	1	1 / 36
3	2	2 / 36
4	3	3 / 36
5	4	4 / 36
6	5	5 / 36
7	6	6 / 36
8	5	5 / 36
9	4	4 / 36
10	3	3 / 36
11	2	2 / 36
12	1	1 / 36

(ii) Is the argument that each sum has probability 1 / 11 correct?

Answer: No, the argument is not correct.

The sums from 2 to 12 are possible, but they are not equally likely.

For example, sum 2 occurs only as (1, 1).

Sum 7 occurs in six ways.

So, each sum cannot have probability 1 / 11.

Q23. A one-rupee coin is tossed 3 times. Hanif wins if all tosses give the same result. Find probability that Hanif loses.

Given:
A coin is tossed 3 times.

Possible outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Total outcomes = 8

Hanif wins if outcomes are HHH or TTT.

Winning outcomes = 2

Losing outcomes = 8 - 2

Losing outcomes = 6

P(Hanif loses) = 6 / 8

P(Hanif loses) = 3 / 4

Final Answer:
Probability that Hanif loses is 3 / 4.

Q24. A die is thrown twice. Find the probabilities.

When a die is thrown twice, total outcomes = 6 × 6 = 36.

(i) Probability that 5 will not come up either time

Outcomes without 5 in one throw = 5

Outcomes without 5 in two throws = 5 × 5

Outcomes without 5 in two throws = 25

P(5 does not come either time) = 25 / 36

Final Answer:
Probability is 25 / 36.

(ii) Probability that 5 will come up at least once

Use complementary event.

P(5 comes at least once) = 1 - P(5 does not come either time)

P(5 comes at least once) = 1 - 25 / 36

P(5 comes at least once) = 11 / 36

Final Answer:
Probability is 11 / 36.

Q25. Which arguments are correct? Give reasons.

(i) If two coins are tossed, there are three outcomes: two heads, two tails or one of each. So each has probability 1 / 3.

Answer: This argument is not correct.

The actual equally likely outcomes are HH, HT, TH and TT.

Two heads has 1 outcome.

Two tails has 1 outcome.

One of each has 2 outcomes: HT and TH.

So, the probabilities are not equal.

(ii) If a die is thrown, there are two outcomes: odd or even. So probability of getting an odd number is 1 / 2.

Answer: This argument is correct.

Odd numbers are 1, 3 and 5.

Even numbers are 2, 4 and 6.

Both have 3 favourable outcomes.

So, P(odd number) = 3 / 6 = 1 / 2.

Probability Class 10 NCERT Solutions: Key Concepts

Class 10 Maths Chapter 14 Probability uses counting to measure chance. These concepts help students solve coin, die, card and marble questions.

Theoretical Probability

Theoretical probability is used when all outcomes are equally likely.

Formula: Probability of event E = Number of favourable outcomes / Total number of possible outcomes

Equally Likely Outcomes

Equally likely outcomes have the same chance of occurring.

A fair coin has head and tail as equally likely outcomes.

Impossible Event

An impossible event cannot happen.

Its probability is 0.

Example: Getting 8 on a normal die.

Sure Event

A sure event will definitely happen.

Its probability is 1.

Example: Getting a number less than 7 on a normal die.

Complementary Events

Complementary events class 10 questions use “not E”.

Formula: P(not E) = 1 - P(E)

Elementary Event

An elementary event has only one outcome.

Example: Getting head in one coin toss.

NCERT Solutions Class 10 Maths Probability: Formula Use

NCERT Solutions Class 10 Maths Probability questions become easier when students identify total and favourable outcomes first. This table shows common formats used in Chapter 14.

Situation	Total Outcomes	Common Favourable Outcomes
Tossing one coin	2	Head or tail
Throwing one die	6	Prime, odd, even or fixed number
Drawing one card	52	Ace, king, spade or face card
Tossing two coins	4	At least one head
Throwing two dice	36	Required sum
Drawing from a box	Total objects	Required colour or type

Coin Questions

Coin questions use head and tail.

For one fair coin, P(head) = 1 / 2.

Die Questions

Die questions use outcomes 1, 2, 3, 4, 5 and 6.

For one die, P(odd number) = 3 / 6 = 1 / 2.

Card Questions

A deck has 52 cards.

It has 4 suits, 13 cards in each suit and 12 face cards.

Marble and Ball Questions

Marble and ball questions use total items in the box or bag.

Favourable outcomes depend on the asked colour or type.

Complement Questions

Complement questions ask for “not red”, “not E” or “at least once”.

Using P(not E) = 1 - P(E) often makes these faster.

Useful Links for Class 10 Maths NCERT Solutions

Section	Useful Links
Class 10 Maths NCERT Solutions	NCERT Solutions for Class 10 Maths
Chapter 1	NCERT Solutions for Class 10 Maths Chapter 1
Chapter 2	NCERT Solutions for Class 10 Maths Chapter 2
Chapter 3	NCERT Solutions for Class 10 Maths Chapter 3
Chapter 4	NCERT Solutions for Class 10 Maths Chapter 4
Chapter 5	NCERT Solutions for Class 10 Maths Chapter 5
Chapter 6	NCERT Solutions for Class 10 Maths Chapter 6
Chapter 7	NCERT Solutions for Class 10 Maths Chapter 7
Chapter 8	NCERT Solutions for Class 10 Maths Chapter 8
Chapter 9	NCERT Solutions for Class 10 Maths Chapter 9
Chapter 10	NCERT Solutions for Class 10 Maths Chapter 10
Chapter 11	NCERT Solutions for Class 10 Maths Chapter 11
Chapter 12	NCERT Solutions for Class 10 Maths Chapter 12
Chapter 13	NCERT Solutions for Class 10 Maths Chapter 13
Chapter 14	NCERT Solutions for Class 10 Maths Chapter 14

Q.1 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	2	3

Which method did you use for finding the mean, and why?

Ans.

$\begin{array}{l} Here, we observe that class marks and frequencies are small \\ quantities. \\ So, we use direct method to compute the mean and proceed \\ as below. \end{array}$

Number of plants	Number of houses (f_i)	x_i	f_ix_i
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	20		162

$\begin{array}{l} Mean = \bar{x} = \frac{\sum f_{i} x_{_{i}}}{\sum f_{i}} = \frac{162}{20} = 8 .1 \\ Therefore, mean number of plants per house is 8.1. \end{array}$

Q.2 Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹)	500-120	520-140	540-160	560-180	580-200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 20 \\ Here, we take assumed mean(a) = 550 and proceed as below. \end{array}$

Daily wages (in ₹)	Number of workers (f_i)	x_i	d_i = x_i – 550	$u_{i} = \frac{x_{i} - 550}{h}$	f_iu_i
500 – 520	12	510	-40	-2	-24
520 – 540	14	530	-20	-1	-14
540 – 560	8	550	0	0	0
560 – 580	6	570	20	1	6
580 – 600	10	590	40	2	20
Total	50				-12

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 550 + (\frac{- 12}{50}) \times 20 = 545 .2 \end{array}$

$\begin{array}{l} Therefore, mean daily wages of the workers is ₹ 545 .20 \end{array}$

Q.3 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Daily pocket allowance

(in ₹)

11-13

13-15

15-17

17-19

18-21

21-23

23-25

Number of children

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 2 \\ Given mean(a) = 18 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}$

Daily pocket allowance (in ₹)	Number of children (f_i)	x_i	d_i = x_i – 18	f_id_i
11 – 13	7	12	-6	-42
13 – 15	6	14	-4	-24
15 – 17	9	16	-2	-18
17 – 19	13	18	0	0
19 – 21	f	20	2	2f
21 – 23	5	22	4	20
23 – 25	4	24	6	24
Total	$\sum f_{i} = 44 + f$			2f – 40

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ or 18 = 18 + \frac{2 f - 40}{44 + f} \\ or f = 20 \\ Therefore, missing frequency f is 20 . \end{array}$

Q.4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 3 \\ Here, we take assumed mean(a) = 75 .5 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Number of heartbeats per minute	Number of women (f_i)	x_i	d_i = x_i – 75.5	$u_{i} = \frac{x_{i} - 75 .5}{h}$	f_iu_i
65-68	2	66.5	-9	-3	-6
68-71	4	69.5	-6	-2	-8
71-74	3	72.5	-3	-1	-3
74-77	8	75.5	0	0	0
77-80	7	78.5	3	1	7
80-83	4	81.5	6	2	8
83-86	2	84.5	9	3	6
Total	30				4

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 75 .5 + (\frac{4}{30}) \times 3 = 75 .9 \\ Therefore, mean heartbeats per minute is 75 .9. \end{array}$

Q.5 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50-52	53-55	56-58	59-61	62-64
Number of boxes	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ans.

$\begin{array}{l} The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limit \\ of each interval. We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 3 \\ Here, we take assumed mean(a) = 57 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Class interval	f_i	x_i	d_i = x_i – 57	$u_{i} = \frac{x_{i} - 57}{h}$	$f_{i} u_{i}$
49.5 – 52.5	15	51	-6	-2	-30
52.5 – 55.5	110	54	-3	-1	-110
55.5 – 58.5	135	57	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	400				25

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 57 + (\frac{25}{400}) \times 3 = 57 .1875 \approx 57 .19 \\ Therefore, mean number of mangoes is 57 .19. \\ We have chosen step deviation method as values of f_{i}, d_{i} are big \\ and there is a common multiple between all d_{i} . \end{array}$

Q.6 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
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 50 \\ Here, we take assumed mean(a) = 225 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Class interval	f_i	x_i	d_i = x_i – 225	$u_{i} = \frac{x_{i} - 57}{h}$	f_iu_i
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
Total					-7

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 225 + (\frac{- 7}{25}) \times 50 = 211 \\ Therefore, mean daily expenditure is ₹ 211 . \end{array}$

Q.7 To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below.

Concentration of SO₂ (in ppm)

Frequency

0.00 – 0.04

0.04– 0.08

0.08 – 0.12

0.12–0.16

0.16–0.20

0.20–0.24

Find the mean concentration of SO₂ in the air.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 0 .04 \\ Here, we take assumed mean(a) = 0 .14 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Concentration of SO₂ (in ppm)	Frequency (f_i)	x_i	d_i = x_i – 0.14	$u_{i} = \frac{x_{i} - 0 .14}{h}$ b	f_iu_i
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	30				-31

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 0 .14 + (\frac{- 31}{30}) \times 0 .04 \\ \approx 0 .099 ppm \\ {Therefore, mean concentration of SO}_{2} in the air is 0 .099 ppm . \end{array}$

Q.8 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	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 17 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}$

Number of days	Number of students f_i	x_i	d_i = x_i – 17	f_id_i
0 – 6	11	3	-14	-154
6 – 10	10	8	-9	-90
10 – 14	7	12	-5	-35
14 – 20	4	17	0	0
20 – 28	4	24	7	28
28 – 38	3	33	16	48
38 – 40	1	39	22	22
Total	40			-181

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ = 17 + \frac{- 181}{40} \\ = 12 .475 \approx 12 .48 \\ Therefore, mean number of days for which a student was absent is 12 .48. \end{array}$

Q.9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 10 \\ Here, we take assumed mean(a) = 70 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Literacy rate (in %)	Number of cities f_i	x_i	d_i = x_i – 70	$u_{i} = \frac{x_{i} - 70}{h}$	f_iu_i
45 – 55	3	50	-20	-2	-6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	35				-2

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 70 + (\frac{- 2}{35}) \times 10 = 69 .43 \\ Therefore, mean literacy rate is 69 .43. \end{array}$

Q.10 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
Number 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.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 30 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}$

Age (in years)	Number of patients (f_i)	Class mark (x_i)	d_i = x_i – 30	f_id_i
5 -15	6	10	-20	-120
15 – 25	11	20	-10	-110
25 – 35	21	30	0	0
35 – 45	23	40	10	230
45 – 55	14	50	20	280
55 – 65	5	60	30	150

$\begin{array}{l} Total = 80 Total = 430 \\ Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ = 30 + \frac{430}{80} \\ = 35 .375 \approx 35 .38 \\ Therefore, mean of the given data is 35 .38 years . \\ Modal class is 35 - 45 . \\ l = 35, f_{1} = 23, f_{0} = 21, f_{2} = 14, h = 10 \end{array}$

$\begin{array}{l} Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 35 + (\frac{23 - 21}{2 \times 23 - 21 - 14}) \times 10 \\ = 36 .8 \\ Mode is 36.8 which represents that maximum number \\ of patients admitted in hospital are of 36.8 years . \end{array}$

$\begin{array}{l} While on average the age of a patient admitted to the \\ hospital is 35 .38 years \end{array}$

Q.11 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components.

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

Determine the modal lifetimes of the components.

Ans.

$\begin{array}{l} From the given data, we have \\ l = 60, f_{1} = 61, f_{0} = 52, f_{2} = 38, h = 20 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 60 + (\frac{61 - 52}{2 \times 61 - 52 - 38}) \times 20 \\ = 65 .625 \\ So, modal lifetime of electrical components is 65.625 hours. \end{array}$

Q.12 The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Expenditure (in ₹)	Number of families
1000 – 1500 1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000	24 40 33 28 30 22 16 7

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 2750 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}$

Expenditure (in ₹)	Number of families (f_i)	x_i	d_i = x_i – 2750	$u_{i} = \frac{x_{i} - 2750}{h}$	f_iu_i
1000 – 1500	24	1250	-1500	-3	-72
1500 – 2000	40	1750	-1000	-2	-80
2000 – 2500	33	2250	-500	-1	-33
2500 – 3000	28	2750	0	0	0
3000 – 3500	30	3250	500	1	30
3500 – 4000	22	3750	1000	2	44
4000 – 4500	16	4250	1500	3	48
4500 – 5000	7	4750	2000	4	28
Total	200				-35

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 2750 + \frac{- 35}{200} \times 500 \\ = 2662 .5 \\ Therefore, mean monthly expenditure is ₹ 2662 .5. \\ From the given data, we have \\ l = 1500, f_{1} = 40, f_{0} = 24, f_{2} = 33, h = 500 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 1500 + (\frac{40 - 24}{2 \times 40 - 24 - 33}) \times 500 \\ = 1847 .826 \approx 1847 .83 \\ So, modal monthly expenditure is ₹ 1847 .83. \end{array}$

Q.13 The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher

Number of states / U .T.

15– 20

20– 25

25–30

30–35

35–40

40–45

45–50

50–55

Ans.

$\begin{array}{l} From the given data, we have \\ l = 30, f_{1} = 10, f_{0} = 9, f_{2} = 3, h = 5 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 30 + (\frac{10 - 9}{2 \times 10 - 9 - 3}) \times 5 \\ = 30 .625 \approx 30 .6 \\ Interpretation: \\ Most of the states/UT have a teacher student ratio as 30.6. \\ Now, we find class mark for each interval by using the \\ following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 32 .5 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}$

Number of students per teacher	Number of states/UT (f_i)	x_i	d_i = x_i – 32.5	$u_{i} = \frac{x_{i} - 32 .5}{h}$ b	f_iu_i
15 – 20	3	17.5	-15	-3	-9
20 – 25	8	22.5	-10	-2	-16
25 – 30	9	27.5	-5	-1	-9
30 – 35	10	32.5	0	0	0
35 – 40	3	37.5	5	1	3
40 – 45	0	42.5	10	2	0
45 – 50	0	47.5	15	3	0
50 – 55	2	52.5	20	4	8
Total	35				-23

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 32 .5 + \frac{- 23}{35} \times 5 \\ = 29 .2 \\ Interpretation: \\ Average teacher student ratio of the states/UT is 29 .2. \end{array}$

Q.14 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored

Number of batsmen

3000 – 4000

4000– 5000

5000– 6000

6000–7000

7000–8000

8000–9000

9000–10000

10000–11000

Find the mode of the data.

Ans.

$\begin{array}{l} From the given data, we have \\ l = 4000, f_{1} = 18, f_{0} = 4, f_{2} = 9, h = 1000 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 4000 + (\frac{18 - 4}{2 \times 18 - 4 - 9}) \times 1000 \\ = 4608 .695 \\ Mode of the given data is 4608 .7 (approx .) \end{array}$

Q.15 A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Number

of cars

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Frequency

Ans.

$\begin{array}{l} From the given data, we have \\ l = 40, f_{1} = 20, f_{0} = 12, f_{2} = 11, h = 10 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 40 + (\frac{20 - 12}{2 \times 20 - 12 - 11}) \times 10 \\ = 44 .7 \\ ∴ Mode of the given data is 44 .7. \end{array}$

Q.16 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)

Number of consumers

65 – 85

85– 105

105– 125

125–145

145–165

165–185

185–205

Ans.

$\begin{array}{l} We find class mark for each interval by using the \\ following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 135 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}$

Monthly consumption (in units)	Number of consumers (f_i)	x_i	d_i = x_i – 135	$u_{i} = \frac{x_{i} - 135}{h}$	f_iu_i
65 – 85	4	75	-60	-3	-12
85 – 105	5	95	-40	-2	-10
105 – 125	13	115	-20	-1	-13
125 – 145	20	135	0	0	0
145 – 165	14	155	20	1	14
165 – 185	8	175	40	2	16
185 – 205	4	195	60	3	12
Total	68				7

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 135 + \frac{7}{68} \times 20 \\ = 137 .058 \\ Modal class = 125 - 145 \\ From the given data, we have \\ l = 125, f_{1} = 20, f_{0} = 13, f_{2} = 14, h = 20 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 125 + (\frac{20 - 13}{2 \times 20 - 13 - 14}) \times 20 \\ = 135 .76 \\ We know that \\ 3 Median = Mode + 2 Mean \\ or Median = \frac{135 .76 + 2 \times 137 .058}{3} = 136 .625 \end{array}$

We observe all these three measures are approximately same.

Q.17 If the median of the distribution given below is 28.5, find the values of x and y.

Class interval

Frequency

0 – 10

10– 20

20– 30

30–40

40–50

50–60

Total

Ans.

We find cumulative frequency of the given data as below.

Class interval

Frequency

Cumulative frequency

0 – 10

10– 20

20– 30

30–40

40–50

50–60

5 + x

25 + x

40 + x

40 + x + y

45 + x + y

Total

Here,
n = 60
or 45 + x + y = 60
or x + y = 15 …(1)
Median of data is given as 28.5 that lies in the interval 20 – 30.
So, median class = 20 – 30

$\begin{array}{l} From the given data, we have \\ l = 20, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 5 + x, \\ Frequency of median class (f) = 20, \\ Class size (h) = 10 \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ or 28 .5 = 20 + (\frac{\frac{60}{2} - 5 - x}{20}) \times 10 \\ or x = 8 \\ On putting this value of x in equation (1), we get \\ x + y = 15 \\ or 8 + y = 15 \\ or y = 15 - 8 = 7 \\ Hence, values of x and y are 8 and 7 respectively. \end{array}$

Q.18 A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Class interval

Frequency

Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Below 55

Below 60

100

Ans.

$\begin{array}{l} We write the class interval, number of policy holders (f_{i}), and cumulative frequency (cf) as below in the following table \\ on the basis of the given information. \end{array}$

Class interval	Number of policy holders (f_i)	Cumulative frequency (cf)
18 – 20	2	2
20 – 25	6 – 2 = 4	6
25 – 30	24 – 6 = 18	24
30 – 35	45 – 24 = 21	45
35 – 40	78 – 45 = 33	78
40 – 45	89 – 78 = 11	89
45 – 50	92 – 89 = 3	92
50 – 55	98 – 92 = 6	98
55 – 60	100 – 98 = 2	100
Total (n)	100

$\begin{array}{l} We have, n = 100 . \\ ∴ \frac{n}{2} = \frac{100}{2} = 50 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 50 is 78 \\ which belongs to interval 35 - 40 . \\ ∴ Median class = 35 - 40 \\ Now, we have \\ l = 35, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 45, \\ Frequency of median class (f) = 33, \end{array}$

$\begin{array}{l} Class size (h) = 5 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 35 + (\frac{\frac{100}{2} - 45}{33}) \times 5 \\ or x = 35 .76 \\ Hence, median age is 35 .76 years. \end{array}$

Q.19 The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm)

Number of leaves

118 – 126

127– 135

136– 144

145–153

154–162

163–171

172–180

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180)

Ans.

$\begin{array}{l} The given class intervals are not continuous. We convert it to continuous class intervals by subtracting 0.5 from each \\ of lower boundary and adding 0.5 in each of upper boundary. \\ We write the class interval, number of leaves (f_{i}), and cumulative frequency (cf) as below in the following table \\ on the basis of the given information. \end{array}$

Class interval	Number of leaves (f_i)	Cumulative frequency (cf)
117.5 – 126.5	3	3
126.5 – 135.5	5	8
13.5 – 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	40

$\begin{array}{l} We have, n = 40 . \\ ∴ \frac{n}{2} = \frac{40}{2} = 20 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 20 is 29 \\ which belongs to interval 144 .5 - 153 .5. \\ ∴ Median class = 144 .5 - 153 .5 \\ Now, we have \\ l = 144 .5, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 17, \\ Frequency of median class (f) = 12, \\ Class size (h) = 9 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 144 .5 + (\frac{\frac{40}{2} - 17}{12}) \times 9 \\ or Median = 146 .75 \\ Hence, median length of leaves is 146 .75 mm. \end{array}$

Q.20 The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000	14 56 60 86 74 62 48

Find the median life time of a lamp.

Ans.

$\begin{array}{l} We write the cumulative frequency (cf) as below in the \\ following table on the basis of the given information. \end{array}$

Life time (in hours)	Number of lamps (f_i)	Cumulative frequency (cf)
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130
3000 – 3500	86	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400
Total (n)	400

$\begin{array}{l} We have, n = 400 . \\ ∴ \frac{n}{2} = \frac{400}{2} = 200 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 200 is 216 \\ which belongs to interval 3000 - 3500 . \\ ∴ Median class = 3000 - 3500 \\ Now, we have \\ l = 3000, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 130, \\ Frequency of median class (f) = 86, \\ Class size (h) = 500 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 3000 + (\frac{\frac{400}{2} - 130}{86}) \times 500 \\ or Median = 3406 .98 \\ Hence, median life time of a lamp is 3406 .98 hours. \end{array}$

Q.21 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Ans.

$\begin{array}{l} We find the cumulative frequency (cf) as below in the \\ following table on the basis of the given information. \end{array}$

Number of letters	Number of surnames (f_i)	Cumulative frequency (cf)
1 – 4	6	6
4 – 7	30	36
7 – 10	40	76
10 – 13	16	92
13 – 16	4	96
16 – 19	4	100

$\begin{array}{l} Total(n) = 100 \\ We have, n = 100 . \\ ∴ \frac{n}{2} = \frac{100}{2} = 50 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 50 is 76 \\ which belongs to interval 7 - 10 . \\ ∴ Median class = 7 - 10 \end{array}$

$\begin{array}{l} Now, we have \\ l = 7, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 36, \\ Frequency of median class (f) = 40, \\ Class size (h) = 3 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 7 + (\frac{\frac{100}{2} - 36}{40}) \times 3 \\ or Median = 8 .05 \\ ∴ Median number of letters in the surnames is 8.05. \\ Now, we find class mark for each interval by using the \\ following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 11 .5 \end{array}$

We proceed as below to find d_i, u_i, f_iu_i .

Number of letters	Number of surnames (f_i)	x_i	d_i = x_i – a	$u_{i} = \frac{x_{i} - a}{h}$	f_iu_i
1 – 4	6	2.5	-9	-3	-18
4 – 7	30	5.5	-6	-2	-60
7 – 10	40	8.5	-3	-1	-40
10 – 13	16	11.5	0	0	0
13 – 16	4	14.5	3	1	4
16 – 19	4	17.5	6	2	8
Total (n)	100				-106

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 11 .5 + \frac{- 106}{100} \times 3 \\ = 8 .32 \\ We know that \\ 3 Median = Mode + 2 Mean \\ 3 \times 8 .05 = Mode + 2 \times 8 .32 \\ or Mode = 7 .51 \\ Hence, mean number of letters in the surnames is 8 .32 \\ and modal size of the surnames is 7 .51. \end{array}$

Q.22 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
Number of students	2	3	8	6	6	3	2

Ans.

$\begin{array}{l} We find the cumulative frequency (cf) as below in the \\ following table on the basis of the given information. \end{array}$

Weight (in kg)	Number of students (f_i)	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 (n)	30

$\begin{array}{l} We have, n = 30 . \\ ∴ \frac{n}{2} = \frac{30}{2} = 15 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 15 is 19 \\ which belongs to interval 55 - 60 . \\ ∴ Median class = 55 - 60 \\ Now, we have \\ l = 55, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 13, \\ Frequency of median class (f) = 6, \\ Class size (h) = 5 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 55 + (\frac{\frac{30}{2} - 13}{6}) \times 5 \\ or Median = 56 .67 \\ Hence, median weight of students is 56 .67 kg. \end{array}$

Q.23 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

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

Ans.

$We find less than type cumulative frequency table as below.$

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

$\begin{array}{l} Now, taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we draw its \\ ogive as following. \end{array}$

Q.24 During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)

Number of students

Less than 38

Less than 40

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

Less than 52

Draw a less than type graph for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Ans.

$Following is the given cumulative frequency distributions of less than type$

$\begin{array}{l} Weight (in kg) \\ upper class limits \end{array}$	$\begin{array}{l} Number of students \\ (Cumulative frequency) \end{array}$
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Now locate 17.5 (half of 35) on y-axis and draw a line parallel to x-axis to cut the graph at a point.
Draw a perpendicular from this point to x-axis. The point where this perpendicular meets the x-axis
determines the median of the data. From the graph we see the median is 46.5.

$\begin{array}{l} Now, we mark the point A(17.5, 46.5). So median of the given data is 46.5. \\ We find class intervals with their respective frequencies as the following. \end{array}$

Weight (in kg)	Frequency (f)	Cumulative Frequency
Less than 38	0	0
38 – 40	3 – 0 = 3	3
40 – 42	5 – 3 = 2	5
42 – 44	9 – 5 = 4	9
44 – 46	14 – 9 = 5	14
46 – 48	28 – 14 = 14	28
48 – 50	32 – 28 = 4	32
50 – 52	35 – 32 = 3	35
Total (n)	35

$\begin{array}{l} \frac{n}{2} = \frac{35}{2} = 17 .5 \\ Cumulative frequency just greater that 17.5 is 28 which belongs \\ to class interval 46 - 48 . \\ ∴ Median class = 46 - 48 \\ Now, we have \\ l = 46, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 14, \\ Frequency of median class (f) = 14, \\ Class size (h) = 2 \end{array}$

$\begin{array}{l} We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 46 + (\frac{\frac{35}{2} - 14}{14}) \times 2 \\ or Median = 46 .5 \\ So median of the given data is 46 .5. \\ Hence, value of the median is verified. \end{array}$

Q.25 The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield

(in kg/ha)

50-55

55-60

60-65

65-70

70-75

75-80

Number of farms

Change the distribution to a more than type distribution, and draw its graph.

Ans.

$Following is the cumulative frequency distributions of more than type of the given data..$

$Production yield (in kg/ha)$	$\begin{array}{l} Number of farms \\ (Cumulative frequency) \end{array}$
More than or equal to 50	100
More than or equal to 55	100 – 2 = 98
More than or equal to 60	98 – 8 = 90
More than or equal to 65	90 – 12 = 78
More than or equal to 70	78 – 24 = 54
More than or equal to 75	54 – 38 = 16

Now, draw the graph by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

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NCERT Solutions for Class 10 Maths Related Chapters

Real Numbers

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

Arithmetic Progressions

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Surface Areas and Volumes

Probability

FAQs (Frequently Asked Questions)

The main formula is Probability = Favourable outcomes / Total possible outcomes. This formula works when all outcomes are equally likely.

Exercise 14.1 has 25 questions. These questions cover coins, dice, cards, marbles, balls, pens and complementary events.

Experimental probability is based on actual trials. Theoretical probability is calculated using possible outcomes and favourable outcomes.

Probability cannot be more than 1 because favourable outcomes cannot exceed total outcomes. So, every probability lies from 0 to 1.

Two-dice sums, complementary events and “not” questions are usually tricky. Students should count total outcomes carefully before solving.

NCERT Solutions Class 10 Maths Chapter 14 Probability

Key Takeaways

NCERT Solutions Class 10 Maths Chapter 14 Structure 2026-27

NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1

Q1. Complete the following statements.

(i) Probability of an event E + Probability of the event ‘not E’ = ___.

(ii) The probability of an event that cannot happen is ___. Such an event is called ___.

(iii) The probability of an event that is certain to happen is ___. Such an event is called ___.

(iv) The sum of probabilities of all elementary events is ___.

(v) The probability of an event is greater than or equal to ___ and less than or equal to ___.

Q2. Which experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

Q3. Why is tossing a coin considered fair for deciding which team gets the ball?

Q4. Which of the following cannot be the probability of an event?

Q5. If P(E) = 0.05, what is the probability of ‘not E’?

Q6. A bag contains lemon flavoured candies only. Find the probability of taking out an orange candy and a lemon candy.

(i) Orange flavoured candy

(ii) Lemon flavoured candy

Q7. Probability of 2 students not having the same birthday is 0.992. Find the probability that they have the same birthday.

Q8. A bag contains 3 red balls and 5 black balls. Find the probability that the ball drawn is red and not red.

(i) Probability of red ball

(ii) Probability of not red

Q9. A box contains 5 red, 8 white and 4 green marbles. Find the probabilities.

(i) Probability of red

(ii) Probability of white

(iii) Probability of not green

Q10. A piggy bank contains 100 fifty-paise coins, 50 one-rupee coins, 20 two-rupee coins and 10 five-rupee coins.

(i) Probability of a 50p coin

(ii) Probability of not getting a Rs 5 coin

Q11. A tank contains 5 male fish and 8 female fish. Find the probability of taking out a male fish.

Q12. A spinner has numbers 1 to 8. Find the required probabilities.

(i) Probability of 8

(ii) Probability of an odd number

(iii) Probability of a number greater than 2

(iv) Probability of a number less than 9

Q13. A die is thrown once. Find the probabilities.

(i) Probability of a prime number

(ii) Probability of a number lying between 2 and 6

(iii) Probability of an odd number

Q14. One card is drawn from a well-shuffled deck of 52 cards.

(i) Probability of a king of red colour

(ii) Probability of a face card

(iii) Probability of a red face card

(iv) Probability of the jack of hearts

(v) Probability of a spade

(vi) Probability of the queen of diamonds

Q15. Five diamond cards are shuffled. Find the probabilities.

(i) Probability that the card is queen

(ii) Queen is drawn and put aside. Find second card probabilities.

(a) Probability that second card is ace

(b) Probability that second card is queen

Q16. 12 defective pens are mixed with 132 good pens. Find the probability of drawing a good pen.

Q17. A lot of 20 bulbs contains 4 defective bulbs.

(i) Probability that one bulb drawn is defective

(ii) First bulb drawn is not defective and not replaced. Find probability that second bulb is not defective.

Q18. A box contains 90 discs numbered 1 to 90.

(i) Probability of a two-digit number

(ii) Probability of a perfect square number

(iii) Probability of a number divisible by 5

Q19. A die has faces A, B, C, D, E, A. Find the probabilities.

(i) Probability of A

(ii) Probability of D

Q20. A die is dropped on a rectangle of 3 m by 2 m. Find probability of landing inside a circle of diameter 1 m.

Q21. A lot has 144 ball pens, of which 20 are defective. Find the probabilities.

(i) Probability that Nuri will buy it

(ii) Probability that Nuri will not buy it

Q22. Complete the probability table for the sum on two dice.

(ii) Is the argument that each sum has probability 1 / 11 correct?

Q23. A one-rupee coin is tossed 3 times. Hanif wins if all tosses give the same result. Find probability that Hanif loses.

Q24. A die is thrown twice. Find the probabilities.

(i) Probability that 5 will not come up either time

(ii) Probability that 5 will come up at least once

Q25. Which arguments are correct? Give reasons.

(i) If two coins are tossed, there are three outcomes: two heads, two tails or one of each. So each has probability 1 / 3.

(ii) If a die is thrown, there are two outcomes: odd or even. So probability of getting an odd number is 1 / 2.

Probability Class 10 NCERT Solutions: Key Concepts

Theoretical Probability

(ii) The probability of an event that cannot happen is _. Such an event is called _.

(iii) The probability of an event that is certain to happen is _. Such an event is called _.

(v) The probability of an event is greater than or equal to _ and less than or equal to _.