Probability Worksheet Class 9 – Ganita Manjari Ch 7, 71 Qs

Class 9 Maths Probability worksheet with answers — 71 questions with step-by-step solutions. Ganita Manjari Ch 7. Free PDF & online practice.

This free Worksheet for CBSE Class IX Maths contains a structured worksheet with MCQs, short answer, case-based and HOTS questions in one place. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.

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First revise the chapter from your notes or textbook.
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The Mathematics of Maybe: Introduction to Probability — Class 09

UNIQUE STUDY POINT BY SUMEET SAHU

The Mathematics of Maybe

Class 09 · Maths (Ganita Manjari) · Introduction to Probability · Practice Worksheet with Solutions

71 Questions 175 Marks 1 hr 30 min Time

Tap any question's "Show Answer" button to reveal the full step-by-step solution.

Section A · Very Short Answer / MCQ / Assertion–Reason (Q1–Q25)

1 Mark each

A card is drawn at random from a pack of 52 cards. The probability that the card drawn is a jack, a queen or a king isMCQ

a) 126
b) 1113
c) 113
d) 313

✓ Correct Answer: (d) 313

Total outcomes = 52.
Favourable outcomes = 4 jacks + 4 queens + 4 kings = 12.
P = 1252 = 313

The probability that a non-leap year has 53 Sundays, isMCQ

a) 57
b) 17
c) 27
d) 67

✓ Correct Answer: (b) 17

A non-leap year has 365 days = 52 weeks + 1 extra day.
This extra day can be any of the 7 days of the week with equal likelihood.
P(53 Sundays) = 17

If P(E) = 0.05, what will be the probability of 'not E'?MCQ

a) 0.59
b) 0.95
c) 0.55
d) 0.095

✓ Correct Answer: (b) 0.95

P(E) + P(not E) = 1
P(not E) = 1 − 0.05 = 0.95

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23, isMCQ

a) 989
b) 445
c) 1090
d) 790

✓ Correct Answer: (b) 445

Primes less than 23: 2, 3, 5, 7, 11, 13, 17, 19 ⇒ 8 favourable outcomes.
P = 890 = 445

What is the primary characteristic of a 'random experiment'?MCQ

a) Its exact outcome cannot be predicted.
b) Its outcome is known before it starts.
c) It can only produce a single outcome.
d) It has a fixed and predictable result.

✓ Correct Answer: (a) Its exact outcome cannot be predicted.

A random experiment is repeatable, and although all possible outcomes are known, the exact result of any single trial cannot be predicted in advance.

A factory produces 200 toys per hour. During a quality check, 10 toys were randomly selected, and 2 were found to be defective. Based on this sample, how many defective toys would be expected in a 4-hour production run?MCQ

a) 32
b) 80
c) 40
d) 160

✓ Correct Answer: (d) 160

Experimental probability of a toy being defective = 210 = 0.20.
Total toys produced in 4 hours = 200 × 4 = 800.
Expected defective toys = 0.20 × 800 = 160

A card is drawn at random from a pack of 52 cards. The probability that the drawn card is not a king isMCQ

a) 113
b) 913
c) 413
d) 1213

✓ Correct Answer: (d) 1213

Number of kings = 4, so cards that are not king = 52 − 4 = 48.
P = 4852 = 1213

The probability that a leap year has 53 Sundays is:MCQ

a) 27
b) 47
c) 37
d) 17

✓ Correct Answer: (a) 27

A leap year has 366 days = 52 weeks + 2 extra days.
The 2 extra days can be: (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon) ⇒ 7 equally likely pairs.
Pairs containing Sunday: (Sat,Sun) and (Sun,Mon) ⇒ 2 favourable.
P(53 Sundays) = 27

A black card is lost from a deck of 52 playing cards. Rest of the cards are shuffled and one card is drawn at random from the available cards. The probability that drawn card is 'king of hearts', isMCQ

a) 126
b) 14
c) 152
d) 151

✓ Correct Answer: (d) 151

Since the lost card is black, the king of hearts (a red card) is still in the deck.
Remaining total cards = 51, and the king of hearts is exactly 1 of them.
P = 151

The probability of getting a chocolate flavoured ice cream at random, in a lot of 600 ice creams is 0.055. The number of chocolate flavoured ice creams in the lot is:MCQ

a) 33
b) 44
c) 55
d) 11

✓ Correct Answer: (a) 33

Number of chocolate ice-creams = 0.055 × 600 = 33

If a word 'MATHEMATICS' is scrambled and arranged, what is the probability that a randomly chosen letter from the original word is 'M'?MCQ

a) 211
b) 15
c) 311
d) 111

✓ Correct Answer: (a) 211

'MATHEMATICS' has 11 letters, and 'M' occurs 2 times.
P('M') = 211

Probability of happening of an event is denoted by p and probability of non-happening of the event is denoted by q. Relation between p and q isMCQ

a) p = 1, q = 1
b) p + q + 1 = 0
c) p = q − 1
d) p + q = 1

✓ Correct Answer: (d) p + q = 1

An event and its complement together cover the whole sample space, so their probabilities always add to 1.

An unbiased die is thrown once. The probability of getting a composite number isMCQ

a) 13
b) 23
c) 25
d) 12

✓ Correct Answer: (a) 13

Composite numbers on a die: {4, 6} ⇒ 2 favourable outcomes out of 6.
P = 26 = 13

Two dice are rolled together and the product (P) of their scores is obtained: Event A: P is 6, Event B: P is an odd number, Event C: P is 35. Which of the following event(s) has probability equal to 0?MCQ

a) Event B
b) Event C
c) Events A and B
d) Events B and C

✓ Correct Answer: (b) Event C

Event A (product = 6): possible, e.g. (1,6), (2,3) — probability ≠ 0.
Event B (odd product): possible, e.g. (1,1), (1,3) — probability ≠ 0.
Event C (product = 35 = 5×7): 7 cannot appear on a die, so no pair gives this product.
P(Event C) = 036 = 0

A coin is tossed 100 times. It lands on heads 58 times. What is the experimental probability of getting tails?MCQ

a) 0.58
b) 0.50
c) 0.60
d) 0.42

✓ Correct Answer: (d) 0.42

Number of tails = 100 − 58 = 42.
P(tails) = 42100 = 0.42

Probability of getting a red king from a deck of well-shuffled 52 cards isMCQ

a) 113
b) 126
c) 115
d) 1352

✓ Correct Answer: (b) 126

Red kings = king of hearts + king of diamonds = 2.
P = 252 = 126

A card is drawn from a pack of well-shuffled cards. The probability of getting either a king or a queen is:MCQ

a) 0
b) 213
c) 313
d) 113

✓ Correct Answer: (b) 213

Kings = 4, Queens = 4 ⇒ favourable = 8.
P = 852 = 213

Two fair coins are tossed together. The probability of getting 2 heads, is:MCQ

a) 14
b) 38
c) 34
d) 12

✓ Correct Answer: (a) 14

Sample space = {HH, HT, TH, TT} ⇒ 4 outcomes; favourable (HH) = 1.
P = 14

A bag contains 14 balls out of which x are white. If 6 more white balls are added to the bag, the probability of drawing a white ball is 12. The value of x is:MCQ

a) 6
b) 10
c) 8
d) 4

✓ Correct Answer: (d) 4

New number of white balls = x + 6; new total balls = 14 + 6 = 20.
x + 620 = 12 ⇒ x + 6 = 10 ⇒ x = 4

Which of the following can be the probability of an event?MCQ

a) −0.04
b) 1823
c) 87
d) 1.004

✓ Correct Answer: (b) 1823

Probability must satisfy 0 ≤ P(E) ≤ 1.
−0.04 is negative, and 87 and 1.004 both exceed 1 — all invalid.
1823 lies between 0 and 1, so it is the only valid probability.

Assertion (A): If a die is thrown, the probability of getting a number less than 3 and greater than 2 is zero.
Reason (R): The probability of an impossible event is zero.Assertion-Reason

a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.

✓ Correct Answer: (a) Both A and R are true and R is the correct explanation of A.

No number can be simultaneously less than 3 and greater than 2, so this is an impossible event.
By definition, an impossible event has probability 0 — so R correctly explains A.

Assertion (A): If a box contains 5 white, 2 red and 4 black marbles, then the probability of not drawing a white marble from the box is 511.
Reason (R): P(ē) = 1 − P(E), where E is any event.Assertion-Reason

a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.

✓ Correct Answer: (d) A is false but R is true.

Total marbles = 5 + 2 + 4 = 11, so P(white) = 511.
P(not white) = 1 − 511 = 611, not 511 — so A is false.
The complement rule P(ē) = 1 − P(E) is a true, standard result — so R is true.

Assertion (A): Probability of getting a perfect square from first 100 even numbers is 530.
Reason (R): There are only 5 even perfect squares out of first 100 natural numbers.Assertion-Reason

a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.

✓ Correct Answer: (d) A is false but R is true.

First 100 even numbers (2 to 200) form a set of size 100, so the denominator 30 in A is incorrect — A is false.
Among natural numbers 1–100, the perfect squares are 1, 4, 9, ..., 100 (10 of them), and the even ones among these are 4, 16, 36, 64, 100 — exactly 5, so R is true.

Assertion (A): If E₁, E₂, ...., E_n are n elementary events associated to a random experiment, then P(E₁) + P(E₂) + ... + P(E_n) = 1.
Reason (R): For any event A associated to an experiment, we have 0 ≤ P(A) ≤ 1.Assertion-Reason

a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.

✓ Correct Answer: (c) A is true but R is false.

The sum of probabilities of all elementary events of a random experiment always equals 1 — A is true.
R is a separate general bound on a single event's probability; it does not by itself explain why the elementary-event probabilities sum to 1.

Assertion (A): In a single toss of two fair coins once, the probability of getting one head is 12.
Reason (R): In a throw of two fair coins, the sample space is HH, HT, TH, TT.Assertion-Reason

a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.

✓ Correct Answer: (a) Both A and R are true and R is the correct explanation of A.

Sample space = {HH, HT, TH, TT} ⇒ 4 outcomes, so R is true.
Favourable outcomes for exactly one head = {HT, TH} = 2, so P = 24 = 12 — A is true and R explains it.

Section B · Short Answer–I (Q26–Q40)

2 Marks each

Case Study — Diwali Fair: A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented below. Prizes are given when a black marble is picked. Shweta plays the game once.

Spinner (numbers 1–10) and the bag of 20 marbles (6 black, 14 white)

What is the probability that she will be allowed to pick a marble from the bag?
Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 marbles out of which 6 are black?

(i) Spinner outcomes = {1, 2, 4, 6, 8, 10} ⇒ n(S) = 6. Even numbers = {2, 4, 6, 8, 10} ⇒ n(favourable) = 5.
P(picking a marble) = 56
(ii) Total marbles = 20, black marbles = 6.
P(prize) = 620 = 310

✓ (i) 56 (ii) 310

The blood groups of 30 students of Class VIII are recorded as: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. Use this data to determine the probability that a student of this class, selected at random, has blood group AB.

Total number of students = 30.
Number of students with blood group AB = 3.
P(AB) = 330 = 110 = 0.1

✓ P(blood group AB) = 110

There are 35 students in a class of whom 20 are boys and 15 are girls. From these students one is chosen at random. What is the probability that the chosen student is a (i) boy, (ii) girl?

Total students = 35.
(i) P(boy) = 2035 = 47
(ii) P(girl) = 1535 = 37

✓ (i) 47 (ii) 37

A particular make and model T.V. set has been on the market for six months. In that time 12,492 sets were sold to consumers. Of those, 327 were identified as needing repair during the warranty period. What is the empirical probability that a given T.V. of this make and model will require repair during the warranty period?

Let E = event that a randomly chosen T.V. requires repair during warranty.
P(E) = 32712492 ≈ 0.026

✓ P(requires repair) ≈ 0.026

One card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting the following:

a spade
the queen of diamonds

(i) Total spade cards = 13.
P(spade) = 1352 = 14
(ii) Only 1 queen of diamonds in the deck.
P(queen of diamonds) = 152

✓ (i) 14 (ii) 152

A coin is tossed once.

Describe the sample space S.
Find the probability of getting a tail.

(i) Sample space S = {H, T}, so n(S) = 2.
(ii) Let E = event of getting a tail; E = {T}, n(E) = 1.
P(tail) = 12

✓ (i) S = {H, T} (ii) 12

Two coins are tossed simultaneously 500 times. If we get two heads 100 times, one head 270 times and no head 130 times, then find the probability of getting one or more than one head.

"One or more heads" = (one head) + (two heads) = 270 + 100 = 370 trials.
P(one or more heads) = 370500 = 3750

✓ P = 3750

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Total pens = 12 + 132 = 144.
P(good pen) = 132144 = 1112

✓ P(good pen) = 1112

The following frequency distribution gives the weights of 40 students of a class. A student from the class is chosen at random. What is the probability that the weight of the chosen student is at least 56 kg?

Weight (kg)	31–35	36–40	41–45	46–50	51–55	56–60	61–65	66–70	71–75
No. of students	10	6	14	4	1	1	2	1	1

Students weighing at least 56 kg fall in the last four classes: 1 + 2 + 1 + 1 = 5.
Total students = 40.
P(weight ≥ 56 kg) = 540 = 18

✓ P = 18

In a single throw of a die, find the probability of getting:

an odd number
a number greater than 5

Sample space S = {1, 2, 3, 4, 5, 6}, n(S) = 6.
(i) Odd numbers = {1, 3, 5}, n = 3.
P(odd) = 36 = 12
(ii) Numbers greater than 5 = {6}, n = 1.
P(greater than 5) = 16

✓ (i) 12 (ii) 16

A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

P₁ (before) = x12.
After adding 6 black balls: total = 18, black = x + 6, so P₂ = x+618.
Given P₂ = 2P₁: x+618 = 2 × x12
Cross-multiplying: 12(x+6) = 36x ⇒ 12x + 72 = 36x ⇒ 24x = 72 ⇒ x = 3.

✓ P(black ball) = x12 and x = 3

A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.

Total outcomes = 8: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Success (same result all 3 tosses) = {HHH, TTT} ⇒ 2 outcomes.
P(success) = 28 = 14
P(losing) = 1 − 14 = 34

✓ P(losing) = 34

At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that?

the first player wins the prize?
the second player wins a prize, if the first has won?

Perfect squares greater than 500 (and ≤ 1000): 23²=529, 24²=576, 25²=625, 26²=676, 27²=729, 28²=784, 29²=841, 30²=900, 31²=961 ⇒ 9 numbers.
(i) P(first player wins) = 91000
(ii) Card not replaced: total now 999, favourable now 8.
P(second player wins) = 8999

✓ (i) 91000 (ii) 8999

Two coins are tossed once. Find the probability of getting:

2 heads
at least 1 tail

Total outcomes = {HH, HT, TH, TT}, n(S) = 4.
(i) Favourable = {HH} = 1.
P(2 heads) = 14
(ii) Favourable = {HT, TH, TT} = 3.
P(at least 1 tail) = 34

✓ (i) 14 (ii) 34

In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it. Find the probability that a lady chosen at random dislikes coffee.

Total ladies = 200; number who dislike coffee = 58.
P(dislikes coffee) = 58200 = 0.29

✓ P(dislikes coffee) = 0.29

Section C · Short Answer–II (Q41–Q55)

3 Marks each

A tyre company records distances before replacement in 1000 cases. Find the probability that a randomly chosen tyre lasts:

Distance (km)	Less than 4000	4001 to 9000	9001 to 14000	More than 14000
Number of cases	20	210	325	445

Less than 4000 km.
Between 4000 and 14000 km.
More than 14000 km.

Total cases = 1000.
(i) P(less than 4000 km) = 201000 = 0.02
(ii) Between 4000 and 14000 km = 210 + 325 = 535.
P = 5351000 = 0.535
(iii) P(more than 14000 km) = 4451000 = 0.445

✓ (i) 0.02 (ii) 0.535 (iii) 0.445

A basket contains 4 identical cards labeled A, B, C, and D. One card is drawn, its letter is noted, and it is put back. Then a second card is drawn and noted.

Write down the complete sample space S of outcomes.
Explain how this sample space differs from a scenario where the first card is laid aside without replacement.

(i) Since the card is replaced, a letter can pair with itself:
S = {AA,AB,AC,AD,BA,BB,BC,BD,CA,CB,CC,CD,DA,DB,DC,DD}, so n(S) = 16.
(ii) With replacement, identical pairs like AA, BB, CC, DD are possible outcomes, giving n(S) = 16. Without replacement, the first card cannot repeat as the second draw, so these 4 identical-letter outcomes disappear and the sample size reduces to n(S) = 12.

✓ n(S) = 16 with replacement; n(S) = 12 without replacement

A bag contains 8 red, 4 white and 3 black balls. One ball is drawn at random. What is the probability that the ball drawn is:

white?
red or white?
neither red nor white?
not red?

Total balls = 8 + 4 + 3 = 15.
(i) P(white) = 415
(ii) Red or white = 8 + 4 = 12.
P(red or white) = 1215 = 45
(iii) Neither red nor white = black = 3.
P(neither red nor white) = 315 = 15
(iv) Not red = white + black = 4 + 3 = 7.
P(not red) = 715

✓ (i) 415 (ii) 45 (iii) 15 (iv) 715

Two different dice are thrown simultaneously. Find the probability of getting:

a number greater than 3 on each dice
an odd number on both dice.

Total outcomes = 6 × 6 = 36.
(i) Favourable = {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)} = 9.
P = 936 = 14
(ii) Favourable = {(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)} = 9.
P = 936 = 14

✓ (i) 14 (ii) 14

Cards with numbers 1, 2, 3, ........, 100 are placed in a box and mixed thoroughly. One card is drawn. What is the probability that the card drawn is

a prime number less than 30?
a multiple of 5 and 7?
a multiple of 5 or 7?

(i) Primes less than 30: 2,3,5,7,11,13,17,19,23,29 ⇒ 10 numbers.
P = 10100 = 110
(ii) A multiple of both 5 and 7 is a multiple of 35: 35, 70 ⇒ 2 numbers.
P = 2100 = 150
(iii) Multiples of 5 = 20, multiples of 7 = 14, multiples of 35 (both) = 2.
By inclusion-exclusion: 20 + 14 − 2 = 32.
P = 32100 = 825

✓ (i) 110 (ii) 150 (iii) 825

All face cards of spades are removed from a pack of 52 playing cards and the remaining pack is shuffled well. A card is then drawn at random from the remaining pack. Find the probability of getting:

a face card
an ace or a jack

Face cards of spades removed = J, Q, K of spades = 3 cards.
Remaining total cards = 52 − 3 = 49.
(a) Original face cards = 12 (J,Q,K of 4 suits); remaining face cards = 12 − 3 = 9.
P(face card) = 949
(b) Aces are unaffected = 4. Jacks remaining = 4 − 1 (spade jack removed) = 3. Favourable = 4 + 3 = 7.
P(ace or jack) = 749 = 17

✓ (a) 949 (b) 17

Cards marked with numbers 5 to 75 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that the number on the card is odd.

Total cards = 75 − 5 + 1 = 71.
Number of odd cards = 36.
P(odd number) = 3671

✓ P(odd) = 3671

In a single throw of a die, find the probability of getting:

a number less than 7
a number divisible by 3

Sample space S = {1,2,3,4,5,6}, n(S) = 6.
(i) Every face (1–6) is less than 7, so favourable = 6.
P = 66 = 1
(ii) Divisible by 3: {3,6} = 2.
P = 26 = 13

✓ (i) 1 (ii) 13

Two fair 6-sided dice are rolled simultaneously. Let the outcome be recorded as ordered pairs (x, y), where x is the score on the first die and y is the score on the second die.

State the sample size n(S).
Write down the elements of the event E: 'The sum of the numbers on the two dice is exactly 10'.

(i) n(S) = 6 × 6 = 36
(ii) Pairs with x + y = 10: (4,6), (5,5), (6,4).
E = {(4,6), (5,5), (6,4)}

✓ n(S) = 36; E has 3 elements

Three different coins are tossed together. Find the probability of getting

exactly two heads
atleast two heads
atleast two tails

Total outcomes = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}, n(S) = 8.
(i) Exactly 2 heads = {HHT,HTH,THH} = 3.
P = 38
(ii) At least 2 heads = {HHT,HTH,THH,HHH} = 4.
P = 48 = 12
(iii) At least 2 tails = {HTT,THT,TTH,TTT} = 4.
P = 48 = 12

✓ (i) 38 (ii) 12 (iii) 12

A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that

it is acceptable to Jimmy?
it is acceptable to Sujatha?

Total shirts = 100.
(i) Jimmy accepts only good shirts = 88.
P = 88100 = 0.88
(ii) Sujatha accepts good + minor-defect shirts = 88 + 8 = 96.
P = 96100 = 0.96

✓ (i) 0.88 (ii) 0.96

The table below shows the number of vehicles owned per family, split by monthly income. Suppose a family is chosen at random. Find the probability that the family chosen is:

Monthly income (₹)	0	1	2	Above 2
Less than 7000	10	160	25	0
7000 – 10000	0	305	27	2
10000 – 13000	1	535	29	1
13000 – 16000	2	469	59	25
16000 or more	1	579	82	88

earning ₹10000 – 13000 per month and owning exactly 2 vehicles.
earning ₹16000 or more per month and owning exactly 1 vehicle.
earning less than ₹7000 per month and does not own any vehicle.
earning ₹13000 – 16000 per month and owning more than 2 vehicles.
not more than 1 vehicle.

Total number of families = 2400.
(i) P = 292400
(ii) P = 5792400
(iii) P = 102400 = 1240
(iv) P = 252400 = 196
(v) "0 vehicles" column total = 10+0+1+2+1 = 14; "1 vehicle" column total = 160+305+535+469+579 = 2048; sum = 2062.
P = 20622400 = 10311200

✓ (i) 292400 (ii) 5792400 (iii) 1240 (iv) 196 (v) 10311200

A tyre manufacturing company monitored the performance lifespan of 1,200 truck tyres under commercial conditions. The records show: lasted less than 30,000 km: 96 tyres; lasted 30,001 to 60,000 km: 432 tyres; lasted 60,001 to 90,000 km: 504 tyres; lasted more than 90,000 km: 168 tyres. Find the experimental probability that a tyre chosen at random will last:

At least 60,001 km.
Less than 30,000 km.

Total tyres = 1200.
(i) At least 60,001 km = 504 + 168 = 672.
P = 6721200 = 1425 = 0.56
(ii) Less than 30,000 km = 96.
P = 961200 = 225 = 0.08

✓ (i) 0.56 (ii) 0.08

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is

not red?
white?

x + 2x + 3x = 24 ⇒ 6x = 24 ⇒ x = 4.
So red = 4, white = 8, blue = 12.
(i) Not red = white + blue = 8 + 12 = 20.
P(not red) = 2024 = 56
(ii) P(white) = 824 = 13

✓ (i) 56 (ii) 13

Two players Niharika and Shreya play a tennis match. It is known that the probability of Niharika winning the match is 0.62. What is the probability of Shreya winning the match?

Niharika winning and Shreya winning are complementary events: P(Niharika) + P(Shreya) = 1.
P(Shreya) = 1 − 0.62 = 0.38

✓ P(Shreya wins) = 0.38

Section D · Case–Based / Data–Based (Q56–Q60)

4 Marks each

Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table. One carton was selected at random. What is the probability that it has? (Answer any TWO of the three parts.)

No. of defective bulbs	0	1	2	3	4	5	6	More than 6
Frequency	400	180	48	41	18	8	3	2

no defective bulb?
defective bulbs from 2 to 6?
defective bulbs less than 4?

Total cartons examined = 700.
(i) No defective bulb: frequency = 400.
P = 400700 = 47
(ii) Defective bulbs 2 to 6: 48+41+18+8+3 = 118.
P = 118700 = 59350
(iii) Defective bulbs less than 4 (0,1,2,3): 400+180+48+41 = 669.
P = 669700

✓ (i) 47 (ii) 59350 (iii) 669700 (any two required)

Case Study: Most people use playing card decks just for fun and to spend time. However, playing cards also improve patience, concentration and memory skills, and help people socialize. Three friends play a game with playing cards. They remove the red and black colour King, Queen and Ace from a pack of 52 playing cards.

Find the probability of a red colour card from the pack. (1)
Find the probability of getting a spade. (1)
Find the probability of getting jack of red colour. (2)
OR — Find the probability of getting a king. (2)

Removing King, Queen and Ace from both red suits and both black suits removes 3 cards × 4 suits = 12 cards. Remaining cards = 52 − 12 = 40 (20 red, 20 black).
(i) P(red card) = 2040 = 12
(ii) Spade cards remaining = 13 − 3 (K,Q,A removed) = 10.
P(spade) = 1040 = 14
(iii) Red jacks remaining = jack of hearts + jack of diamonds = 2.
P(jack of red colour) = 240 = 120
OR: All kings have been removed from the pack, so there are 0 kings left.
P(king) = 040 = 0

✓ (i) 12 (ii) 14 (iii) 120 OR 0

Case Study — Eight Ball: Eight Ball is a game played on a pool table with 15 balls numbered 1 to 15 and a cue ball that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8, and seven are striped balls numbered 9 to 15. The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random.

What is the probability that the drawn ball bears number 8? (1)
What is the probability that the drawn ball bears an even number? (2)
OR — What is the probability that the drawn ball bears a number which is a multiple of 3? (2)
What is the probability that the drawn ball is solid coloured and bears an even number? (1)

Total balls = 15.
(i) P(number 8) = 115
(ii) Even-numbered balls: 2,4,6,8,10,12,14 ⇒ 7.
P(even) = 715
OR: Multiples of 3: 3,6,9,12,15 ⇒ 5.
P(multiple of 3) = 515 = 13
(iii) Solid balls are numbered 1–8; even numbers among these: 2,4,6,8 ⇒ 4.
P(solid and even) = 415

✓ (i) 115 (ii) 715 OR 13 (iii) 415

Case Study: The school library conducted a lucky draw for 50 students who returned books on time. Their names were put in a box: 20 from Grade 9, 18 from Grade 8, and 12 from Grade 7. One name is drawn at random. Separately, a student Meena observed that in the last 200 school days, it rained on 50 days. She wants to use experimental probability to predict if it will rain today.

Grade-wise distribution of the 50 lucky-draw entries

Write the sample space for the lucky draw in terms of grades. Find the theoretical probability that the winner is from Grade 9. (1)
Find the experimental probability that it will rain today, based on Meena's data. Place this probability on the probability scale and describe its likelihood. (1)
The principal says "The chance of a Grade 8 student winning is the same as a Grade 7 student winning." Is the principal correct? Compare P(Grade 8) and P(Grade 7) and justify. (2)
OR — Two names are drawn one after the other with replacement (each name is put back before the next draw). Find the probability that both draws give a Grade 9 student's name. (2)

(i) Sample space S = {Grade 9, Grade 8, Grade 7} (weighted by 20, 18, 12), n(S) = 50.
P(Grade 9) = 2050 = 25 = 0.4
(ii) P(rain) = 50200 = 14 = 0.25
Since 0 < 0.25 < 0.5, this sits closer to 0 than to 1 on the probability scale — it's an unlikely (less likely) event, though not impossible.
(iii) P(Grade 8) = 1850 = 925 = 0.36; P(Grade 7) = 1250 = 625 = 0.24.
Since 0.36 ≠ 0.24, the principal is incorrect — a Grade 8 student has a higher chance of winning than a Grade 7 student.
OR: With replacement, each draw is independent: P(Grade 9 on 1st) = 25, P(Grade 9 on 2nd) = 25.
P(both Grade 9) = 25 × 25 = 425 = 0.16

✓ (i) 25 (ii) 0.25 (unlikely) (iii) Not correct OR 425

Fifty seeds were selected at random from each of 5 bags of seeds and kept under standardised conditions favourable to germination. After 20 days, the number of seeds that had germinated in each collection were counted and recorded below. What is the probability of germination of: (Answer any TWO of the three parts.)

Bag	1	2	3	4	5
Number of seeds germinated	40	48	42	39	41

More than 40 seeds in a bag?
49 seeds in a bag?
More than 35 seeds in a bag?

Total bags = 5.
(i) Bags with more than 40 seeds germinated: 48, 42, 41 ⇒ 3 bags.
P = 35 = 0.6
(ii) No bag had exactly 49 seeds germinate ⇒ 0 bags.
P = 05 = 0
(iii) All 5 bags (40,48,42,39,41) germinated more than 35 seeds.
P = 55 = 1

✓ (i) 0.6 (ii) 0 (iii) 1 (any two required)

Section E · Long Answer (Q61–Q71)

5 Marks each

Fill in the blanks:

The probability of an event that cannot happen is . Such an event is called . (1)
The probability of an event is greater than or equal to and less than or equal to . (1)
Probability of an event E + Probability of the event 'not E' = . (1)
The set of all possible outcomes of a random experiment is called the . (1)
The sum of the probabilities of all the elementary events of an experiment is . (1)

(a) 0; an impossible event.
(b) greater than or equal to 0 and less than or equal to 1.
(c) 1.
(d) sample space.
(e) 1.

✓ (a) 0, impossible event (b) 0, 1 (c) 1 (d) sample space (e) 1

A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

a two digit number,
a number divisible by 5.

Total discs = 90.
(i) Two-digit numbers run from 10 to 90: 90 − 10 + 1 = 81.
P = 8190 = 910
(ii) Numbers divisible by 5 (5, 10, 15, ..., 90): 18 numbers.
P = 1890 = 15

✓ (i) 910 (ii) 15

A box contains 5 identical cards numbered 1 to 5. An experimenter draws one card at random, notes its value, and without replacing it, draws a second card and notes its value.

Construct the complete sample space using a tree diagram list format.
State the value of n(S).
List the elements of the event E₁: 'The first card drawn is an odd number and the second card drawn is an even number'.
List the elements of the event E₂: 'The product of the two numbers drawn is greater than or equal to 12'.

(a) Since the first card is not replaced, no pair can repeat the same number:
S = {(1,2),(1,3),(1,4),(1,5),(2,1),(2,3),(2,4),(2,5),(3,1),(3,2),(3,4),(3,5),(4,1),(4,2),(4,3),(4,5),(5,1),(5,2),(5,3),(5,4)}
(b) n(S) = 5 × 4 = 20
(c) Odd numbers = {1,3,5}, even numbers = {2,4}. Pairing first-odd with second-even:
E₁ = {(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)}
(d) Checking all pairs whose product ≥ 12:
E₂ = {(3,4),(3,5),(4,3),(4,5),(5,3),(5,4)}

✓ n(S) = 20; n(E₁) = 6; n(E₂) = 6

A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:

white
neither red nor white.

Total balls = 7 + 8 + 5 = 20.
(i) P(white) = 520 = 14
(ii) Neither red nor white = green = 8.
P(neither red nor white) = 820 = 25

✓ (i) 14 (ii) 25

Two fair coins are tossed together. Can you calculate the probability of getting one head and one tail?

A tree diagram gives all outcomes: HH, HT, TH, TT ⇒ total outcomes = 4.
Favourable outcomes (exactly one head and one tail) = {HT, TH} = 2.
P(one head and one tail) = 24 = 12

✓ P = 12 (a 50% / even chance)

A student rolls a standard 6-sided die 300 times as part of a probability laboratory assignment. The frequencies of the appearing outcomes are tabulated below.

Outcome	1	2	3	4	5	6
Frequency	42	56	48	62	38	54

Calculate the experimental probability of rolling an odd number.
Calculate the experimental probability of rolling a number greater than 4.
Calculate the theoretical probability of rolling an odd number, assuming a perfectly balanced die. Explain why the experimental results show a higher relative frequency for the outcome '4' than '5'.

Total trials = 300.
(i) Odd outcomes (1,3,5): 42+48+38 = 128.
P(odd) = 128300 = 3275 ≈ 0.427
(ii) Outcomes greater than 4 (5,6): 38+54 = 92.
P(>4) = 92300 = 2375 ≈ 0.307
(iii) Theoretically, odd faces {1,3,5} = 3 out of 6 equally likely outcomes.
P(odd, theoretical) = 36 = 12 = 0.5
The higher frequency for '4' (62) compared to '5' (38) is simply natural random variability over a moderate number of trials — it doesn't necessarily indicate a biased die, since individual trial sequences are inherently unpredictable.

✓ (i) ≈0.427 (ii) ≈0.307 (iii) 0.5 (theoretical)

Two different dice are thrown together. Find the probability that the numbers obtained have

even sum
even product.

Total outcomes = 6 × 6 = 36.
(i) Pairs with even sum (both same parity): 18 such pairs.
P(even sum) = 1836 = 0.5
(ii) Product is even whenever at least one die shows an even number: 27 such pairs.
P(even product) = 2736 = 0.75

✓ (i) 0.5 (ii) 0.75

A die is thrown 500 times and the outcomes are noted as given below. If a die is thrown at random, find the probability of getting

Outcome	1	2	3	4	5	6
Frequency	95	80	84	68	70	103

Total throws = 500. Each probability = frequency500.
(i) P(1) = 95500 = 19100 = 0.19
(ii) P(2) = 80500 = 16100 = 0.16
(iii) P(3) = 84500 = 0.168
(iv) P(4) = 68500 = 0.136
(v) P(5) = 70500 = 750 = 0.14
(vi) P(6) = 103500 = 0.206

✓ 0.19, 0.16, 0.168, 0.136, 0.14, 0.206

A specialized automated machine creates mechanical components. A quality assurance test samples three components sequentially from the production line, classifying each as either Acceptable (A) or Rejected (R).

Construct the complete sample space for this three-step inspection experiment and determine the sample size n(S).
List the elements of the event E₁: 'Exactly one component is rejected'.
List the elements of the event E₂: 'The first component inspected is acceptable'.
List the elements of the event E₃: 'All three components share an identical classification status'.

(i) Each of the 3 stages has 2 outcomes {A, R}:
S = {AAA,AAR,ARA,ARR,RAA,RAR,RRA,RRR}, n(S) = 8.
(ii) Outcomes with exactly one 'R':
E₁ = {AAR, ARA, RAA}
(iii) Outcomes starting with 'A':
E₂ = {AAA, AAR, ARA, ARR}
(iv) Outcomes that are all-A or all-R:
E₃ = {AAA, RRR}

✓ n(S) = 8; n(E₁) = 3; n(E₂) = 4; n(E₃) = 2

A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that the product of x and y is less than 16.

Total outcomes (x,y) = 4 × 4 = 16.
Checking each: x=1 gives products 1,4,9,16 (3 valid, i.e. <16); x=2 gives 2,8,18,32 (2 valid); x=3 gives 3,12,27,48 (2 valid); x=4 gives 4,16,36,64 (1 valid).
Total favourable = 3 + 2 + 2 + 1 = 8.
P(xy < 16) = 816 = 12

✓ P(product < 16) = 12

An electronics distributor receives a massive consignment of 5,000 lithium-ion batteries. To evaluate quality, a random sample of 250 batteries is selected and continuously tested under high workload conditions. The lab tracks operational hours before failure: failed under 200 hours: 15 batteries; failed between 200 and 500 hours: 55 batteries; failed between 501 and 1000 hours: 130 batteries; operated safely past 1000 hours: 50 batteries.

What is the experimental probability that a battery chosen at random from this sample will fail before reaching 501 hours?
What is the experimental probability that a battery will work reliably for more than 500 hours?
Based on the sample results, estimate the total number of batteries in the entire shipment of 5,000 that are expected to survive past 1000 operational hours.

Sample size = 250.
(i) Failure before 501 hours = 15 + 55 = 70.
P(fail < 501 hrs) = 70250 = 725 = 0.28
(ii) Survival beyond 500 hours = 130 + 50 = 180.
P(survive > 500 hrs) = 180250 = 1825 = 0.72
(iii) P(survive past 1000 hrs) = 50250 = 15 = 0.20.
Estimated batteries in shipment = 0.20 × 5000 = 1,000

✓ (i) 0.28 (ii) 0.72 (iii) ≈ 1,000 batteries

📋 Details

Class	Class IX (CBSE / NCERT)
Subject	Maths
Resource Type	Worksheet
Last Updated	05 September 2026
Session	2026-27 (Latest NCERT Syllabus)
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Prepared by	Sumeet Sahu, Unique Study Point, Indore
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Probability Worksheet Class 9 – Ganita Manjari Ch 7, 71 Qs

📌 How to use this Worksheet

The Mathematics of Maybe

Class 09 · Maths (Ganita Manjari) · Introduction to Probability · Practice Worksheet with Solutions

Section A · Very Short Answer / MCQ / Assertion–Reason (Q1–Q25)

Section B · Short Answer–I (Q26–Q40)

Section C · Short Answer–II (Q41–Q55)

Section D · Case–Based / Data–Based (Q56–Q60)

Section E · Long Answer (Q61–Q71)

📋 Details

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