Class 9 Mathematics • Ganita Manjari • Chapter 7

The Mathematics of Maybe: Introduction to Probability — Advanced Question Bank

A difficult, source-aligned practice resource covering randomness, the probability scale, subjective and objective estimates, experimental and theoretical probability, relative frequency, sampling, Law of Large Numbers, Gambler’s Fallacy, fair and unbiased experiments, sample spaces, events, tree diagrams, replacement versus non-replacement, multi-step experiments and introductory geometric probability.

30 MCQs15 One-Mark15 Two-Mark 15 Three-Mark10 Four-Mark5 Case Studies 15 Special Exemplar20 Olympiad/HOTS10 High-Level Thinking

Probability Revision Sheet

Probability Scale

0 ≤ P(E) ≤ 1
0 = impossible
1/2 = even chance
1 = certain

Experimental Probability

Experimental P(E) = frequency of E / total number of trials

Theoretical Probability

For equally likely outcomes:
P(E)=n(E)/n(S)

Sample Space & Event

S lists all possible outcomes once.
An event E is a subset of S.

Multi-step Experiments

Use systematic listing or tree diagrams. Each complete path represents one ordered outcome.

Long-run Thinking

Experimental probability may differ in small samples but tends toward the theoretical value over many fair trials.

With replacement → probabilities reset; without replacement → probabilities change after each draw
Independent trials do not “remember” earlier outcomes — avoid Gambler’s Fallacy
For uniform geometric selection: Probability = favourable area / total area
Difficulty design: the main bank is above routine textbook level. The Exemplar, Olympiad/HOTS and High-Level Thinking sections add conditional reasoning, complements, counting, replacement effects, sample-space design, geometric probability and deeper interpretation of randomness.

Visual Learning — HTML Canvas

JavaScript is limited to these four diagrams. All solution accordions are native HTML <details>.

Canvas 1: Probability scale from impossible to certain.
Canvas 2: Tree diagram for two fair coin tosses.
Canvas 3: Experimental relative frequency moving toward theoretical probability as trials increase.
Canvas 4: Geometric probability — circular target inside a rectangle.

Section A — 30 MCQs

Conceptual, computational, source-inspired and reasoning-based.

1Core

Probability is primarily a measure of:

  1. A. length
  2. B. likelihood of an event
  3. C. area
  4. D. number of outcomes only
View answer / solution

Answer: B
Probability measures how likely an event is to occur.

2Core

Which of the following best describes a random experiment?

  1. A. Its outcome is known in advance
  2. B. It cannot be repeated
  3. C. Its possible outcomes may be known but the actual outcome is uncertain
  4. D. It always has two outcomes
View answer / solution

Answer: C
A random experiment is repeatable, but the exact outcome of a trial cannot be predicted with certainty.

3Core

The probability of an impossible event is:

  1. A. −1
  2. B. 0
  3. C. 1/2
  4. D. 1
View answer / solution

Answer: B
Impossible events have probability 0.

4Core

The probability of a certain event is:

  1. A. 0
  2. B. 1/4
  3. C. 1/2
  4. D. 1
View answer / solution

Answer: D
Certain events have probability 1.

5Source-based

If P(E)=0.75, then E is:

  1. A. impossible
  2. B. less likely than not
  3. C. more likely than not
  4. D. certain
View answer / solution

Answer: C
0.75 is greater than 0.5, so the event is more likely than not.

6Core

Which statement is correct for every event E?

  1. A. P(E)<0
  2. B. 0≤P(E)≤1
  3. C. P(E)>1
  4. D. P(E) is always rational
View answer / solution

Answer: B
The probability scale runs from 0 to 1 inclusive.

7Source-based

A die is rolled 50 times and a 4 appears 8 times. The experimental probability of 4 is:

  1. A. 1/6
  2. B. 4/25
  3. C. 8/42
  4. D. 42/50
View answer / solution

Answer: B
Experimental probability=8/50=4/25=0.16.

8Source-based

Theoretical probability is computed directly from favourable/total outcomes only when:

  1. A. outcomes are equally likely
  2. B. there is experimental data
  3. C. the sample is large
  4. D. the event is certain
View answer / solution

Answer: A
The chapter explicitly requires equally likely outcomes for the theoretical formula.

9Core

The sample space for tossing two fair coins is:

  1. A. {H,T}
  2. B. {HH,HT,TT}
  3. C. {HH,HT,TH,TT}
  4. D. {0,1,2}
View answer / solution

Answer: C
The ordered outcomes are HH, HT, TH and TT.

10Source-based

When three coins are tossed and only the number of heads is recorded, a suitable sample space is:

  1. A. {H,T}
  2. B. {HHH,HHT,HTH,THH}
  3. C. {0,1,2,3}
  4. D. {1,2,3}
View answer / solution

Answer: C
The recorded variable is the count of heads, which can be 0,1,2 or 3.

11Core

An event is:

  1. A. always one outcome only
  2. B. a subset of the sample space
  3. C. the entire experiment
  4. D. the number of trials
View answer / solution

Answer: B
An event may contain one or several outcomes and is a subset of S.

12Source-based

For S={1,2,3,4,5,6}, the event 'number greater than 4' is:

  1. A. {4,5,6}
  2. B. {5,6}
  3. C. {1,2,3,4}
  4. D. {6}
View answer / solution

Answer: B
Only 5 and 6 are greater than 4.

13Core

A tree diagram is most useful for:

  1. A. measuring length
  2. B. listing outcomes of multi-step experiments
  3. C. showing only impossible events
  4. D. finding experimental frequency without trials
View answer / solution

Answer: B
Tree diagrams list paths/outcomes in multi-step experiments.

14Source-based

Two fair coin tosses give P(HH)=

  1. A. 1/2
  2. B. 1/3
  3. C. 1/4
  4. D. 3/4
View answer / solution

Answer: C
There are four equally likely outcomes and only HH is favourable.

15Source-based

For two fair coin tosses, P(exactly one head)=

  1. A. 1/4
  2. B. 1/2
  3. C. 3/4
  4. D. 1
View answer / solution

Answer: B
HT and TH are favourable out of four outcomes.

16Source-based

A fair die shows six 6s in a row. P(6 on the next roll) is:

  1. A. 0
  2. B. 1/36
  3. C. 1/6
  4. D. greater than 1/6
View answer / solution

Answer: C
Each fair die roll is independent; previous rolls do not change the next-roll probability.

17Source-based

Believing that tails is 'due' after six consecutive heads is called:

  1. A. sampling bias
  2. B. Gambler's Fallacy
  3. C. relative frequency
  4. D. sample-space error
View answer / solution

Answer: B
The chapter names this misconception Gambler's Fallacy.

18Source-based

As the number of fair repeated trials becomes large, experimental probability tends to:

  1. A. 0
  2. B. 1
  3. C. the theoretical probability
  4. D. become impossible to calculate
View answer / solution

Answer: C
This is the Law of Large Numbers described in the chapter.

19Source-based

A fair coin is called unbiased because:

  1. A. it always alternates H,T
  2. B. neither face has a reason to occur more often
  3. C. it has no mass
  4. D. it cannot land on an edge
View answer / solution

Answer: B
Symmetry gives no reason for one side to be favoured.

20Source-based

A sample has 20 mango-lovers among 50 students. The relative frequency is:

  1. A. 0.2
  2. B. 0.4
  3. C. 0.5
  4. D. 2.5
View answer / solution

Answer: B
20/50=0.4.

21Source-based

Using the 0.4 sample estimate for a school of 1500, the estimated number preferring mango is:

  1. A. 400
  2. B. 500
  3. C. 600
  4. D. 750
View answer / solution

Answer: C
0.4×1500=600.

22Concept

A box has 5 green and 7 red balls. If one ball is drawn and only colour is recorded, the sample space is:

  1. A. {G,R}
  2. B. {G,G,G,G,G,R,…}
  3. C. {1,…,12}
  4. D. {5,7}
View answer / solution

Answer: A
The possible recorded outcomes are simply Green or Red; they are not equally likely.

23HOTS

A bag has 3 red and 7 blue marbles. Which statement is correct?

  1. A. Red and blue are equally likely because there are two colours
  2. B. P(red)=3/10 and P(blue)=7/10
  3. C. P(red)=1/2
  4. D. Theoretical probability cannot be found
View answer / solution

Answer: B
Each marble is equally likely, but colour outcomes are not; probabilities depend on counts.

24Source-based

The letters of PEACE are on five identical cards. P(not E) is:

  1. A. 1/5
  2. B. 2/5
  3. C. 3/5
  4. D. 4/5
View answer / solution

Answer: C
There are 3 non-E cards: P,A,C.

25Source-based

A spinner has equally likely outcomes 1 to 8. P(multiple of 3) is:

  1. A. 1/8
  2. B. 1/4
  3. C. 3/8
  4. D. 1/2
View answer / solution

Answer: B
Multiples of 3 are 3 and 6: 2/8=1/4.

26Exemplar-level

Two dice are rolled. The probability that the sum is 7 is:

  1. A. 1/12
  2. B. 1/9
  3. C. 1/6
  4. D. 5/36
View answer / solution

Answer: C
Six ordered pairs sum to 7 out of 36 equally likely pairs.

27Source-based

Two balls are drawn without replacement from 4 red and 5 blue. P(red then blue) is:

  1. A. 4/9×5/9
  2. B. 4/9×5/8
  3. C. 5/9×4/8
  4. D. 20/81
View answer / solution

Answer: B
After a red is removed, 8 balls remain, 5 blue.

28Source-based

From the same basket, P(two blue) is:

  1. A. 5/9×5/9
  2. B. 5/9×4/8
  3. C. 5/8×4/7
  4. D. 4/9
View answer / solution

Answer: B
Without replacement: 5/9 then 4/8.

29Source-Olympiad

A dye is dropped uniformly at random on a 3 m by 2 m rectangle containing a circle of diameter 1 m. The probability it lands in the circle is:

  1. A. π/24
  2. B. π/12
  3. C. π/6
  4. D. 1/6
View answer / solution

Answer: A
Geometric probability=area circle/area rectangle=[π(1/2)²]/6=π/24.

30Source-Olympiad

Three multiple-choice questions have 4 options each, one correct. A student guesses. P(exactly 2 correct) is:

  1. A. 3/64
  2. B. 9/64
  3. C. 27/64
  4. D. 1/16
View answer / solution

Answer: B
Choose which 2 are correct: C(3,2)(1/4)²(3/4)=3×1/16×3/4=9/64.

Section B — 15 One-Mark Questions

Core terminology, sample spaces and quick probabilities.

1Core

Define probability in one sentence.

View answer / solution

Probability is a measurement of the likelihood of an event.

2Core

What is randomness?

View answer / solution

A situation in which possible outcomes may be known but the exact outcome of an individual trial cannot be predicted in advance.

3Core

Write the probability scale.

View answer / solution

0≤P(E)≤1.

4Core

What is a sample space?

View answer / solution

The set/list of all possible outcomes of a random experiment.

5Core

What is an event?

View answer / solution

An event is one outcome or a group of outcomes, i.e. a subset of the sample space.

6Core

Write the sample space for one fair coin toss.

View answer / solution

{H,T}.

7Core

Write the sample space for one standard die roll.

View answer / solution

{1,2,3,4,5,6}.

8Core

Write the sample space for two coin tosses.

View answer / solution

{HH,HT,TH,TT}.

9Core

Write the experimental probability formula.

View answer / solution

Number of times the event occurred / total number of trials.

10Core

Write the theoretical probability formula for equally likely outcomes.

View answer / solution

P(E)=number of favourable outcomes / total number of possible outcomes.

11Source-based

A die is rolled 12 times and a 3 appears 3 times. Find the experimental probability.

View answer / solution

3/12=1/4.

12Core

For a fair die, find P(3).

View answer / solution

1/6.

13Source-based

A fair coin gives 6 heads in succession. Find P(tail on the next toss).

View answer / solution

1/2.

14Source-based

In a sample of 50 students, 15 like football. Find the relative frequency.

View answer / solution

15/50=3/10=0.3.

15Source-based

If S={HH,HT,TH,TT}, write the event 'at least one head'.

View answer / solution

{HH,HT,TH}.

Section C — 15 Two-Mark Questions

Short experimental, theoretical and sample-space applications.

1Core

Differentiate experimental probability from theoretical probability.

View answer / solution

Experimental probability is based on observed relative frequency from actual trials/data. Theoretical probability is calculated by reasoning under equally likely outcomes in an ideal fair setting.

2Source-based

A teacher samples 30 sweets: 10 red, 8 green, 7 yellow, 5 blue. Find the experimental probability of green.

View answer / solution

P(green)=8/30=4/15.

3Source-based

Using the same sample, estimate how many yellow sweets there are in a bag of 600 sweets.

View answer / solution

Estimated yellow proportion=7/30. Estimated count=600×7/30=140.

4Source-based

A sample of 40 students has 11 Arts Club preferences. Find the estimated probability and expected count in a school of 800.

View answer / solution

Probability=11/40=0.275. Estimated count=800×11/40=220.

5Core

A fair die is rolled. Find P(even number).

View answer / solution

Favourable={2,4,6}:3 outcomes out of 6. P=1/2.

6Core

A die is rolled. Find P(number greater than 4).

View answer / solution

Favourable={5,6}; P=2/6=1/3.

7Source-based

Two coins are tossed. Find P(at least one head).

View answer / solution

S={HH,HT,TH,TT}; favourable={HH,HT,TH}. P=3/4.

8Source-based

Three coins are tossed. Find P(exactly two heads).

View answer / solution

Favourable={HHT,HTH,THH}; total=8. P=3/8.

9Source-based

Ten cards numbered 1 to 10 are equally likely. Find P(even card).

View answer / solution

Even cards={2,4,6,8,10}:5 of 10, so P=1/2.

10Source-based

A bag contains 3 red,2 blue,1 green ball. Find P(not red).

View answer / solution

Non-red=3 balls out of 6, so P=1/2.

11Source-based

A random letter is chosen from PROBABILITY. Find P(B).

View answer / solution

PROBABILITY has 11 letters and 2 Bs, so P(B)=2/11.

12Source-based

The letters of PEACE are on identical cards. Find P(P,E or C).

View answer / solution

P,E,E,C are favourable:4 cards out of 5, so P=4/5.

13Source-based

A spinner has equally likely numbers 1 to 8. Find P(odd) and P(number>2).

View answer / solution

Odd={1,3,5,7}:4/8=1/2. Number>2={3,4,5,6,7,8}:6/8=3/4.

14Source-based

Explain why getting a 6 three times in a row does not change P(6) on the next fair die roll.

View answer / solution

Each die roll is independent. The die has no memory, so every roll retains P(6)=1/6.

15HOTS

A box has 5 green and 7 red balls. Explain why S={G,R} does not mean P(G)=P(R).

View answer / solution

Sample space lists possible recorded outcomes, not their probabilities. Since there are 5 green and 7 red equally likely balls, P(G)=5/12 and P(R)=7/12.

Section D — 15 Three-Mark Questions

Tree diagrams, surveys, dice, replacement and multi-step probability.

1Source-based

Explain the Law of Large Numbers in the context of rolling a fair die.

View answer / solution

For a fair die, theoretical P(4)=1/6. In a small number of rolls, the observed relative frequency of 4 may differ noticeably. As the number of independent rolls becomes large, the experimental proportion tends to get closer to 1/6.

2Source-based

A die is rolled 12 times and a 3 appears 3 times. Compare experimental and theoretical probabilities and explain the difference.

View answer / solution

Experimental P=3/12=1/4. Theoretical P=1/6. They differ because 12 trials are few and random variation is expected. With many more rolls, the experimental value should tend toward 1/6.

3Source-extension

Construct the sample space for rolling one die and tossing one coin together, and find P(even number and H).

View answer / solution

S={(1,H),(1,T),…,(6,H),(6,T)}, so n(S)=12. Favourable={(2,H),(4,H),(6,H)}:3 outcomes. P=3/12=1/4.

4Source-based

A village fair offers 3 snacks (Samosa,Pakora,Bhaji) and 2 drinks (Chai,Lassi). List the sample space and find P(Samosa) if each combination is equally likely.

View answer / solution

There are 6 combinations: SC,SL,PC,PL,BC,BL. Samosa event={SC,SL}, so P=2/6=1/3.

5Source-based

Two fair coins are tossed. Use a tree-diagram argument to find P(one head and one tail).

View answer / solution

First toss branches H,T each with 1/2; each branches again H,T. Terminal outcomes HH,HT,TH,TT each have probability 1/4. One head and one tail={HT,TH}, so P=1/2.

6Source-based

Basket A has one apple and two oranges; Basket B has one banana and one mango. One fruit is selected randomly from each basket. Find P(apple and banana).

View answer / solution

P(apple from A)=1/3 and P(banana from B)=1/2. The selections are independent, so P=1/3×1/2=1/6.

7Source-extension

A box has 3 red,4 black,2 green pens. A pen is picked, replaced, and another is picked. Find P(same colour).

View answer / solution

P(RR)=(3/9)²=1/9; P(BB)=(4/9)²=16/81; P(GG)=(2/9)²=4/81. Total=9/81+16/81+4/81=29/81.

8Source-based

A tyre company's 1000 records are 20 (<4000), 210 (4001–9000), 325 (9001–14000),445 (>14000). Find the three probabilities requested in the chapter.

View answer / solution

P(<4000)=20/1000=0.02. P(4000–14000)=(210+325)/1000=0.535. P(>14000)=445/1000=0.445.

9Source-based

A basket has 4 red and 5 blue balls. Two are drawn without replacement. Find P(R then B) and P(BB).

View answer / solution

P(R then B)=4/9×5/8=5/18. P(BB)=5/9×4/8=5/18.

10Source-based

Two fair dice are rolled. Find P(sum is a prime greater than 5).

View answer / solution

Possible prime sums >5 are 7 and 11. Sum 7 has 6 ordered pairs; sum 11 has 2. Total favourable=8 of 36, so P=2/9.

11Source-based

A bag has 4 red,3 green,2 blue balls. Two are drawn without replacement. Find P(different colours).

View answer / solution

Total unordered pairs=C(9,2)=36. Same-colour pairs=C(4,2)+C(3,2)+C(2,2)=6+3+1=10. Different=26, so P=26/36=13/18.

12Source-based

Three fair coins are tossed. Find P(first coin is H and exactly two heads occur in total).

View answer / solution

The favourable ordered outcomes are HHT and HTH. Total outcomes=8. Therefore P=2/8=1/4.

13Source-based

A four-digit number is formed using 1,2,3,4 without repetition. Find P(number is even).

View answer / solution

Total permutations=4!=24. An even number must end in 2 or 4:2 choices. Remaining digits arrange in 3!=6 ways. Favourable=12. P=12/24=1/2.

14Source-based

A student guesses answers to 3 questions, each with 4 options. Find P(exactly 2 correct).

View answer / solution

Choose which 2 are correct: C(3,2)=3. Probability for a chosen pattern=(1/4)²(3/4). Thus total=3×1/16×3/4=9/64.

15Source-based

A dye is dropped uniformly on a 3m×2m rectangle. A circle of diameter 1m lies inside it. Find the probability of landing in the circle.

View answer / solution

Rectangle area=6 m². Circle radius=1/2 m; area=π/4. Under uniform landing, probability=(π/4)/6=π/24.

Section E — 10 Four-Mark Questions

Full reasoning, probability models, independence and geometric probability.

1Source-based

A box has balls numbered 1 to 4. Two draws are made with replacement. Write the sample space and its size; then find P(the two numbers are different).

View answer / solution

With replacement, S={(i,j):i,j∈{1,2,3,4}}, so n(S)=16. Equal-number outcomes are (1,1),(2,2),(3,3),(4,4):4. Different outcomes=12. P=12/16=3/4.

2Source-based

Repeat the previous experiment without replacement. Write the sample space size and find P(the second number is greater than the first).

View answer / solution

Without replacement there are 4×3=12 ordered outcomes. For each unordered pair {a,b}, exactly one order has second>first. There are C(4,2)=6 such favourable outcomes. P=6/12=1/2.

3Source-extension

A bag contains 4 red and 5 blue balls. Two are drawn without replacement. Find P(one of each colour), P(same colour), and verify the probabilities sum to 1.

View answer / solution

One of each=P(RB)+P(BR)=4/9·5/8+5/9·4/8=10/18=5/9. Same colour=P(RR)+P(BB)=4/9·3/8+5/9·4/8=12/72+20/72=4/9. Sum=5/9+4/9=1.

4Source-based

Two fair dice are rolled. Find the probability that the sum is a prime greater than 5 and compare it with the probability that the sum is 7.

View answer / solution

Prime sums >5 are 7 and 11:8 favourable pairs out of 36, so 2/9. Sum 7 has 6 favourable pairs, so 1/6. Difference=2/9−1/6=1/18; prime>5 is more likely.

5HOTS

A fair coin is tossed 8 times and gives 8 heads. A student says tails is now more likely. Identify the fallacy and give the correct next-toss probability. Then distinguish this from the probability of getting 9 heads in a row from the start.

View answer / solution

The claim is Gambler's Fallacy. The next toss is independent, so P(T)=1/2. But before any tosses, P(9 heads in a row)=(1/2)^9=1/512. Conditional on already observing 8 heads, only the ninth toss remains, so P(H next)=1/2.

6Concept

A biased-looking bag contains 3 red and 7 blue marbles. Explain why the colour outcomes are not equally likely, yet individual marble outcomes may be considered equally likely. Use this to find P(red).

View answer / solution

If the draw is fair, each of 10 individual marbles is equally likely. But the colour event Red contains 3 elementary outcomes while Blue contains 7, so the colour outcomes are not equally likely. Hence P(red)=3/10.

7Source-based

A random sample of 50 students gives mango preference 20, apple 15, banana 10, grapes 5. Estimate counts for a school of 1500 and explain one limitation.

View answer / solution

Estimated proportions are 0.4,0.3,0.2,0.1. Estimated counts:600 mango,450 apple,300 banana,150 grapes. Limitation: the sample may be too small or not representative; a larger, less biased sample can improve confidence.

8Source-HOTS

For three fair coins, compare these two experiments: A records the full sequence; B records only number of heads. Write each sample space and explain why the four outcomes of B are not equally likely.

View answer / solution

A: {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}, 8 equally likely outcomes. B:{0,1,2,3} heads. Their probabilities are 1/8,3/8,3/8,1/8 respectively, so B's four outcomes are not equally likely. Thus one cannot use '1/4 each' merely because the sample space has four recorded values.

9Source-based

A spinner has equally likely labels 1 to 8. Find P(odd), P(>2), P(<9), P(multiple of 3), and explain which event is certain.

View answer / solution

P(odd)=4/8=1/2. P(>2)=6/8=3/4. P(<9)=8/8=1, so this event is certain. P(multiple of 3)=2/8=1/4.

10Source-Olympiad

A rectangle is 3m×2m and contains a circle of diameter 1m. If a point is selected uniformly at random, derive the geometric probability of landing inside the circle and outside the circle.

View answer / solution

Rectangle area=6. Circle area=π(1/2)²=π/4. P(inside)=π/24. P(outside)=1−π/24=(24−π)/24.

Section F — 5 Case Studies

Sampling, long-run frequency, non-replacement, guessing and geometric probability.

Case 1Source-based Statistical Probability

Case Study 1 — School Fruit Survey

A class sample of 50 students gives preferences: mango 20, apple 15, banana 10 and grapes 5. The school has 1500 students.

  1. a) Find the estimated probability of mango preference.
  2. b) Estimate the number preferring mango in the school.
  3. c) Estimate the number preferring banana.
  4. d) State one way to improve the reliability of the estimate.
View case-study solutions

a) Find the estimated probability of mango preference. 20/50=0.4.

b) Estimate the number preferring mango in the school. 0.4×1500=600.

c) Estimate the number preferring banana. 10/50×1500=300.

d) State one way to improve the reliability of the estimate. Use a larger and more representative sample across different classes/grades.

Case 2Law of Large Numbers

Case Study 2 — Fair Die Experiment

A fair die is rolled 60 times. The number 4 appears 7 times.

  1. a) Find the experimental probability of 4.
  2. b) Find the theoretical probability of 4.
  3. c) Are the two values required to be exactly equal?
  4. d) What is expected if the experiment is repeated for very many trials?
View case-study solutions

a) Find the experimental probability of 4. 7/60.

b) Find the theoretical probability of 4. 1/6.

c) Are the two values required to be exactly equal? No. Random variation can make experimental probability differ, especially in a finite sample.

d) What is expected if the experiment is repeated for very many trials? The experimental relative frequency should tend to move closer to 1/6.

Case 3Source-based Tree Diagram

Case Study 3 — Two Draws Without Replacement

A basket contains 4 red and 5 blue balls. One ball is drawn and laid aside; then a second is drawn.

  1. a) Find P(R then B).
  2. b) Find P(B then R).
  3. c) Find P(one of each colour).
  4. d) Why do second-draw probabilities change?
View case-study solutions

a) Find P(R then B). 4/9×5/8=5/18.

b) Find P(B then R). 5/9×4/8=5/18.

c) Find P(one of each colour). 10/18=5/9.

d) Why do second-draw probabilities change? The first ball is not replaced, so the composition and total number of balls change.

Case 4Special Olympiad

Case Study 4 — Multiple-Choice Guessing

A student guesses on 3 questions. Each question has 4 options and exactly one correct option.

  1. a) Find P(correct on one question).
  2. b) Find P(wrong on one question).
  3. c) Find P(exactly two correct).
  4. d) Find P(all three correct).
View case-study solutions

a) Find P(correct on one question). 1/4.

b) Find P(wrong on one question). 3/4.

c) Find P(exactly two correct). C(3,2)(1/4)²(3/4)=9/64.

d) Find P(all three correct). (1/4)³=1/64.

Case 5Source-based HOTS

Case Study 5 — Geometric Probability

A dye is dropped uniformly at random on a 3 m by 2 m rectangular sheet. A circular target of diameter 1 m lies completely inside it.

  1. a) Find the area of the rectangle.
  2. b) Find the area of the target.
  3. c) Find P(hit the target).
  4. d) Find P(miss the target).
View case-study solutions

a) Find the area of the rectangle. 6 m².

b) Find the area of the target. π/4 m².

c) Find P(hit the target). π/24.

d) Find P(miss the target). 1−π/24=(24−π)/24.

Special Exemplar Challenge — 15 Questions

Non-routine complements, counting, dice, coins and replacement problems.

1Special Exemplar

A fair die is rolled once. Find P(prime number), P(composite number), and P(neither prime nor composite).

View worked solution

Prime={2,3,5}:1/2. Composite={4,6}:1/3. Neither={1}:1/6.

2Special Exemplar

Two coins are tossed. Find P(at least one tail) without listing favourable outcomes directly.

View worked solution

Use complement: P(at least one tail)=1−P(HH)=1−1/4=3/4.

3Special Exemplar

Three coins are tossed. Find P(at most one head).

View worked solution

Outcomes with 0 or 1 head: TTT,HTT,THT,TTH =4 of 8. P=1/2.

4Special Exemplar

A number is selected from 1 to 20. Find P(multiple of 3 or 5).

View worked solution

Multiples of 3:6; of 5:4; common multiples of 15:1. Favourable=6+4−1=9, so P=9/20.

5Special Exemplar

A number is selected from 1 to 30. Find P(prime).

View worked solution

Primes:2,3,5,7,11,13,17,19,23,29 →10. P=10/30=1/3.

6Special Exemplar

Two dice are rolled. Find P(sum=8).

View worked solution

Favourable ordered pairs:(2,6),(3,5),(4,4),(5,3),(6,2):5. P=5/36.

7Special Exemplar

Two dice are rolled. Find P(sum≥10).

View worked solution

Sums 10,11,12 have 3+2+1=6 outcomes. P=6/36=1/6.

8Special Exemplar

Two dice are rolled. Find P(product is even).

View worked solution

Complement: product odd only when both dice are odd:3×3=9 outcomes. P(even product)=1−9/36=3/4.

9Special Exemplar

From PEACE, find P(vowel) and P(E | the chosen card is a vowel) conceptually.

View worked solution

Vowel cards are E,A,E:3/5. Conditional among vowels, two of the three are E, so 2/3.

10Special Exemplar

A bag has 2 red,3 green,5 blue balls. One is drawn. Find P(not blue).

View worked solution

Non-blue=5 out of 10, so P=1/2.

11Special Exemplar

A bag has 4 red,3 green,2 blue balls. Two are drawn without replacement. Find P(both same colour).

View worked solution

[C(4,2)+C(3,2)+C(2,2)]/C(9,2)=(6+3+1)/36=5/18.

12Special Exemplar

Two draws are made with replacement from balls numbered 1,2,3,4. Find P(sum=5).

View worked solution

There are 16 ordered pairs; favourable (1,4),(2,3),(3,2),(4,1):4. P=1/4.

13Special Exemplar

Two draws are made without replacement from 1,2,3,4. Find P(sum=5).

View worked solution

There are 12 ordered outcomes; favourable same four ordered pairs:4. P=1/3.

14Special Exemplar

A point is selected uniformly from a 10 cm by 10 cm square. A circle of radius 5 cm is inscribed. Find P(point lies in the circle).

View worked solution

Area ratio=25π/100=π/4.

15Special Exemplar

A student guesses 4 true/false questions. Find P(exactly 3 correct).

View worked solution

C(4,3)(1/2)^4=4/16=1/4.

Extra Olympiad / HOTS Challenge — 20 Questions

Conditional reasoning, advanced counting, repeated trials and geometric probability.

1Special Olympiad / HOTS

Three fair coins are tossed. Given that at least one head occurs, find the probability of exactly two heads.

View worked solution

Conditioned sample excludes TTT, leaving 7 equally likely full-sequence outcomes. Exactly two heads has 3 outcomes, so probability=3/7.

2Special Olympiad / HOTS

Two dice are rolled. Given that the sum is 8, find P(at least one die is 3).

View worked solution

Sum-8 outcomes are (2,6),(3,5),(4,4),(5,3),(6,2). Two contain a 3, so probability=2/5.

3Special Olympiad / HOTS

Two dice are rolled. Find P(the larger number is 4).

View worked solution

Ordered outcomes with max=4 are all pairs from {1,2,3,4} minus all pairs from {1,2,3}:16−9=7. P=7/36.

4Special Olympiad / HOTS

Two dice are rolled. Find P(the numbers are different).

View worked solution

Complement of doubles. P=1−6/36=5/6.

5Special Olympiad / HOTS

Two dice are rolled. Find P(sum is prime).

View worked solution

Prime sums possible:2,3,5,7,11. Counts=1+2+4+6+2=15. P=15/36=5/12.

6Special Olympiad / HOTS

Three coins are tossed. Let X be number of heads. Find P(X is even).

View worked solution

X=0 or 2. Counts=1+C(3,2)=1+3=4 of 8, so P=1/2.

7Special Olympiad / HOTS

Four fair coins are tossed. Find P(exactly two heads).

View worked solution

C(4,2)/16=6/16=3/8.

8Special Olympiad / HOTS

Four fair coins are tossed. Find P(at least three heads).

View worked solution

[C(4,3)+C(4,4)]/16=(4+1)/16=5/16.

9Special Olympiad / HOTS

A number is chosen uniformly from 1 to 100. Find P(it is divisible by 2 or 5 but not both).

View worked solution

Divisible by 2:50; by 5:20; by both(10):10. Exactly one=50+20−2·10=50. Probability=1/2.

10Special Olympiad / HOTS

A number is chosen from 1 to 60. Find P(it is divisible by 4 or 6).

View worked solution

Multiples of 4=15; of 6=10; of lcm12=5. Union=20. P=20/60=1/3.

11Special Olympiad / HOTS

A bag has 5 red,4 blue,3 green balls. Two are drawn without replacement. Find P(at least one red).

View worked solution

Use complement: no red means both from 7 non-red balls. P=1−C(7,2)/C(12,2)=1−21/66=45/66=15/22.

12Special Olympiad / HOTS

From the same bag, find P(the two balls have different colours).

View worked solution

Total pairs=C(12,2)=66. Same-colour=C(5,2)+C(4,2)+C(3,2)=10+6+3=19. Different=47. P=47/66.

13Special Olympiad / HOTS

A box contains 2 red and 3 blue balls. Two balls are drawn with replacement. Find P(exactly one red).

View worked solution

P(RB)+P(BR)=2/5·3/5+3/5·2/5=12/25.

14Special Olympiad / HOTS

The same box is used without replacement. Find P(exactly one red).

View worked solution

P=2/5·3/4+3/5·2/4=6/20+6/20=3/5.

15Special Olympiad / HOTS

A student guesses 5 multiple-choice questions with 4 choices each. Find P(exactly one correct).

View worked solution

C(5,1)(1/4)(3/4)^4=5·1/4·81/256=405/1024.

16Special Olympiad / HOTS

A student guesses 5 such questions. Find P(at least one correct).

View worked solution

Complement all wrong:1−(3/4)^5=1−243/1024=781/1024.

17Special Olympiad / HOTS

A point is chosen uniformly inside a square of side 2r. A circle of radius r is inscribed. Find P(point is outside the circle).

View worked solution

Square area=4r², circle=πr². P(outside)=(4−π)/4=1−π/4.

18Special Olympiad / HOTS

A point is uniformly selected in an annulus between concentric radii r and 2r. Find probability it lies within distance 3r/2 of the centre.

View worked solution

Desired annular area=π[(3r/2)²−r²]=5πr²/4. Total annulus=π(4r²−r²)=3πr². Probability=(5/4)/3=5/12.

19Special Olympiad / HOTS

A fair coin is tossed repeatedly until the first head. Find P(first head occurs on toss 3).

View worked solution

The sequence must be TTH. Probability=(1/2)^3=1/8.

20Special Olympiad / HOTS

A fair die is rolled repeatedly until a 6 appears. Find P(first 6 occurs on roll 4).

View worked solution

First three must be non-6 and fourth is 6: (5/6)^3(1/6)=125/1296.

High-Level Thinking Laboratory — 10 Questions

Students analyse assumptions, sample-space design, independence, bias and long-run reasoning.

1High-Level Thinking

Why does a sample space need enough detail to match the question being asked?

View worked solution

A coarse sample space can merge outcomes that the question needs to distinguish. For example, {Rain,No Rain} is enough for 'Will it rain?' but not for comparing drizzle, light rain and heavy rain.

2High-Level Thinking

Why can two outcomes listed in a sample space fail to be equally likely?

View worked solution

A sample space records what can happen, not necessarily equal probabilities. For example, colour outcomes {R,B} from a bag with 3 red and 7 blue are not equally likely.

3High-Level Thinking

Why is 'experimental probability should equal theoretical probability after 12 trials' an incorrect claim?

View worked solution

Experimental probability fluctuates randomly. The Law of Large Numbers describes long-run tendency, not exact equality after a fixed small number of trials.

4High-Level Thinking

A coin lands heads 10 times in a row. Explain the difference between 'the probability of 11 heads from the start' and 'the probability the next toss is head now'.

View worked solution

Before tossing, P(11 consecutive heads)=1/2^11. After 10 heads have already occurred, the next fair toss is independent, so P(head next)=1/2.

5High-Level Thinking

Why is P(E)=favourable/total dangerous if applied without checking equally likely outcomes?

View worked solution

Counting outcomes works only if elementary outcomes have equal probabilities. If recorded categories have unequal weights, simple counting can give the wrong answer.

6High-Level Thinking

Why does replacement matter in multi-step probability?

View worked solution

Replacement restores the original composition, making successive draw probabilities unchanged. Without replacement, totals and category counts change after each draw.

7High-Level Thinking

Why is a larger sample usually more useful but not automatically unbiased?

View worked solution

A larger sample reduces random fluctuation, but a systematically unrepresentative selection method can remain biased regardless of size.

8High-Level Thinking

Why can geometric probability be treated as an area ratio in the dye-drop question?

View worked solution

Uniform random landing means equal-area regions are equally likely. Therefore probability is proportional to area, so favourable area/total area gives the probability.

9High-Level Thinking

Why is {0,1,2,3} a valid sample space for 'number of heads in three tosses' even though these four outcomes are not equally likely?

View worked solution

A sample space only needs to list all possible recorded outcomes once. Equal likelihood is not a requirement for being a sample space.

10High-Level Thinking

How does a tree diagram help prevent omitted or duplicated outcomes?

View worked solution

Each complete root-to-leaf path corresponds to one ordered outcome, so systematic branching makes it easier to enumerate every possible sequence exactly once.

Exam Strategy & Common Traps

TopicBest first moveCommon trap
Sample spaceIdentify exactly what the experiment records.Using an over-detailed or under-detailed sample space.
Theoretical probabilityCheck that elementary outcomes are equally likely before counting.Assuming listed categories are equally likely.
Experimental probabilityUse observed frequency ÷ total trials.Replacing observed frequency with theoretical expectation.
Tree diagramsLabel every branch and multiply probabilities along a path.Adding probabilities along one path instead of multiplying.
Without replacementUpdate both numerator and denominator after the first draw.Reusing first-draw probabilities.
At least / notConsider a complement when it is simpler.Omitting cases.
Gambler’s FallacyAsk whether trials are independent.Believing an outcome is “due”.
SamplingCheck both sample size and representativeness.Thinking a large biased sample must be reliable.

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