HOTS Questions on Class 9 Maths Chapter 7 ‘The Mathematics of Maybe: Introduction to Probability’ with Answers

This Higher-Order Thinking Skills (HOTS) question set for Chapter 7: ‘The Mathematics of Maybe: Introduction to Probability’ for Class 9 is designed to help students go beyond basic concepts and develop strong probability reasoning skills. Aligned with CBSE and NCERT aims, these questions emphasise conceptual understanding and creative application of probability ideas such as outcomes and sample spaces, experimental vs theoretical probability, simple event computation, complement and conditional thinking, and modelling chance in everyday contexts. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Quick Concept Review of Chapter 7

• Experiment: Any action with an uncertain outcome. Example: Tossing a coin, rolling a die, drawing a card.

• Sample Space (S): The complete set of all possible outcomes of an experiment.

• One coin: S = {H, T} → n(S) = 2

• Two coins: S = {HH, HT, TH, TT} → n(S) = 4

• One die: S = {1, 2, 3, 4, 5, 6} → n(S) = 6

• Two dice: S = 36 outcomes 

• A standard deck of cards: n(S) = 52

• Probability Formula:

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

• Key Rules:

• 0 ≤ P(E) ≤ 1 always

• P(E) + P(E') = 1, where E' is the complement (the event NOT happening)

• P(impossible event) = 0

• P(certain event) = 1

• Equally Likely Outcomes: Each outcome has the same chance of occurring.
  
  

Solved HOTS Questions on Probability

Question 1: Three unbiased coins are tossed simultaneously. Find the probability of getting:

(a) Exactly two heads

(b) At least one head

(c) At most two tails

(d) Exactly one tail

Solution:

Sample Space: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total outcomes: n(S) = 8

(a) Exactly two heads: {HHT, HTH, THH} 

⇒ n(E) = 3

P(exactly two heads) = 3/8

(b) At least one head (one or more heads): {HHH, HHT, HTH, THH, HTT, THT, TTH} 

⇒ n(E) = 7

P(at least one head) = 7/8

Shortcut: P(at least one head) = 1 − P(no heads) = 1 − 1/8 = 7/8

(c) At most two tails (0, 1, or 2 tails): {HHH, HHT, HTH, THH, HTT, THT, TTH} ⇒  n(E) = 7

P(at most two tails) = 7/8

(d) Exactly one tail: {HHT, HTH, THH} 

⇒ n(E) = 3

P(exactly one tail) = 3/8

Question 2: Two dice are thrown. Given that the sum of the numbers on the two dice is 7, what is the probability that one of the dice shows a 2?

Solution:

Outcomes where sum = 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} ⇒  6 outcomes

Among these, outcomes where at least one die shows 2: {(2,5),(5,2)} ⇒  2 outcomes

P(one die shows 2 | sum is 7) = 2/6 = ⅓

Question 3: When two dice are thrown together, the product of the numbers appearing on them lies between 8 and 13 (exclusive). Find the probability.

Solution:

We need products: 9, 10, 11, 12 (strictly between 8 and 13)

• Product = 9: (3,3) → 1 way

• Product = 10: (2,5),(5,2) → 2 ways

• Product = 11: Not possible (11 is prime, and dice only go to 6)

• Product = 12: (2,6),(6,2),(3,4),(4,3) → 4 ways

Total: 1+2+4 = 7 outcomes

P = 7/36

Question 4: All queens, jacks, and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and one is drawn. Find the probability that the card drawn is:

(a) A king

(b) A face card

(c) A card of clubs

Solution:

Cards removed: 4 queens + 4 jacks + 4 aces = 12 cards

Remaining cards: 52 − 12 = 40

(a) Kings remaining: 4 kings (not removed)

P = 4/40 = 1/10

(b) Face cards remaining: After removing queens and jacks, only kings remain as face cards = 4

P = 4/40 = 1/10

(c) Clubs remaining: 13 clubs originally − 1 queen of clubs − 1 jack of clubs − 1 ace of clubs = 10

P = 10/40 = 1/4

Question 5: From a deck of 52 cards, one card is drawn at random. If it is known that the drawn card is red, what is the probability that it is a king?

Solution:

Red cards in a deck: 26 (13 hearts + 13 diamonds)

Red kings: 2 (king of hearts + king of diamonds)

P(king | card is red) = 2/26 = 1/13

Question 6: A box contains cards numbered 6 to 55. A card is drawn at random. Find the probability that the number on the card is:

(a) A perfect square

(b) Divisible by 5

(c) A prime number less than 20

(d) A multiple of both 3 and 7

Solution:

Cards: 6, 7, 8, ..., 55 

⇒ Total = 55 − 6 + 1 = 50 cards

(a) Perfect squares from 6 to 55: 9, 16, 25, 36, 49 

⇒ 5 numbers

P = 5/50 = 1/10

(b) Divisible by 5: 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 

⇒ 10 numbers

P = 10/50 = 1/5

(c) Prime numbers less than 20, in range 6–55: 7, 11, 13, 17, 19 

⇒ 5 numbers

P = 5/50 = 1/10

(d) Multiple of both 3 and 7 = multiple of 21: 21, 42 

⇒ 2 numbers

P = 2/50 = 1/25

Question 7: In a survey of 200 students, it was found that 60 like cricket, 80 like football, 40 like both, and the rest like neither. A student is chosen at random. Find the probability that the student:

(a) Likes cricket only

(b) Likes football only

(c) Likes neither

(d) Likes at least one sport

Solution:

Students who like cricket only = 60 − 40 = 20

Students who like football only = 80 − 40 = 40

Students who like both = 40

Students who like neither = 200 − (20 + 40 + 40) = 100

(a) P(cricket only) = 20/200 = 1/10

(b) P(football only) = 40/200 = 1/5

(c) P(neither) = 100/200 = 1/2

(d) P(at least one sport) = 1 − 1/2 = ½

Questions PDF with worked-out examples for Class 9 Chapter 7: The Mathematics of Maybe: Introduction to Probability, perfect for last-minute CBSE exam revision.

Class 9 Chapter 7: The Mathematics of Maybe: Introduction to Probability HOTS PDF