Statistics And Probability Chapter 13: NCERT Exemplar for Class 10

The NCERT Exemplar Class 10 Math Chapter 13 Statistics and Probability is provided here for students to prepare for CBSE exams. These problems and solutions are prepared by our subject matter experts in acceptance with CBSE’s latest syllabus of 2024-25 respectively to help students score better. Students can use the NCERT exemplars for class 10th for studying as well as practicing questions. The exemplar also contains solved questions relevant to the exercise problems present in the textbook. it will allow students to have a thorough revision of the entire chapter and prepare them to face the board exam.

Chapter 13 has problems based on the different statistical measures like mean, mode and median. Students will learn about how to solve these problems and also explain the concept of cumulative frequency curves and more. Additionally, students will also gains an experimental approach to probability. Furthermore, they will explore concepts like the multiplication rule of probability, Bayes’ theorem, and the independence of events. These Exemplars problems and solutions cover the following topics in the chapter Statistics and Probability. These all concepts can be consider as new to students but this chapter is a glimpse or advance version of earlier chapter and can be crack easily with highest scoring segment.

With proper technique and practice,boards can be clear with flying colors.

Scroll down below to access the solutions!!

Question 1: An event is not likely to happen. Its probability is closest to

(A) 0.0001

(B) 0.001

(C) 0.01

(D) 0.1

Answer 1: (A) 0.0001

Explanation:

The probability for the event which is not likely to happen is closest to zero, and from the given alternative, 0.0001 is closest to zero.

So, the correct answer is an option (A).

 

Question 2. When an event cannot occur, its probability is

(A)1

(B) ¾

(C) ½

(D) 0

Answer 2: (D) 0

Explanation:

The event which cannot occur is called an impossible event.

The probability of an impossible event is zero.Therefore, the probability is 0.

So, option (D) is correct.

 

Question 3. Which among the following cannot be the probability of an event?

(A)1/3

(B) 0.1

(C) 3%

(D)17/16

Answer 3: (D)17/16

Explanation:

The probability for the event always lies between 0 and 1.

Probability for the event cannot be more than 1 or negative as (17/16) > 1

So, option (D) is correct

 

Question 4: When the probability for the event is p, the probability of the complementary event will be

(A) p – 1

(B) p

(C) 1 – p

(D) 1-1/p

Answer 4: (C) 1 – p

Explanation:

As the probability for the event + probability for the complimentary event = 1

It can be written as

Probability for the complimentary event = (1 – Probability for the an event) = 1 – p

Thus, the correct answer is an option (C).

 

Question 5: The probability expressed as the percentage for the particular occurrence can never be

(A) less than 100

(B) less than 0

(C) greater than 1

(D) anything but the whole number

Answer 5: (B) less than 0

Explanation:

We are aware that the probability expressed as a percentage always lies between 0 and 100. Thus, it cannot be less than 0.

Thus, the correct answer is an option (B).

 

Question 6: When you P(A) denotes the probability for the event A, then

(A) P(A) < 0

(B) P(A) > 1

(C) 0 ≤ P(A) ≤ 1

(D) –1 ≤ P(A) ≤1

Answer 6: (C) 0 ≤ P(A) ≤ 1

Explanation:

The probability for the event always lies between 0 and 1.

we conclude that a correct answer is an option (C).

Question 7: A card has been selected from a deck of 52 cards. The probability for the being the red face card is

(A) 3/26

(B)3/13

(C) 2/13

(D) 1/2

Answer 7: (A) 3/26

Explanation:

In a deck for the 52 cards, there are 12 face cards that are 6 red and 6 black cards.

Hence, probability for getting a red face card = 6/52 = 3/26.

we conclude a correct answer is an option (A).

 

Question 8: The probability, which is a non-leap year selected at random, will contain 53 Sundays is

(A) 1/7

(B) 2/7

(C)3/7

(D) 5/7

Answer 8: (A) 1/7

Explanation:

A non-leap year has 365 days and thus 52 weeks and 1 day. This 1 day may be Sunday, Monday, Tuesday, Wednesday or Thursday or Friday, or Saturday.

Hence, out of 7 possibilities, 1 favourable event is the event that the one day is Sunday.

The required probability = 1/7

So, the correct answer is an option (A).

 

Question 9: When the die is thrown, the probability for the getting an odd number less than 3 is

(A) 1/6

(B) 1/3

(C) 1/2

(D) 0

Answer 9: (A) 1/6

Explanation :

If the die is thrown, then the total number of outcomes = 6

The odd number less than 3 is 1 only. Number of possible outcomes = 1

The required probability =1/6

we conclude the correct answer is an option (A).

 

Question 10: A card is drawn for the deck of 52 cards. The event E so that card is not the ace for the hearts. The number for the outcomes favourable to E is

(A) 4

(B) 13

(C) 48

(D) 51

Answer 10: (D) 51

Explanation :

In the deck of 52 cards, there are 13 cards of the heart, and 1 is the ace of the heart.

thus, the number of outcomes favourable to E = 52 − 1 = 51

Therefore, the correct answer is an option (D).

 

Question 11: The probability for getting a bad egg in a lot for the 400 is 0.035. The number for the bad eggs in the lot is

(A) 7

(B) 14

(C) 21

(D) 28

Answer 11: (B) 14

Explanation:

The total number of eggs = 400

Probability for the getting a bad egg = 0.035

Probability for the getting bad egg = Number for the bad eggs / Total number of eggs

⇒ Probability for the getting bad egg = Number for the bad eggsTotal number of eggs

⇒0.035 = Number for the bad eggs / 400

Number for the bad eggs = 0.035 × 400 = 14

Thus, the correct answer is option (B).

 

Question 12: A girl calculates that the probability of winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?

(A) 40

(B) 240

(C) 480

(D) 750

Answer 12: (C) 480

Explanation: Given that,

The total number for sold tickets = 6000

Assume that girl bought X tickets.

probability for winning the first prize is given as,

X / 6000=0.08

X = 480

Thus, the correct answer is option (C).

 

Question 13: One ticket was drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

(A) 1/5

(B) 3/5

(C)4/5

(D) 1/3

Answer 13: (A) 1/5

Explanation:

The number of total outcomes = 40

Multiples of 5 between 1 to 40 = 5, 10, 15, 20, 25, 30, 35, 40

Total number for the possible outcomes = 8

Required probability = 8/40 = ⅕

Therefore, the correct answer is Option (A).

 

Question 14: Someone is asked to take the number from 1 to 100. The probability which is a prime is

(A) 1/5

(B) 6/25

(C) 1/4

(D) 13/50

Answer 14: (C) 1/4

Explanation:

Total numbers for the outcomes = 100

Thus, the prime numbers for the 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89 and 97.

Total number of possible outcomes = 25

Required probability =25/100 = ¼

Therefore, the probability that it is a prime is ¼

we conclude a correct answer is an option (C).

 

Question 15: The school has five houses A, B, C, D, and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D, and the rest from house E. A single student is selected at random to be the class monitor. The probability of the selected student is not among A, B, and C is

(A) 4/23

(B) 6/23

(C) 8/23

(D) 17/23

Answer 15: (B) 6/23

Explanation:

Total number for students = 23

Number for the students in houses A, B and C = 4 + 8 + 5 = 17

Remaining students = 23 − 17 = 6

Thus, probability that the selected student is not from houses A, B and C = 6/23

Therefore, the correct answer is option (B).

 

Question 16: In a family that has three children, there might be no girl, one girl, two girls, or three girls. Hence, the probability for each is 1/4. Is this correct? Justify your answer.

Answer 16:

No, the probability for each is not 1/4.

Let the boys be B and girls be G.

Outcomes could be {BBB, GGG, BBG, BGB, GBB, GGB, GBG, BGG}.

The total number for the outcomes = 8

Thus,

P(No girl) = 1/8

P(1 girl) = 3/8

P(2 girls) = 3/8

P(3 girls) = 1/8

 

Question 17: Apoorv throws two dice once as well as computes the product for the numbers appearing on the dice. Peehu throws one die as well as squares the number which appears on it. Who has the better chance of getting the number 36? Why?

Answer 17:

Apoorv has thrown two dice once.

Thus, the total number for the outcomes is 62 = 36

The number for the outcomes for getting product 36 = 1 (6 × 6)

Probability for Apoorv = 1/36

Peehu threw one die.

Thus, the total number for the outcomes is 6

The number for the outcomes for getting square 36 = 1 (62)

Probability of Peehu = 1/6

T Therefore, Peehu has a better chance of getting the number 36.

 

Question 18: When we toss the coin, there are two possible outcomes – Head or tail. Thus, the probability of each outcome is 1/2. Justify your answer.

Answer 18:

Yes, the probability of each outcome is ½ as the head and tail both are equally likely events.

 

Question 19: A student says that when you throw a die, it will show up 1 or not 1. Hence, the probability for getting 1 and the probability for getting ‘not 1’ each is equal to 1/2. Is this correct? State reason.

Answer 19:

No, this is not correct.

Suppose we throw the die, then the total number for the outcomes = 6

For the possible outcomes = 1 or 2 or 3 or 4 or 5 or 6

Probability for the getting 1 = ⅙

Then, the probability for the getting, not 1 = 1 – Probability for getting 1

= 1-1/6

= ⅚

 

Question 20: I toss the three coins together. The possibilities for the outcomes are no heads, 1 head, 2 heads as well as 3 heads. So, I say that the probability for the no heads is 1/4. What is wrong with the given conclusion?

Answer 20:

Total number for the outcomes = 23 = 8

Possible outcomes are (HHH), (HTT), (THT), (TTH), (HHT), (THH), (HTH), and (TTT).

Then, the probability for the getting no head = ⅛

So, the given conclusion is wrong as the probability for the no head is 1/8.

 

Question 21: When you toss a coin 6 times, and it comes down heads on each occasion tossed, can you say that the probability for getting the head is 1? State reasons.

Answer 21:

No, when we toss a coin, the possible outcomes are head or tail.

Both the events are equally likely.

Hence, the probability is 1/2.

When we toss the coin six times, the probability will be the same in each case.

Thus, the probability of getting the head is not 1.

 

Question 22: A bag has slips numbered from 1 to 100. When Fatima chooses the slip at random from the bag, it will either have an odd number or an even number. Since the situation has only two possible outcomes, therefore, the probability for each is 1/2. Give justification.

Answer 22 :

We have numbers between 1 to 100. Half the numbers are even, as well as half the numbers are odd is 50 numbers that are (2, 4, 6, 8, …, 96, 98, 100) even, as well as 50 numbers that are (1, 3, 5, 7, …, 97, 99) odd.

Thus, both the events are equally likely.

Hence, the probability for the getting an even number = 50/100 = ½

Also, probability for the getting odd number = 50/100 = ½

Therefore, the probability of each is 1/2.

 

Question 23: The coin is tossed two times. Find out the probability for getting at most one head.

Answer 23:

The possibility of outcomes, when a coin is tossed 2 times are

S = {(HH), (TT), (HT), (TH)}

n(S) = 4

Let E = Event for the getting at most one head = {(TT), (HT), (TH)}

n(E) = 3

Therefore, required probability = n(E)/n(S)=3/4.

 

Question 24. Complete the following sentences:

(i) Probability for the event E + Probability for the event ‘not E’ = ___________.

(ii) The probability for the event that cannot happen is __________. Such an event is known as________.

(iii) The probability for the event that is certain to happen is _________. Such an event is known as _________.

(iv) The sum for the probabilities of all the elementary events for the experiment is __________.

(v) The probability for the event is greater than or same as ___ and less than or same as__________.

Answer 24:

(i) Probability for the event E + Probability for the event ‘not E’ = 1.

(ii) The probability for the event that cannot happen is 0. Such an event is known as an impossible event.

(iii) The probability for the event that is certain to happen is 1. Such an event is known as a sure or certain event.

(iv) The sum for the probabilities of all the elementary events for the experiment is 1.

(v) The probability for the event is greater than or same as 0 and less than or same as 1.

 

Question 25. Why is tossing the coin considered to be a fair way for deciding which team shall bring the ball at the beginning of the football game?

Answer 25:

Tossing for the coin is a fair way of deciding as the number for the possible outcomes is only 2, that is, either head or tail. As these two outcomes are equally likely outcomes, tossing is unpredictable as well as considered to be completely unbiased.

 

Question 26. Which among the following cannot be the probability for the event?

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

Answer 26:

The probability for the event (E) always lies between 0 and 1, i.e., 0 ≤ P(E) ≤ 1. Thus, from the above alternatives, option (B) -1.5 cannot be the probability for the event.

 

Question 27. When P(E) = 0.05, what is the probability for the ‘not E’?

Answer 27:

We have,

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

Given that, P(E) = 0.05

Thus, P(not E) = 1-P(E)

And P(not E) = 1-0.05

P(not E) = 0.95

 

Question 28. A bag consists of lemon-flavoured candies only. Malini removes one candy without looking into the bag. What is the probability which she takes out

(i) an orange-flavoured candy?

(ii) a lemon-flavoured candy?

Answer 28:

(i) We have a bag that only contains lemon-flavoured candies.

Therefore, the no. of orange-flavoured candies = 0

The probability for the taking out orange-flavoured candies = 0/1 = 0

(ii) Since there are only lemon-flavoured candies, P(lemon-flavoured candies) = 1 (or 100%)

 

Question 29. It is given that in a group for the 3 students, the probability for the 2 students who do not have the same birthday is 0.992. What is the probability when the 2 students have the same birthday?

Answer 29:

Assume the event where 2 students have the same birthday be E

Given, P(E) = 0.992

We have,

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

And P(not E) = 1–0.992 = 0.008

The probability for the 2 students have the same birthday is 0.008

 

Question 30. A bag consists of 3 red balls and 5 black balls. A ball is removed at random from the bag. What is the probability for the ball drawn is

(i) red?

(ii) not red?

Answer 30:

The total number for the balls = No. for the red balls + No. of black balls

So, the total no. for the balls = 5+3 = 8

We know that the probability for the event is the ratio between the no. of favourable outcomes and the total number of outcomes.

P(E) = (Number for the favourable outcomes/ Total number for the outcomes)

(i) Probability for the drawing red balls = P (red balls) = (no. for the red balls/total no. of balls) = 3/8

(ii) Probability for the drawing black balls = P (black balls) = (no. of black balls/total no. of balls) = ⅝

 

Question 31. A game consists of tossing the one rupee coin three times as well as noting the outcome each time. Hanif wins when all the tosses give the same result, that is, three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer 31:

The total number for the outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes for which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (lose the game) = 6/8 = ¾ = 0.75

Therefore, the probability that Hanif will lose the game is 0.75

 

Question 32. A bag consists of 5 red balls and some blue balls. When the probability for drawing a blue ball

is double for the red ball, determine the number for the blue balls in the bag.

Answer 32:

It is given that the total number for the red balls = 5

Assume the total number for the blue balls = x

Hence, the total no. of balls = x+5

P(E) = (Number for the favourable outcomes/ Total number of outcomes)

P (drawing the blue ball) = [x/(x+5)] ——–(i)

Same as,

P (drawing the red ball) = [5/(x+5)] ——–(i)

From the equation (i) and (ii)

x = 10

Hence, the total number for blue balls are = 10

 

Question 33. A jar consists of 24 marbles, some are green as well as others are blue. When marble is drawn at random from the jar, the probability for the green is ⅔. Find out the number for the blue balls in the jar.

Answer 33:

The Total marbles = 24

Assume the total green marbles = x

Thus, the total blue marbles = 24-x

P(getting green marble) = x/24

From the given question, x/24 = ⅔

Hence, the total green marbles = 16

Therefore, the total blue marbles = 24-16 = 8

Question 35. A die is thrown once. Find out the probability for getting a prime number.

Answer 35: Total possible events if a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

P(E) = (Number for the favourable outcomes/ Total number of outcomes)

Total number for the prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½

 

Question 34. Five cards are the ten, jack, queen, king, and an ace for the diamonds are well-shuffled with their face downwards. One card is picked up at random. When the queen is taken out and put aside, what is the probability that the second card picked up is a queen?

Answer 34:

Total numbers for the cards = 5

P(E) = (Number for the favourable outcomes/ Total number of outcomes)

When the queen is taken out and put aside, the total numbers for the cards left is (5-4) = 4

Total numbers for the queen = 0

P (picking a queen) = 0/4 = 0

 

Question 35. 12 defective pens are accidentally mixed by 132 good ones. It is not possible to just look for the pen and tell if or not it is defective. One pen is removed at random from the lot. Finding out the probability that the pen is taken out is a good one.

Answer 35: Numbers for the pens = Numbers for the defective pens + Numbers for the good pens

∴ Total number for the pens = 132+12 = 144 pens

P(E) = (Number for the favourable outcomes/ Total number for the outcomes)

P (pickup a good pen) = 132/144 = 11/12 = 0.916

 

Question 36. A lot consists of 144 ball pens, in which 20 are defective, and the others are good. Nuri will buy a pen when it is good but will not buy it when it is defective. The shopkeeper draws one pen at random to give it to her. What is the probability that she would not buy it?

Answer 36: The total numbers having outcomes, i.e., pens = 144

Given that numbers having defective pens = 20

∴ The numbers having non-defective pens = 144-20 = 124

P(E) = (Number having favourable outcomes/ Total number of outcomes)

Total numbers of events for which she will not buy them = 20

Thus, P for (not buying) = 20/144 = 5/36

 

Question 37. When two coins are tossed simultaneously, there are three possible outcomes, i.e., two heads, two tails, or one for each. Hence, for each of the outcomes, the probability is ⅓

Answer 37: The possibility for the events are (H,H); (H,T); (T,H) and (T,T)

Thus, P for (getting two heads) = ¼

also,

P for (getting one of the each) = 2/4 = ½

The statement is incorrect.

 

Question 38. When a die is thrown, there are two possible outcomes that are an odd number or an even number. Thus, the probability for getting an odd number is 1/2

Answer 38:

As the two outcomes are equally likely, the statement is correct.

 

Question 39. The probability for the event that will surely happen is. This event is called.

Answer 39: The probability for the event that is certain to happen is 1. This event is called a sure or a certain event.

 

Question 40. A box consists of 5 red marbles, 8 white marbles, and 4 green marbles. One marble is taken out from the box randomly. What is the probability that the marble removed will be red?

Answer 40:

The Total no. for balls = 5+8+4 = 17

P(E) = (Number for favourable outcomes/ Total number for outcomes)

Total number for the red balls = 5

P for (red ball) = 5/17 = 0.29

 

Question 41. A piggy bank consists of hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins, and ten ₹5 coins. When it is equally likely for one of the coins, it will fall out if the bank is turned upside down. What will be the probability that the coin will be a 50 p coin?

Answer 41: Total no. for coins = 100+50+20+10 = 180

P(E) = (Number for favourable outcomes/ Total number for outcomes)

Total number for 50 p coin = 100

P for (50 p coin) = 100/180 = 5/9

 

Question 42. Two coins are tossed simultaneously. The student argues that there are 11 possibilities of outcomes for the sum for the numbers on two dice: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Thus, each of them has a probability of 1/11. Do you agree with the argument?

Answer 42: The student’s argument that there are 11 possible outcomes is correct, but the outcomes are not equally likely. Hence, each of the outcomes would not have an equal probability for 11/36.

As a result, the student’s argument is incorrect.

 

Question 43. For a family of 3 children, find out the probability of having at least one boy.

Answer 43:

The probability of each child being a boy will be 1/2.

The probability of each child being a girl will be 1/2.

Probability of no boys = 1/2 × 1/2 × 1/2 = 1/8

Probability of at least 1 boy = 1 – Probability of no boys = 1 – 1/8 = 7/8

or

The number of the possibilities is as given below:

• Probability of 0 boys = 1/8
• Probability of 1 boy = 3 × 1/8 = 3/8
• Probability of 2 boys = 3 × 1/8 = 3/8
• Probability of 3 boys = 1/8

So the probability of at least 1 boy is equal to the probability of 1 or 2 or 3 boys.

i.e. = 3/8 + 3/8 + 1/8 = 7/8.