Table of Contents

## Probability Class 10 Extra Questions Maths Chapter 15

Welcome to Infinity Learn, your ultimate destination for comprehensive exam preparation. In the realm of Class 10 Mathematics, Chapter 15 – Probability holds paramount significance. To enhance your understanding and fortify your exam readiness, we present a curated selection of Class 10 Extra Questions of Maths Chapter 15. These meticulously crafted queries are designed to sharpen your grasp on the intricate concepts of probability, ensuring that you are well-equipped to tackle any challenges that may arise during your examination. Also check out other resources of class 10 that can be helpful for your exam preparation **CBSE Class 10 Maths Notes**, and **Class 10 Maths previous year paper with solutions**. These valuable resources provide comprehensive study materials and practice papers designed to help you excel in your academic journey.

### NCERT Solutions for Class 10 All Subjects

**NCERT Solutions for Class 10 Maths****NCERT Solutions for Class 10 Science****NCERT Solutions for Class 10 Social Science****NCERT Solutions for Class 10 English**

**Question 1.**

**A number is chosen at random from the numbers -3, -2, -1,0,1,2,3. What will be the probability that square of this number is less than or equal to 1 ? [CBSE Delhi 2017]**

**Answer:**

Here total outcomes = 7, favourable outcomes are -1, 0, 1

Reqd. probability = No. of favourable outcomes Total no. of outcomes =37

**Question 2.**

**The probability that it will rain today is 0.75. /£) What is the probability that it will not rain today?**

**Answer:**

P(rain today) = 0.75

P(not rain today) = 1 – P(rain today)

= 1 – 0.75

= 0.25

**Question 3.**

**The probability of getting a bad egg in a lot of 500 eggs is 0.028. Find the number of good eggs in the lot.**

**Answer:**

Total eggs = 500

Let no. of bad eggs = x

P(bad eggs) = x500

0.028 = x500 ⇒ x = 14

Number of good eggs = 500 – 14

= 486

Hence, number of good eggs is 486.

**Question 4.**

**A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen. ****[CBSE Outside Delhi 2016]**

**Answer:**

Let E be the event of getting neither red card nor queen

No. of red cards = 26 (including 2 queens)

Remaining (black) queens = 2

Neither red nor queen

= 52 – (26 + 2) = 52 – 28 = 24 24 6

P(E) = 2452=613

**Also Check: Some Applications of Trigonometry Class 10 Extra Questions Maths Chapter 9**

**Question 5.**

**A card is selected at random from a deck of 52 cards. Find the probability that the selected card is red face card.**

**Answer:**

Total cards = 52

Red face cards = 6

P(a red face card) = 652=326

**Question 6.**

**A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.**

**[CBSE Outside Delhi 2016]**

**Answer:**

Let E be the event of getting neither red card nor queen

No. of red cards = 26 (including 2 queens)

Remaining (black) queens = 2

Neither red nor queen

= 52 – (26 + 2) = 52 – 28 = 24 24 6

P(E) = 2452=613

**Question 7.**

**In a well shuffled pack of cards, a card is © drawn at random. Find the probability of getting a black queen.**

**Answer:**

Total cards = 52

Total no. of black queens = 2

∴ P(a black queen) = 252=126

**Question 8.**

**Two coins are tossed simultaneously. What is the probability of getting exactly one ‘. head?**

**Answer:**

Here, sample space S = {HH, HT, TH, TT}

∴ n(S) = 4

and E = {HT, TH}

∴ n(E) = 2

∴ Reqd. Prob. = P(E) = 24=12

**Question 9.**

**In a lottery there are 13 prizes and 117 blanks. What is the probability of not winning a prize?**

**Answer:**

No. of ways of not winning a prize

= No. of blanks = 117

Total cases = 13 + 117 = 130

P(not winning) = 117130=910

**Question 10.**

**What is the probability of getting “a black¬face card’ from a well shuffled deck of 52 playing cards?**

**Answer:**

There are 6 black face cards.

(2 red kings, 2 red queens and 2 jacks)

Total cards = 52

∴ Required Probability = 652=326

**Question 11.**

**Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the bulb drawn is non-defective.**

**Answer:**

No. of non-defective bulbs = 400 – 15 = 385

Total bulbs = 400

∴ P(a non-defective bulb) = 385400=7780

Question 16.

If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons.

Answer:

No, the outcomes ‘head’ and ‘tail’ are equally likely every time regardless of what you get in a few tosses.

**Also Check: Polynomials Class 10 Extra Questions Maths Chapter 2**

**Question 12.**

**Cards marked with number 3,4,5……………. ,50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. [CBSE 2016]**

**Answer:**

It is given that the box contains cards marked with numbers 3,4,5, ………… , 50.

∴ Total number of outcomes 48

There are six perfect squares, i.e., 4, 9, 16, 25, 36 and 49.

∴ Number of favourable outcomes = 6

Probability that a card drawn at random bears a perfect square

= Number of favourable outcomes Total number of outcomes =648=18

**Question 13.**

**What is the probability that a doublet occurs on throwing two dice?**

**Answer:**

Total outcomes = 36

Doublets = 6

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

P(a doublets) = 636=16

**Question 14.**

**Two unbiased coins are tossed. What is the probability of getting at most one head?**

**Answer:**

Here, S = {HH, HT, TH, TT).

Let E = event of getting at most one head.

∴ E = {TT, HT, TH}.

∴ P(E) = n(E)n( S)=34

**Question 15.**

**An unbiased die is tossed. Find the probability of getting a multiple of 3?**

**Answer:**

Here, S = {1, 2, 3, 4, 5, 6}

Then, E = {3, 6}

∴ P(E) = n(E)n( S)=26=13

**Also Check: Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13**

### Short Answer Type Questions

**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 34. 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 34: 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 35. For a family of 3 children, find out the probability of having at least one boy.**

Answer 35:

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.

**Question 36. A letter of the English alphabet is chosen at random. Find out the probability that the chosen letter is a consonant.**

Answer 36:

Total English alphabets are = 26

The number of consonants is = 21

So, P for (letter is a consonant) = 21/26

**Question 37. A card is drawn at random in a well-shuffled pack of 52 playing cards. Find out the probability of getting neither a red card nor a queen.**

Answer 37:

For S = 52

P for (neither a red card nor a queen)

= 1 – P for (red card or a queen)

[Red card = 26, Red Queen = 2, Queen = 4]= 1 – [(26+4+2)/52]

= 1 – (28/52)

= 24/52

= 6/ 13

**Question 38. A box consists of cards numbered 6 to 50. A card is drawn randomly from the box. The probability with the drawn card has a number that is a complete square is.**

Answer 38:

Total number for cards = 50 – 6 + 1 = 45

complete square numbers are 9, 16, 25, 36, 49, that is

5 numbers

Thus, P for (a perfect square) = 5/45 =1/9

**Question 39: The probability for selecting a rotten apple at random from a heap of 900 apples is 0.18. What is the number of rotten apples in a heap?**

Answer 39:

Given,

Total number having apples in the heap = n(S) = 900

Let E be the event for selecting a rotten apple from the heap.

Number having outcomes favourable to E = n(E)

P(E) = n(E)/n(S)

0.18 = n(E)/900

⇒ n(E) = 900 × 0.18

⇒ n(E) = 162

Thus, the number having rotten apples in a heap = 162

**Question 40. A die is thrown with repetition until a six comes up. What is the sample space of the experiment? HINT: A={6}, B={1,2,3,4,5,}**

Answer 50: The sample space of the given experiment is

{A, BA, BBA, BBBA, BBBBA)