Introduction to Probability for Class 9: Important Questions and Answers on The Mathematics of Maybe

Important questions on Introduction to Probability for Class 9 are provided in this Maths article to help students understand the basic concepts of probability and improve problem-solving skills. These questions cover important topics such as experiments, outcomes, events, and probability calculations with solved examples and exam-oriented explanations based on the CBSE syllabus and NCERT textbook. Whether you follow NCERT or Ganita Manjari, this resource supports quick revision and effective exam preparation.

Table of Contents

• Key Concepts on The Mathematics of Maybe: Introduction to Probability for Class 9
• Important Questions on Introduction to Probability for Class 9

Key Concepts on The Mathematics of Maybe: Introduction to Probability for Class 9

All of Class 9 probability rests on one formula.

P(E) = Number of times event E occurred ÷ Total number of trials

Also written as: P(E) = Favourable Outcomes / Total Outcomes

And the complementary probability rule, which is tested very frequently:

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

So P(not E) = 1 − P(E)

Important Questions on Introduction to Probability for Class 9

Q1: If P(E) = 0.44, find P(not E).

Solution: Given: P(E) = 0.44

Formula: P(not E) = 1 − P(E)

P(not E) = 1 − 0.44 = 0.56

∴ P(not E) = 0.56

Q2: Can the probability of an event be 1.3? Give a reason.

Answer: No. Probability must always lie between 0 and 1 (both inclusive). A value of 1.3 is greater than 1, which is impossible. 0 ≤ P(E) ≤ 1.

Q3: In a factory of 364 workers, 91 are married. Find the probability of selecting a worker who is (i) married, and (ii) not married.

Solution: Total workers = 364

Married = 91

Not married = 364 − 91 = 273

(i) P(married) = 91/364 = 1/4 = 0.25

(ii) P(not married) = 273/364 = 3/4 = 0.75

Q4: A die is rolled and the outcome recorded. The results of 600 rolls are:

Find the probability of getting (i) an outcome less than 4, (ii) outcome 6.

Solution: Total trials = 600

Outcomes (< 4) = 80 + 75 + 110 = 265

Outcome (= 6) = 120

(i) P(outcome < 4) = 265/600 = 53/120 ≈ 0.44

(ii) P(outcome = 6) = 120/600 = 1/5 = 0.2

Q5: A cricket batsman plays 50 innings and scores runs as recorded below. A match is chosen at random. What is the probability that the batsman scored (i) a duck (0 runs), (ii) more than 50 runs, (iii) between 25 and 75 runs (inclusive)?

Solution: Total innings = 50

(i) P(duck) = 5/50 = 1/10 = 0.1

(ii) Innings with > 50 runs: 9 + 5 + 3 = 17

P(>50 runs) = 17/50 = 0.34

(iii) ‘Between 25 and 75’ covers 26–50 and 51–75 

⇒16 + 9 = 25

⇒ P(25 to 75) = 25/50 = 1/2 = 0.5

Q6: A class has 30 students. On checking attendance records for 60 school days, it was found that student Priya was absent 12 times, student Rahul was absent 18 times, and the rest had perfect attendance. A school inspector randomly visits on one day. What is the probability that both Priya and Rahul are present?

Solution: P(Priya present) = (60−12)/60 = 48/60 = 4/5

P(Rahul present) = (60−18)/60 = 42/60 = 7/10

P(Both present) = (4/5) × (7/10) = 28/50 = 14/25

∴ P(Both Priya and Rahul present) = 14/25 = 0.56

Q7: In a class of 40 students, 10 like only cricket, 15 like only football, 8 like both, and the rest like neither. If one student is chosen at random, find the probability that the student likes at least one sport.

Solution: Students who like at least one sport = 10 + 15 + 8 = 33

Total students = 40

P(at least one sport) = 33/40

∴ P(likes at least one sport) = 33/40 = 0.825

Q8: A survey of 200 people found: 45 people jog daily, 80 people walk daily, 30 people do yoga daily, and 45 people do no exercise. If a person is selected at random, find the probability that they do NOT jog.

Solution: Total surveyed = 200

Joggers = 45

Non-joggers = 200 − 45 = 155

P(not a jogger) = 155/200 = 31/40

∴ P(person does NOT jog) = 31/40 = 0.775

Q9: Two coins are tossed simultaneously 360 times. The number of times ‘2 Tails’ appeared was three times the number of 'No Tail' outcomes. ‘1 Tail’ appeared twice as often as 'No Tail'. Find the probability of getting ‘Two Tails’.

Solution: Let ‘No Tail’ occur x times

Then ‘2 Tails’ = 3x, ‘1 Tail’ = 2x

Total= x + 2x + 3x = 360 

⇒ 6x = 360 

⇒ x = 60

2 Tails occurred: 3 × 60 = 180 times

P(Two Tails) = 180/360 = 1/2

∴ P(Two Tails) = 1/2 = 0.5

Q10: 1,500 families with 2 children were surveyed. The data is below. Compute probabilities for a randomly chosen family having (i) 2 girls, (ii) 1 girl, (iii) no girls. Also verify that the probabilities add up to 1.

Solution: Total families = 1500

(i) P(2 girls) = 475/1500 = 19/60 ≈ 0.317

(ii) P(1 girl) = 814/1500 = 407/750 ≈ 0.543

(iii) P(0 girls) = 211/1500 = 211/1500 ≈ 0.141

Q11: A bag contains 50 bolts and 150 nuts. Half of both are rusted. One item is picked at random. Find the probability that it is (i) rusted, (ii) not rusted, (iii) a bolt that is not rusted.

Solution: Total items = 50 + 150 = 200

Rusted bolts = 25  and Rusted nuts = 75 

⇒ Total rusted: 100

Unrusted bolts: 25 

Total unrusted: 100

(i) P(rusted) = 100/200 = 1/2

(ii) P(not rusted) = 100/200 = 1/2

(iii) P(unrusted bolt) = 25/200 = ⅛

Q12: From a bag containing red and blue balls, the probability of picking a red ball is x/2. If the probability of picking a blue ball is 2/3, find x.

Solution: P(red) + P(blue) = 1

⇒ x/2 + 2/3 = 1

⇒ 3x + 4 = 6 

⇒ 3x = 2 

⇒ x = 2/3

∴ x = 2/3

Q13: A coin is tossed 200 times, and head appears 120 times. Find P(Head).

Solution: Favourable outcomes = 120 (Heads)

Total trials = 200

P(Head) = 120 / 200 = 3/5 = 0.6

Q14: From a set of 10 cards numbered 6, 5, 3, 9, 7, 6, 4, 2, 8, 2 . One card is picked at random. Find: (i) P(card value > 5), and (ii) P(even number).

Solution: Total cards = 10

Cards > 5: {6, 9, 7, 6, 8} 

Number of favourable outcomes =  5 

Even cards: {6, 4, 2, 8, 6, 2} 

Number of favourable outcomes =  6 

(i) P(value > 5) = 5/10 = 0.5

(ii) P(even) = 6/10 = 3/5 = 0.6