SK Tuitions • Class 10 Mathematics
Probability — Virtual Teacher + Solved Question Bank
Learn the ideas first, practise them visually, then solve a progressively challenging bank covering board-style, competency, HOTS and foundation reasoning.
30 MCQs15 × 2 Marks15 × 3 Marks20 × 4 Marks10 × 5 Marks5 Case Studies
Scope note: Classical/theoretical probability and standard Class 10 applications form the core. Content marked Enrichment / Foundation / Olympiad is included for stronger reasoning practice rather than presented as compulsory board content.
Virtual Teacher
1. Probability from the Ground Up
Probability measures how likely an event is to happen. It lies between 0 and 1.
Think before calculating
- The Sun rising tomorrow is treated as a certain event in an elementary model.
- A fair coin does not definitely show heads; H and T are equally likely.
- A standard die cannot show 8; that event is impossible.
- Drawing an ace from a 52-card deck is possible but not certain.
Theoretical Probability
For a finite experiment with equally likely outcomes:
P(E) = Number of favourable outcomesTotal number of equally likely outcomes
P(E) = n(E)n(S)
S is the sample space and E is the event.
Random Experiment → Outcome → Sample Space → Event
Random experiment
A process whose possible outcomes are known but whose exact result cannot be predicted beforehand.
Examples: toss, roll, draw, spin.Outcome
One possible result.
Example: 4 on a die.Sample space S
The set of all possible outcomes.
Coin: S={H,T}Event E
A collection of outcomes satisfying a stated condition.
Even die result: E={2,4,6}Why “2 + 3” is not a random experiment: its result is fixed at 5. No uncertainty is involved.
Favourable Outcomes
Favourable outcomes are exactly those outcomes that satisfy the event condition.
Example: Roll one fair die and obtain a prime number.
S = {1,2,3,4,5,6}
E = {2,3,5}
P(E)=36=12
Common trap: 1 is neither prime nor composite.
Equally Likely Outcomes
The formula n(E)n(S) is used when the elementary outcomes are equally likely.
- Fair coin: H and T are equally likely.
- Fair die: 1,2,3,4,5,6 are equally likely.
- Two dice: the 36 ordered pairs are equally likely, but the sums 2 to 12 are not equally likely.
Range, Impossible and Certain Events
0 ≤ P(E) ≤ 1
Impossible
P(E)=0
Example: rolling 7 on a standard die.
Between 0 and 1
0<P(E)<1
Possible but not certain.
Certain
P(E)=1
Example: rolling a number less than 7 on a standard die.
Probability scale: 0 means impossible, 1/2 means an even chance, and 1 means certain.
Complementary Event — the Most Useful Shortcut
If E is an event, “not E” is its complement, written E′ or E̅.
P(E) + P(not E) = 1
P(not E) = 1 − P(E)
Example
Not getting 6 on one die:
1−16=56
When complement is especially useful
Words such as not, does not, at least one, neither and not equal to often signal a faster complement route.
Exactly, At Least, At Most
Exactly one
One and only one.
At least one
One or more.
At most one
Zero or one.
Two coins: S={HH,HT,TH,TT}.
- Exactly one head: {HT,TH} → 24=12
- At least one head: {HH,HT,TH} → 34
- At most one head: {HT,TH,TT} → 34
- No head: {TT} → 14
Two-toss sample space: HH, HT, TH and TT. HT and TH are different ordered outcomes.
Coins
One fair coin
S={H,T}
P(H)=P(T)=12
Three coins — Enrichment / Foundation
Enrichment / Foundation2³=8 outcomes
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
A fair coin has two equally likely outcomes: heads and tails.
One Die and the Number Theory You Need
S={1,2,3,4,5,6}
- Even: {2,4,6}
- Odd: {1,3,5}
- Prime: {2,3,5}
- Composite: {4,6}
- Factors of 6: {1,2,3,6}
Quick number revision
Prime: exactly two positive factors.
Composite: more than two positive factors.
1: neither prime nor composite.
Also revise factors, multiples, perfect squares, cubes and divisibility.
A standard fair die has six equally likely faces numbered 1 to 6.
Two Dice — 36 Ordered Outcomes
n(S)=6×6=36
Represent each outcome as (first die, second die). For example, (2,5) and (5,2) are distinct.
The 6×6 grid contains all 36 ordered outcomes. Outcomes whose sum is 7 are highlighted by the drawing.
| Sum | Number of outcomes |
|---|---|
| 2 | 1 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
| 6 | 5 |
| 7 | 6 |
| 8 | 5 |
| 9 | 4 |
| 10 | 3 |
| 11 | 2 |
| 12 | 1 |
Key insight: the frequency pattern 1,2,3,4,5,6,5,4,3,2,1 is symmetric about 7. This is why all sums are not equally likely.
Playing Cards — Complete Class 10 Reference
52 cards
4 suits × 13 cards.
26 red
Hearts + Diamonds.
26 black
Clubs + Spades.
12 face cards
J, Q, K in each suit.
4 aces, 4 kings, 4 queens, 4 jacks
6 red face cards, 6 black face cards
36 number cards
Cards 2 through 10: 9 per suit × 4.
Ace is not a face card in the standard school definition. If cards are removed, always update both the total number of cards and the number of favourable cards.
A deck has four suits: hearts, diamonds, clubs and spades; 13 cards in each suit.
Numbers, Letters, Spinners and Bags
Number selection
If one integer is chosen from a through b inclusive, the number of integers is:
b−a+1
Then list or count primes, multiples, factors, squares, cubes or other required properties.
Letter selection
When a letter-position is selected from a written word, repeated letters represent separate positions. Count every position.
Equal-sector spinner
If all sectors are equal, each sector is equally likely. If sectors are unequal, do not assume equal probability.
Balls in a bag
For a single random draw from physically identical balls, total balls form the equally likely selections. Replacement matters only in multi-stage draws and must be stated.
Equal-sector spinner numbered 1 to 8.
Illustrative bag containing coloured balls; probability is based on the stated counts.
Experimental vs Theoretical Probability
Conceptual EnrichmentTheoretical
P(E)=n(E)n(S)
Based on a mathematical model with equally likely outcomes.
Experimental
P_exp(E)≈number of times E occursnumber of trials
Based on observed results. It need not equal the theoretical value exactly in a small number of trials.
Probability is not a guarantee: P(H)=1/2 does not mean that every pair of coin tosses must contain exactly one head.
Foundation Vocabulary
Enrichment / Foundation VocabularyEvents that cannot occur together are often called mutually exclusive. You do not need advanced probability laws to solve the questions here; elementary sample-space counting is sufficient.
Geometric / Visual Probability
Enrichment / FoundationFor a figure divided into equal-area elementary regions, one may count regions:
P(shaded)=number of shaded equal regionstotal equal regions
This module avoids continuous geometric probability formulas beyond Class 10 level.
Revision Sheet
2. Probability Formula & Method Revision Sheet
Event
P(E)=n(E)n(S)
Complement
P(not E)=1−P(E)
Range
0≤P(E)≤1
Impossible
P(E)=0
Certain
P(E)=1
Event + Complement
P(E)+P(not E)=1
Two coins
n(S)=4
Two dice
n(S)=36
Standard deck
52 total • 26 red • 26 black • 12 face • 4 aces
The 6-Step Probability Method
- Understand the experiment. What is being tossed, rolled, drawn, selected or spun?
- Determine S. Write the full sample space when small; otherwise find n(S).
- Define E. Translate the words into the required event.
- Count favourable outcomes. List systematically where necessary.
- Apply the formula. P(E)=n(E)/n(S).
- Simplify and check. Your final probability must lie from 0 to 1.
Usually write the full sample space
- Two coins
- Small spinners
- Small number sets
- Simple combined experiments
Usually count instead of listing everything
- One die
- Standard deck
- Two dice after understanding 36 ordered pairs
- Large number ranges
Interactive Enrichment
3. Probability Lab
These simulations enrich understanding; none is required to access the theory or questions.
Experiment with Coin Tosses
Compare experimental frequency with the theoretical value 1/2.
Choose a trial count to simulate coin tosses.
Interactive Dice Explorer
Use the controls to explore die outcomes.
Equal-Sector Spinner
Spinner sectors are numbered 1 to 8 and are equally likely.
Solved Practice
4. Complete Question Bank — 95 Main Entries
Difficulty progresses from foundation to board-standard, competency, HOTS and Olympiad-style elementary counting.
Solutions Viewed: 0 / 95
Section A — 30 MCQs 30 × 1 mark
A1
1 Mark
Level 1 — Foundation
NCERT Type
Random Experiment
Which of the following is a random experiment?
Solution
Correct option: B
Reasoning
A random experiment has known possible outcomes but its exact result cannot be predicted beforehand. A die can show any one of 1, 2, 3, 4, 5, 6.
A2
1 Mark
Level 1 — Foundation
NCERT Type
Sample Space
The sample space for one toss of a fair coin is:
Solution
Correct option: C
Sample space
S = {H, T}
Answer
Both possible outcomes must be included.
A3
1 Mark
Level 1 — Foundation
NCERT Type
Impossible Event
A standard six-sided die is rolled. The probability of obtaining 7 is:
Solution
Correct option: A
Favourable outcomes
There is no face numbered 7 on a standard die, so n(E) = 0.
Probability
P(E) = 06 = 0
A4
1 Mark
Level 1 — Foundation
NCERT Type
Die
A fair die is rolled once. What is the probability of obtaining a prime number?
Solution
Correct option: B
Sample space
S = {1, 2, 3, 4, 5, 6}, n(S)=6
Prime outcomes
E = {2, 3, 5}, n(E)=3
Probability
P(E) = 36 = 12
Common trap: 1 is neither prime nor composite.
A5
1 Mark
Level 1 — Foundation
NCERT Type
Complement
If P(E) = 0.37, then P(not E) is:
Solution
Correct option: B
Complement rule
P(not E) = 1 − P(E)
Calculation
= 1 − 0.37 = 0.63
A6
1 Mark
Level 1 — Foundation
NCERT Type
Two Coins
Two fair coins are tossed. The probability of getting exactly one head is:
Solution
Correct option: B
Sample space
S = {HH, HT, TH, TT}
Favourable outcomes
E = {HT, TH}
Probability
P(E) = 24 = 12
A7
1 Mark
Level 2 — Standard
CBSE/PYQ Pattern
Two Coins
Two fair coins are tossed. The probability of getting at least one head is:
Solution
Correct option: C
Interpretation
“At least one head” means one or two heads.
Favourable outcomes
E = {HH, HT, TH}
Probability
P(E) = 34
A8
1 Mark
Level 1 — Foundation
NCERT Type
Cards
One card is drawn from a well-shuffled standard deck of 52 cards. The probability that it is red is:
Solution
Correct option: B
Deck fact
There are 26 red cards: 13 hearts and 13 diamonds.
Probability
P(red) = 2652 = 12
A9
1 Mark
Level 1 — Foundation
NCERT Type
Cards
The probability of drawing a face card from a standard deck is:
Solution
Correct option: B
Face-card count
J, Q and K are face cards: 3 per suit × 4 suits = 12.
Probability
P(face) = 1252 = 313
Common trap: Ace is not a face card in the standard school convention.
A10
1 Mark
Level 1 — Foundation
NCERT Type
Cards
A card is drawn from a standard deck. The probability of drawing a red king is:
Solution
Correct option: B
Favourable cards
There are 2 red kings: king of hearts and king of diamonds.
Probability
P(red king) = 252 = 126
A11
1 Mark
Level 1 — Foundation
NCERT Type
Number Selection
An integer is chosen at random from 1 to 20. The probability that it is a multiple of 3 is:
Solution
Correct option: B
Total outcomes
n(S)=20.
Multiples of 3
{3, 6, 9, 12, 15, 18}; n(E)=6
Probability
P(E)=620=310
A12
1 Mark
Level 2 — Standard
NCERT Exemplar Type
Letters
One letter-position is chosen at random from the word MATHEMATICS. The probability of choosing a vowel is:
Solution
Correct option: B
Count positions
MATHEMATICS has 11 letter-positions.
Vowel positions
A, E, A, I → 4 positions.
Probability
P(vowel)=411
Why? Repeated letters count separately because a position is being selected.
A13
1 Mark
Level 1 — Foundation
NCERT Type
Spinner
A spinner has 8 equal sectors numbered 1 to 8. The probability that the pointer stops on a prime number is:
Solution
Correct option: B
Prime sectors
{2, 3, 5, 7}
Probability
P(prime)=48=12
A14
1 Mark
Level 1 — Foundation
NCERT Type
Bag of Balls
A bag contains 5 red, 3 blue and 2 green balls. One ball is chosen at random. The probability that it is not blue is:
Solution
Correct option: B
Total balls
5 + 3 + 2 = 10.
Not blue
Red or green = 5 + 2 = 7 balls.
Probability
P(not blue)=710
A15
1 Mark
Level 1 — Foundation
NCERT Type
Range
Which of the following cannot be the probability of an event?
Solution
Correct option: B
Probability range
0 ≤ P(E) ≤ 1
Conclusion
54 = 1.25 is greater than 1, so it cannot be a probability.
A16
1 Mark
Level 2 — Standard
CBSE/PYQ Pattern
Two Dice
Two fair dice are thrown. The probability that their sum is 7 is:
Solution
Correct option: B
Total outcomes
n(S)=6×6=36
Favourable ordered pairs
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Probability
P(sum 7)=636=16
A17
1 Mark
Level 2 — Standard
NCERT Exemplar Type
Two Dice
Two fair dice are thrown. The probability of getting equal numbers on both dice is:
Solution
Correct option: A
Favourable outcomes
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
Probability
P(doubles)=636=16
A18
1 Mark
Level 3 — Advanced
Competency Based
Two Dice
Two fair dice are thrown. The probability that their product is odd is:
Solution
Correct option: A
Key idea
A product is odd only when both factors are odd.
Odd faces
Each die has 3 odd faces: 1, 3, 5.
Favourable outcomes
3 × 3 = 9.
Probability
P(odd product)=936=14
A19
1 Mark
Level 3 — Advanced
Competency Based
Two Dice
Two fair dice are thrown. The probability that at least one die shows 6 is:
Solution
Correct option: C
Complement
No die shows 6: each die then has 5 allowed outcomes.
No 6
P(no 6)=2536
At least one 6
1−2536=1136
A20
1 Mark
Level 3 — Advanced
CBSE/PYQ Pattern
Two Dice
Two fair dice are thrown. The probability that neither die shows 1 is:
Solution
Correct option: A
Allowed faces
For each die: {2,3,4,5,6}, so 5 choices.
Favourable outcomes
5 × 5 = 25.
Probability
P(neither 1)=2536
A21
1 Mark
Level 3 — Advanced
Assertion–Reason
Concept
Assertion (A): The probability of an event in a finite equally likely sample space cannot exceed 1.
Reason (R): The number of favourable outcomes cannot exceed the total number of outcomes.
Reason (R): The number of favourable outcomes cannot exceed the total number of outcomes.
Solution
Correct option: A
Reasoning
P(E)=n(E)n(S)
Comparison
Since n(E) ≤ n(S), the fraction cannot exceed 1. Therefore R correctly explains A.
A22
1 Mark
Level 3 — Advanced
Error Analysis
Two Dice
A student says, “The possible sums of two dice are 2 to 12, so each sum has probability 1/11.” Which statement is correct?
Solution
Correct option: B
Key idea
The 36 ordered pairs are equally likely, but the 11 sums are not.
Example
Sum 2 occurs in 1 way, while sum 7 occurs in 6 ways. Therefore assigning 1/11 to each sum is incorrect.
A23
1 Mark
Level 3 — Advanced
Competency Based
Cards
One king is removed from a standard deck. A card is then drawn from the remaining cards. The probability of drawing a king is:
Solution
Correct option: B
New deck size
52 − 1 = 51.
Remaining kings
4 − 1 = 3.
Probability
P(king)=351=117
A24
1 Mark
Level 3 — Advanced
NCERT Exemplar Type
Cards
All four aces are removed from a standard deck. The probability of drawing a face card from the remaining deck is:
Solution
Correct option: A
New total
52 − 4 = 48 cards.
Face cards
No face card was removed because aces are not face cards; 12 remain.
Probability
P(face)=1248=14
A25
1 Mark
Level 3 — Advanced
Competency Based
Number Selection
An integer is selected at random from 10 to 30 inclusive. The probability that it is a perfect square is:
Solution
Correct option: B
Total integers
30 − 10 + 1 = 21
Perfect squares
16 and 25 → 2 favourable integers.
Probability
221
A26
1 Mark
Level 3 — Advanced
Competency Based
Number Selection
An integer is selected at random from 1 to 50. The probability that it is divisible by both 2 and 3 is:
Solution
Correct option: A
Interpretation
Divisible by both 2 and 3 means divisible by 6.
Multiples of 6
6, 12, 18, 24, 30, 36, 42, 48 → 8 numbers.
Probability
850=425
A27
1 Mark
Level 4 — HOTS
Foundation
Three Coins
Enrichment / Foundation Three fair coins are tossed. The probability of getting exactly two heads is:
Solution
Correct option: B
Sample space size
2³ = 8.
Exactly two heads
{HHT, HTH, THH}
Probability
38
A28
1 Mark
Level 3 — Advanced
Competency Based
Spinner
An equal-sector spinner is labelled A, A, B, C, C, C. The probability of landing on C is:
Solution
Correct option: C
Equal sectors
There are 6 equally likely sectors.
C sectors
3 of the 6 sectors are labelled C.
Probability
36=12
A29
1 Mark
Level 4 — HOTS
HOTS
Complement
If P(E) = 2P(not E), then P(E) equals:
Solution
Correct option: C
Let
P(not E) = x, so P(E)=2x.
Complement sum
2x + x = 1 ⇒ 3x = 1 ⇒ x = 1/3
Therefore
P(E)=2×13=23
A30
1 Mark
Level 5 — Olympiad Challenge
Olympiad
Two Dice
Two fair dice are thrown. The probability that the sum is a prime number is:
Solution
Correct option: A
Prime sums
Possible prime sums are 2, 3, 5, 7, 11.
Ways
1 + 2 + 4 + 6 + 2 = 15 favourable ordered pairs.
Probability
1536=512
Section B — 15 Two-Mark Questions 15 × 2 marks
B1
2 Marks
Level 1 — Foundation
NCERT Type
Coin
A fair coin is tossed once. Find the probability of getting a tail.
Solution
Sample space
S = {H, T}, n(S)=2
Event
E = {T}, n(E)=1
Probability
P(T)=12
Answer: 1/2
B2
2 Marks
Level 1 — Foundation
NCERT Type
Die
A fair die is rolled once. Find the probability of obtaining a factor of 6.
Solution
Sample space
S={1,2,3,4,5,6}, n(S)=6
Factors of 6
E={1,2,3,6}, n(E)=4
Probability
P(E)=46=23
B3
2 Marks
Level 1 — Foundation
NCERT Type
Die
A fair die is rolled. Find the probability of getting a number that is not composite.
Solution
Composite faces
{4,6}
Not composite
E={1,2,3,5}, n(E)=4
Probability
P(E)=46=23
Remember: 1 is neither prime nor composite.
B4
2 Marks
Level 1 — Foundation
CBSE/PYQ Pattern
Complement
If the probability of an event E is 712, find the probability that E does not occur.
Solution
Complement rule
P(not E)=1−P(E)
Calculation
P(not E)=1−712=512
B5
2 Marks
Level 1 — Foundation
NCERT Type
Cards
One card is drawn from a well-shuffled standard deck. Find the probability that the card is an ace.
Solution
Total cards
n(S)=52.
Aces
There are 4 aces.
Probability
P(ace)=452=113
B6
2 Marks
Level 2 — Standard
NCERT Exemplar Type
Cards
One card is drawn from a standard deck. Find the probability that it is not a face card.
Solution
Face cards
12 face cards.
Not face cards
52−12=40 cards.
Probability
P(not face)=4052=1013
B7
2 Marks
Level 2 — Standard
NCERT Type
Number Selection
An integer is chosen at random from 1 to 25. Find the probability that it is a perfect square.
Solution
Total outcomes
n(S)=25.
Perfect squares
{1,4,9,16,25}; n(E)=5
Probability
P(E)=525=15
B8
2 Marks
Level 2 — Standard
Competency Based
Letters
One letter-position is selected at random from the word PROBABILITY. Find the probability that the selected letter is a vowel.
Solution
Count positions
PROBABILITY has 11 letter-positions.
Vowels
O, A, I, I → 4 vowel positions.
Probability
P(vowel)=411
B9
2 Marks
Level 2 — Standard
Competency Based
Spinner
A spinner has 10 equal sectors numbered 1 to 10. Find the probability of landing on an even number.
Solution
Total sectors
n(S)=10.
Even sectors
{2,4,6,8,10}; n(E)=5
Probability
P(even)=510=12
B10
2 Marks
Level 2 — Standard
NCERT Type
Bag of Balls
A bag contains 6 red, 4 blue and 5 green balls. One ball is drawn at random. Find the probability that it is green.
Solution
Total balls
6+4+5=15.
Green balls
5.
Probability
P(green)=515=13
B11
2 Marks
Level 2 — Standard
CBSE/PYQ Pattern
Number Selection
An integer is selected at random from 20 to 40 inclusive. Find the probability that it is a multiple of 5.
Solution
Total integers
40−20+1=21
Multiples of 5
{20,25,30,35,40}; n(E)=5
Probability
P(E)=521
B12
2 Marks
Level 2 — Standard
NCERT Exemplar Type
Cards
A card is drawn at random from a standard deck. Find the probability that it is a number card from 2 through 10.
Solution
Number cards per suit
2 through 10 gives 9 cards per suit.
Across four suits
9×4=36 number cards.
Probability
P(number card)=3652=913
B13
2 Marks
Level 2 — Standard
NCERT Type
Die
A fair die is rolled once. Find the probability of getting a number greater than 4.
Solution
Sample space
S={1,2,3,4,5,6}
Favourable outcomes
E={5,6}, n(E)=2
Probability
P(E)=26=13
B14
2 Marks
Level 2 — Standard
Competency Based
Spinner
An equal-sector spinner has 8 sectors, of which 3 are blue. Find the probability that the pointer does not stop on blue.
Solution
Blue probability
P(blue)=38
Complement
P(not blue)=1−38=58
B15
2 Marks
Level 2 — Standard
Conceptual Enrichment
Experimental Probability
Conceptual Enrichment A game has only two outcomes: win or lose. If the theoretical probability of winning is 0.2, find the probability of losing and explain whether this means a player must lose exactly 8 times in every 10 plays.
Solution
Complement
P(lose)=1−0.2=0.8
Interpretation
No. Probability describes likelihood over repeated trials; it does not guarantee an exact short-run pattern in every group of 10 plays.
Section C — 15 Three-Mark Questions 15 × 3 marks
C1
3 Marks
Level 2 — Standard
NCERT Type
Two Coins
Two fair coins are tossed simultaneously. Find the probability of getting at most one head.
Solution
Sample space
S={HH,HT,TH,TT}, n(S)=4
At most one head
Zero or one head: E={TT,HT,TH}, so n(E)=3.
Probability
P(E)=34
C2
3 Marks
Level 2 — Standard
NCERT Exemplar Type
Two Coins
Two fair coins are tossed. Find the probability that both coins show the same face.
Solution
Sample space
S={HH,HT,TH,TT}
Same face
E={HH,TT}, n(E)=2
Probability
P(E)=24=12
C3
3 Marks
Level 2 — Standard
CBSE/PYQ Pattern
Two Dice
Two fair dice are thrown. Find the probability that the sum of the numbers obtained is 9.
Solution
Total outcomes
n(S)=6×6=36
Favourable ordered pairs
(3,6),(4,5),(5,4),(6,3)
Count
n(E)=4.
Probability
P(sum 9)=436=19
Why this step? (3,6) and (6,3) are distinct because the first and second dice are distinguishable by order.
C4
3 Marks
Level 3 — Advanced
NCERT Exemplar Type
Two Dice
Two fair dice are thrown. Find the probability that the absolute difference between the two numbers is 2.
Solution
Total outcomes
n(S)=36
Favourable pairs
(1,3),(2,4),(3,1),(3,5),(4,2),(4,6),(5,3),(6,4)
Count
n(E)=8.
Probability
P(|difference|=2)=836=29
C5
3 Marks
Level 2 — Standard
NCERT Type
Cards
A card is drawn from a standard deck. Find the probability that it is a black face card.
Solution
Black suits
Clubs and spades.
Face cards per black suit
J, Q, K → 3 each; total black face cards = 6.
Probability
P(black face)=652=326
C6
3 Marks
Level 3 — Advanced
Competency Based
Cards
One queen is removed from a standard deck. From the remaining cards, find the probability of drawing a card that is not a queen.
Solution
New total
52−1=51.
Remaining queens
4−1=3.
Not queens
51−3=48.
Probability
P(not queen)=4851=1617
C7
3 Marks
Level 2 — Standard
CBSE/PYQ Pattern
Number Selection
An integer is chosen at random from 1 to 30. Find the probability that it is prime.
Solution
Total outcomes
n(S)=30.
Prime numbers
{2,3,5,7,11,13,17,19,23,29}
Count
n(E)=10.
Probability
P(prime)=1030=13
C8
3 Marks
Level 3 — Advanced
Competency Based
Number Selection
An integer is chosen at random from 1 to 40. Find the probability that it is a perfect square or a perfect cube.
Solution
Perfect squares
{1,4,9,16,25,36}
Perfect cubes
{1,8,27}
Avoid double-counting
1 belongs to both lists, so the union is {1,4,8,9,16,25,27,36}: 8 numbers.
Probability
P(E)=840=15
C9
3 Marks
Level 2 — Standard
CBSE/PYQ Pattern
Two Coins
Two fair coins are tossed. Using the complement method, find the probability of getting at least one head.
Solution
Complement event
The complement of “at least one head” is “no head”, i.e. TT.
Probability of no head
P(TT)=14
Required probability
P(at least one H)=1−14=34
C10
3 Marks
Level 2 — Standard
Competency Based
Table
A box contains balls as shown: Red 5, Blue 7, Green 3. One ball is selected at random. Find the probability that it is blue or green.
| Colour | Number |
|---|---|
| Red | 5 |
| Blue | 7 |
| Green | 3 |
Solution
Total balls
5+7+3=15.
Blue or green
7+3=10 favourable balls.
Probability
P(blue or green)=1015=23
C11
3 Marks
Level 3 — Advanced
Competency Based
Missing Frequency
A bag contains x red balls, 6 blue balls and 4 green balls. If the probability of drawing a red ball is 1/2, find x.
Solution
Total balls
x+6+4=x+10.
Use probability
xx+10=12
Solve
2x=x+10 ⇒ x=10
Answer
The bag contains 10 red balls.
C12
3 Marks
Level 3 — Advanced
NCERT Exemplar Type
Two Dice
Two fair dice are thrown. Find the probability that both numbers obtained are prime.
Solution
Prime faces on one die
{2,3,5} → 3 choices
Favourable ordered pairs
3×3=9.
Total outcomes
36.
Probability
P(both prime)=936=14
C13
3 Marks
Level 3 — Advanced
CBSE/PYQ Pattern
Two Dice
Two fair dice are thrown. Find the probability that their sum is greater than 9.
Solution
Possible sums
10, 11, 12.
Number of ways
Sum 10: 3 ways; sum 11: 2 ways; sum 12: 1 way. Total = 6.
Probability
P(sum>9)=636=16
C14
3 Marks
Level 3 — Advanced
Competency Based
Letters
One letter-position is selected at random from the word STATISTICS. Find the probabilities of selecting (i) a vowel and (ii) a consonant.
Solution
Total positions
STATISTICS has 10 letters.
Vowels
A, I, I → 3 vowel positions.
(i) Vowel
P(vowel)=310
(ii) Consonant
P(consonant)=1−310=710
C15
3 Marks
Level 3 — Advanced
Competency Based
Spinner
A spinner has 12 equal sectors numbered 1 to 12. Find the probability that the number obtained is (i) a factor of 12 and (ii) not a multiple of 3.
Solution
Total sectors
12.
(i) Factors of 12
{1,2,3,4,6,12} ⇒ P=612=12
(ii) Multiples of 3
{3,6,9,12} → 4 sectors, so not a multiple of 3 → 8 sectors.
Probability
P(not multiple of 3)=812=23
Section D — 20 Four-Mark Questions 20 × 4 marks
D1
4 Marks
Level 3 — Advanced
CBSE/PYQ Pattern
Two Dice
Two fair dice are thrown. Find the probability that the sum is a prime number.
Solution
Total outcomes
n(S)=6×6=36
Prime sums
2, 3, 5, 7 and 11.
Number of ways
1 + 2 + 4 + 6 + 2 = 15.
Probability
P(prime sum)=1536=512
D2
4 Marks
Level 3 — Advanced
NCERT Exemplar Type
Two Dice
Two fair dice are thrown. Find the probability that their product is even.
Solution
Use complement
Product is not even only when both numbers are odd.
Odd faces
{1,3,5}: 3 choices on each die, so 3×3=9 odd-product outcomes.
Odd product probability
936=14
Required probability
P(even product)=1−14=34
D3
4 Marks
Level 3 — Advanced
Competency Based
Two Dice
Two fair dice are rolled. Find the probability that at least one die shows 6.
Solution
Total outcomes
36.
Complement
If neither die shows 6, each die has 5 possible faces.
No 6
P(no 6)=2536
At least one 6
1−2536=1136
Exam Tip: “At least one” is often fastest by complement.
D4
4 Marks
Level 3 — Advanced
Competency Based
Two Dice
Two fair dice are thrown. Find the probability that neither die shows 1 or 6.
Solution
Allowed faces on each die
{2,3,4,5} → 4 choices
Favourable ordered pairs
4×4=16.
Total outcomes
36.
Probability
P(E)=1636=49
D5
4 Marks
Level 3 — Advanced
CBSE/PYQ Pattern
Cards
One card is drawn from a standard deck. Find the probability that it is either a red card or a face card.
Solution
Red cards
26.
Black face cards
Among the 12 face cards, 6 are black. Count these in addition to all red cards to avoid double-counting red face cards.
Favourable cards
26+6=32.
Probability
P(red or face)=3252=813
D6
4 Marks
Level 3 — Advanced
Competency Based
Cards
All four aces are removed from a standard deck. One card is drawn from the remaining cards. Find the probability that it is a black face card.
Solution
New total
52−4=48.
Black face cards
J, Q, K of clubs and spades = 6. No black face card was removed.
Probability
P(black face)=648=18
D7
4 Marks
Level 3 — Advanced
Competency Based
Cards
The king of hearts and queen of spades are removed from a standard deck. One card is drawn from the remaining cards. Find (i) the probability of drawing a face card and (ii) the probability of drawing a red card.
Solution
New total
52−2=50.
(i) Face cards
Two face cards were removed, so 12−2=10 remain.
Probability
P(face)=1050=15
(ii) Red cards
Only the king of hearts is red, so 26−1=25 red cards remain.
Probability
P(red)=2550=12
D8
4 Marks
Level 3 — Advanced
Competency Based
Number Selection
An integer is selected at random from 1 to 60. Find the probability that it is a multiple of 4 or a multiple of 6.
Solution
Multiples of 4
⌊60/4⌋ = 15.
Multiples of 6
⌊60/6⌋ = 10.
Overlap
Numbers divisible by both are multiples of 12: 5 of them.
Favourable count
15+10−5=20.
Probability
P(E)=2060=13
D9
4 Marks
Level 4 — HOTS
HOTS
Number Selection
An integer is selected at random from 1 to 100. Find the probability that it is a perfect square or a perfect cube.
Solution
Perfect squares
1² to 10² → 10 numbers.
Perfect cubes
1³ to 4³ → 4 numbers.
Overlap
A number that is both a square and a cube is a sixth power. Up to 100 these are 1 and 64 → 2 overlaps.
Favourable count
10+4−2=12.
Probability
P(E)=12100=325
D10
4 Marks
Level 3 — Advanced
Competency Based
Spinner
A spinner has 12 equal sectors numbered 1 to 12. Find the probability of obtaining a prime number or a multiple of 4.
Solution
Prime numbers
{2,3,5,7,11} → 5 sectors
Multiples of 4
{4,8,12} → 3 sectors
Overlap
There is no number in 1–12 that is both prime and a multiple of 4.
Probability
P(E)=812=23
D11
4 Marks
Level 3 — Advanced
Competency Based
Letters
One letter-position is selected at random from the word MISSISSIPPI. Find (i) the probability of selecting a vowel and (ii) the probability of selecting a letter that occurs exactly twice in the word.
Solution
Total positions
MISSISSIPPI has 11 letter-positions.
(i) Vowels
Only I is a vowel here and it occurs 4 times.
Probability
P(vowel)=411
(ii) Letter occurring exactly twice
P occurs exactly twice; selecting P means 2 favourable positions.
Probability
P(letter occurring exactly twice)=211
D12
4 Marks
Level 3 — Advanced
Conceptual Enrichment
Experimental Probability
Conceptual Enrichment A coin is tossed 100 times and heads occurs 56 times. Find the experimental probability of heads, compare it with the theoretical probability for a fair coin, and explain the difference.
Solution
Experimental probability
P_exp(H)=56100=0.56
Theoretical probability
P_theory(H)=12=0.50
Difference
0.56−0.50=0.06
Interpretation
A finite experiment need not match the theoretical value exactly. With many trials, the experimental proportion often tends to move closer to the theoretical probability.
D13
4 Marks
Level 3 — Advanced
Competency Based
Grid
A point is chosen by first selecting one of the 20 equal squares in the grid at random. Seven squares are shaded. Find (i) the probability of selecting a shaded square and (ii) the probability of selecting an unshaded square.
1234567891011121314151617181920
Solution
Equal elementary regions
20 equal squares are equally likely.
(i) Shaded
7 shaded squares.
Probability
P(shaded)=720
(ii) Unshaded
20−7=13 squares.
Probability
P(unshaded)=1320
D14
4 Marks
Level 4 — HOTS
Error Analysis
Two Dice
A student claims that the probability of obtaining a sum of 8 when two dice are thrown is 1/11 because the possible sums are 2 through 12. Identify the error and calculate the correct probability.
Solution
Error
The 11 possible sums are not equally likely. The equally likely elementary outcomes are the 36 ordered pairs.
Sum 8 outcomes
(2,6),(3,5),(4,4),(5,3),(6,2)
Count
n(E)=5.
Correct probability
P(sum 8)=536
D15
4 Marks
Level 4 — HOTS
HOTS
Two Dice
Two fair dice are thrown. Find the probability that their sum is divisible by 4.
Solution
Possible sums divisible by 4
4, 8, 12.
Ways for sum 4
(1,3),(2,2),(3,1) → 3 ways.
Ways for sum 8
5 ways; for sum 12: 1 way.
Total favourable
3+5+1=9.
Probability
P(E)=936=14
D16
4 Marks
Level 4 — HOTS
HOTS
Two Dice
Two fair dice are thrown. Find the probability that the product of the two numbers is a multiple of 3.
Solution
Complement idea
The product is not a multiple of 3 only if neither die shows a multiple of 3.
Faces not divisible by 3
{1,2,4,5} → 4 choices per die
Complement probability
P(no factor divisible by 3)=1636
Required probability
1−1636=2036=59
D17
4 Marks
Level 4 — HOTS
HOTS
Two Dice
Two fair dice are thrown. Find the probability that exactly one of the two numbers obtained is prime.
Solution
Prime faces
{2,3,5} → 3
Non-prime faces
{1,4,6} → 3
Exactly one prime
Prime on first and non-prime on second: 3×3=9. Non-prime on first and prime on second: another 9.
Favourable outcomes
18.
Probability
1836=12
D18
4 Marks
Level 4 — HOTS
Foundation
Three Coins
Enrichment / Foundation Three fair coins are tossed. Find the probability of getting at least two heads.
Solution
Sample space
S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
At least two heads
E={HHH,HHT,HTH,THH}
Count
n(S)=8, n(E)=4.
Probability
P(E)=48=12
D19
4 Marks
Level 5 — Olympiad Challenge
Olympiad
Number Selection
One numbered token is selected at random from 1 to 20. Find the probability that its number is either a factor of 24 or a prime number.
Solution
Factors of 24 within 1–20
{1,2,3,4,6,8,12} → 7
Primes within 1–20
{2,3,5,7,11,13,17,19} → 8
Overlap
2 and 3 are in both sets, so count them once.
Favourable count
7+8−2=13.
Probability
P(E)=1320
D20
4 Marks
Level 5 — Olympiad Challenge
Olympiad
Combined Experiment
Enrichment / Foundation A fair coin is tossed and an integer from 1 to 9 is selected at random. Find the probability that either (i) the coin shows heads and the number is prime, or (ii) the coin shows tails and the number is even.
Solution
Total outcomes
2 coin outcomes × 9 number outcomes = 18 equally likely combined outcomes.
Heads and prime
Primes in 1–9 are {2,3,5,7}: 4 outcomes.
Tails and even
Evens in 1–9 are {2,4,6,8}: 4 outcomes.
Disjoint cases
A single outcome cannot have both H and T, so total favourable = 4+4=8.
Probability
P(E)=818=49
Section E — 10 Five-Mark Questions 10 × 5 marks
E1
5 Marks
Level 4 — HOTS
HOTS
Two Dice
Two fair dice are thrown. Find the probability that the sum is at least 10 or that the two dice show equal numbers.
Solution
Total outcomes
n(S)=36
Sum at least 10
Sums 10, 11, 12 occur in 3+2+1=6 ways.
Equal numbers
Doubles: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) → 6 ways.
Overlap
(5,5) and (6,6) are already counted among sums at least 10 → 2 overlapping outcomes.
Favourable count
6+6−2=10.
Probability
P(E)=1036=518
E2
5 Marks
Level 4 — HOTS
Competency Based
Combined Experiment
Enrichment / Foundation A fair coin is tossed and a fair die is rolled. Find the probability of getting either (i) a head with a prime number on the die or (ii) a tail with a number greater than 4.
Solution
Combined sample space
There are 2×6=12 equally likely outcomes: H1,…,H6,T1,…,T6.
Head with prime
H2, H3, H5 → 3 outcomes.
Tail with number >4
T5, T6 → 2 outcomes.
Total favourable
3+2=5; the two cases are disjoint.
Probability
P(E)=512
E3
5 Marks
Level 3 — Advanced
CBSE/PYQ Pattern
Cards
One card is drawn from a standard deck. Find the probability that it is either a red face card or an ace.
Solution
Red face cards
J, Q, K in hearts and diamonds → 3×2=6.
Aces
4 aces.
Overlap
An ace is not a face card, so there is no overlap.
Favourable cards
6+4=10.
Probability
P(E)=1052=526
E4
5 Marks
Level 4 — HOTS
Competency Based
Missing Card
One card is lost from a standard deck. From the remaining 51 cards, the probability of drawing a king is 1/17. Determine whether the lost card was a king. Then find the probability of drawing a non-king from the remaining deck.
Solution
Use the given probability
remaining kings51=117
Remaining kings
remaining kings = 51×117 = 3
Inference
A full deck has 4 kings. Since only 3 remain, the lost card must have been a king.
Non-kings remaining
51−3=48.
Probability
P(non-king)=4851=1617
E5
5 Marks
Level 4 — HOTS
HOTS
Number Selection
An integer is selected at random from 1 to 120. Find the probability that it is either a multiple of 6 or a perfect square.
Solution
Multiples of 6
⌊120/6⌋=20.
Perfect squares
1² through 10² → 10 squares.
Overlap
A square divisible by 6 must have its square root divisible by both 2 and 3. Among 1 to 10, only 6 qualifies, giving 36. Hence overlap = 1.
Favourable count
20+10−1=29.
Probability
P(E)=29120
E6
5 Marks
Level 4 — HOTS
Competency Based
Probability Table
A box contains four types of tokens A, B, C and D. Their frequencies are x, 12, 8 and 10 respectively. If P(A)=2/5, find (i) x, (ii) P(B or D), and (iii) P(not C).
| Token | A | B | C | D |
|---|---|---|---|---|
| Frequency | x | 12 | 8 | 10 |
Solution
Total tokens
x+12+8+10=x+30.
Use P(A)
xx+30=25
Solve
5x=2x+60 ⇒ 3x=60 ⇒ x=20
New total
20+30=50.
(ii) B or D
P(B or D)=12+1050=2250=1125
(iii) Not C
P(not C)=50−850=4250=2125
E7
5 Marks
Level 4 — HOTS
Competency Based
Spinner
A spinner has 12 equal sectors numbered 1 to 12. Find (i) the probability of obtaining a number that is prime or a factor of 12 and (ii) the probability of obtaining neither.
Solution
Prime numbers
{2,3,5,7,11}
Factors of 12
{1,2,3,4,6,12}
Union
{1,2,3,4,5,6,7,11,12} → 9 distinct sectors.
(i) Probability
P(prime or factor)=912=34
(ii) Neither
1−34=14
E8
5 Marks
Level 5 — Olympiad Challenge
Foundation
Three Coins + Die
Enrichment / Foundation Three fair coins are tossed and a fair die is rolled. Find the probability of getting exactly two heads and an even number on the die.
Solution
Coin outcomes
Three fair coins have 2³=8 equally likely outcomes.
Exactly two heads
{HHT,HTH,THH} → 3 outcomes
Die outcomes
A die has 6 outcomes; even faces are {2,4,6} → 3 outcomes.
Combined sample space
8×6=48 outcomes.
Favourable combined outcomes
3×3=9.
Probability
P(E)=948=316
E9
5 Marks
Level 5 — Olympiad Challenge
Olympiad
Two Dice
Two fair dice are thrown. Find the probability that the sum is odd or the product is a multiple of 5.
Solution
Sum odd
One die must be odd and the other even: 3×3+3×3=18 outcomes.
Product multiple of 5
At least one die must show 5. Outcomes with a 5 on first die =6, on second die =6, subtract (5,5) counted twice → 11.
Overlap
For sum odd with a 5 present, the other die must be even: (5,2),(5,4),(5,6),(2,5),(4,5),(6,5) → 6.
Favourable union
18+11−6=23.
Probability
P(E)=2336
E10
5 Marks
Level 5 — Olympiad Challenge
Olympiad
Number Selection
An integer is selected at random from 1 to 90. Find the probability that it is divisible by exactly one of 2 and 3.
Solution
Multiples of 2
⌊90/2⌋=45.
Multiples of 3
⌊90/3⌋=30.
Multiples of both
Multiples of 6: ⌊90/6⌋=15.
Exactly one condition
Divisible by 2 but not 3: 45−15=30. Divisible by 3 but not 2: 30−15=15.
Favourable count
30+15=45.
Probability
P(E)=4590=12
Section F — 5 Case Studies 4 sub-questions each
F1
Case Study
Level 3 → 5
Competency Based
Spinner
School Lucky-Draw Spinner
At the school mathematics fair, a spinner is divided into 8 equal sectors numbered 1 to 8. Every sector is equally likely.
- Write the sample space and state its size.
- Find the probability that the pointer stops on a prime number.
- Find the probability that the pointer does not stop on a multiple of 3.
- A prize is awarded when the number is even or prime. Find the probability of winning the prize.
Visual representation for this case study. The numerical information is also stated in text, so the problem remains fully accessible without Canvas.
Solution
(a) Sample space
S={1,2,3,4,5,6,7,8}, n(S)=8
(b) Prime number
Prime sectors={2,3,5,7} ⇒ P=48=12
(c) Not a multiple of 3
Multiples of 3 are {3,6}. Therefore 6 sectors are not multiples of 3.
P(not multiple of 3)=68=34
(d) Even or prime
Even={2,4,6,8}; Prime={2,3,5,7}. The union is {2,3,4,5,6,7,8}: 7 sectors.
P(win)=78
F2
Case Study
Level 3 → 5
Competency Based
Two Dice
Board-Game Dice Challenge
In a board game, a player throws two fair standard dice. The result is recorded as an ordered pair (first die, second die).
- How many equally likely ordered outcomes are possible?
- Find the probability that the sum is 7.
- Find the probability that at least one die shows 6.
- Find the probability that the sum is 8 or the two dice show equal numbers.
Visual representation for this case study. The numerical information is also stated in text, so the problem remains fully accessible without Canvas.
Solution
(a) Sample-space size
n(S)=6×6=36
(b) Sum 7
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1): 6 ways.
P(sum 7)=636=16
(c) At least one 6
Use complement: no 6 gives 5×5=25 outcomes.
P(at least one 6)=1−2536=1136
(d) Sum 8 or doubles
Sum 8 has 5 outcomes. Doubles have 6 outcomes. (4,4) lies in both sets, so favourable count =5+6−1=10.
P=1036=518
F3
Case Study
Level 3 → 5
Competency Based
Cards
Card-Game Probability
A standard deck of 52 cards is well shuffled. Recall that each suit has 13 cards and the face cards are J, Q and K.
- How many red cards and how many black cards are in the deck?
- Find the probability of drawing a black card.
- Find the probability of drawing a card that is not a face card.
- Suppose one ace is removed before the draw. Find the probability of drawing an ace from the remaining deck.
Visual representation for this case study. The numerical information is also stated in text, so the problem remains fully accessible without Canvas.
Solution
(a) Colour counts
Red: 26; Black: 26.
(b) Black card
P(black)=2652=12
(c) Not a face card
There are 12 face cards, so 52−12=40 non-face cards.
P(not face)=4052=1013
(d) One ace removed
New total=51; remaining aces=3.
P(ace)=351=117
F4
Case Study
Level 3 → 5
Competency Based
Number Selection
Numbered Tokens
A bag contains 30 identical tokens numbered 1 to 30. One token is drawn at random.
- State the number of equally likely outcomes.
- Find the probability that the number is a multiple of 5.
- Find the probability that the number is prime.
- Find the probability that the number is prime or a multiple of 5.
Solution
(a) Total outcomes
n(S)=30.
(b) Multiples of 5
{5,10,15,20,25,30}: 6 numbers.
P=630=15
(c) Primes
{2,3,5,7,11,13,17,19,23,29}: 10 numbers.
P=1030=13
(d) Prime or multiple of 5
The two sets overlap at 5 only. Favourable count=10+6−1=15.
P=1530=12
F5
Case Study
Level 3 → 5
Competency Based
Bag/Table
Quality-Control Colour Sampling
A quality-control tray contains 20 identical markers: 6 red, 5 blue, 4 green and 5 yellow. One marker is selected at random.
- How many possible marker selections are there by position?
- Find the probability of selecting a blue marker.
- Find the probability of not selecting a green marker.
- A marker earns a ‘common-colour’ label if its colour occurs at least 5 times in the tray. Find the probability that a randomly selected marker has a common colour.
Solution
(a) Total markers
20 equally likely physical markers.
(b) Blue
P(blue)=520=14
(c) Not green
Green markers=4; not green=16.
P(not green)=1620=45
(d) Common colours
Red (6), blue (5), yellow (5) qualify. Favourable markers=6+5+5=16.
P(common colour)=1620=45
Exam Protection
5. 12 Common Mistakes in Probability
1. Outcome vs sample space
An outcome is one result; the sample space contains all possible outcomes.
2. Wrong favourable count
Translate the event condition before counting.
3. Forgetting the total outcomes
Always establish n(S) before using n(E)/n(S).
4. Probability greater than 1
Any answer outside 0≤P(E)≤1 is impossible.
5. Treating HT and TH as the same
For sequential tosses they are distinct ordered outcomes.
6. Treating dice sums as equally likely
The 36 ordered pairs are equally likely; sums have different frequencies.
7. Counting 1 as prime
1 is neither prime nor composite.
8. Counting ace as a face card
Face cards are J, Q and K.
9. Forgetting removed cards
Update both the deck total and favourable-card count.
10. Misreading “at least” and “at most”
At least one means one or more; at most one means zero or one.
11. Ignoring repeated letters
If a letter-position is chosen, every written position counts separately.
12. Assuming unequal outcomes are equally likely
Use n(E)/n(S) only when the elementary outcomes justify equal likelihood.
How to Score Full Marks in Probability
- State the sample space or total number of equally likely outcomes.
- Identify favourable outcomes clearly.
- Write the probability formula before substitution when method marks matter.
- For two dice, remember ordered pairs.
- Use complement when it shortens “not”, “neither” or “at least one” problems.
- Use correct deck facts: 52 cards, 12 face cards, 4 aces.
- Update totals when cards or objects have been removed.
- Simplify the fraction.
- Check 0≤P(E)≤1.
- Box or clearly state the final probability.
Self-Check
6. Can You Solve These Without Looking at the Probability Sheet?
Five additional ungraded challenge questions. These are separate from the 95 main entries.
Challenge 1
Ungraded
Level 5 — Olympiad Challenge
Self-Check
Two Dice
Two fair dice are thrown. Find the probability that their sum is divisible by 3.
Solution
Divisible-by-3 sums
3, 6, 9 and 12.
Number of ways
Sum 3 → 2 ways; sum 6 → 5 ways; sum 9 → 4 ways; sum 12 → 1 way.
Total favourable
2+5+4+1=12.
Probability
1236=13
Challenge 2
Ungraded
Level 5 — Olympiad Challenge
Self-Check
Cards
All red aces are removed from a standard deck. From the remaining cards, find the probability of drawing a black card that is not a face card.
Solution
New total
52−2=50.
Black cards
No black card was removed, so 26 black cards remain.
Black face cards
6.
Black non-face cards
26−6=20.
Probability
2050=25
Challenge 3
Ungraded
Level 5 — Olympiad Challenge
Self-Check
Number Selection
An integer is selected at random from 1 to 100. Find the probability that it is divisible by neither 2 nor 5.
Solution
Multiples of 2
50.
Multiples of 5
20.
Overlap
Multiples of 10 = 10.
Divisible by 2 or 5
50+20−10=60.
Neither
100−60=40.
Probability
40100=25
Challenge 4
Ungraded
Level 5 — Olympiad Challenge
Self-Check
Combined Experiment
Enrichment / Foundation A fair coin is tossed and an equal-sector spinner numbered 1 to 6 is spun. Find the probability of getting a head with an even number or a tail with a prime number.
Solution
Total outcomes
2×6=12.
H and even
H2,H4,H6 → 3 outcomes.
T and prime
T2,T3,T5 → 3 outcomes.
Probability
612=12
Challenge 5
Ungraded
Level 5 — Olympiad Challenge
Self-Check
Equal-Area Grid
Geometric Probability — Enrichment A 5×5 board has 25 equal squares. All squares on either diagonal are shaded. If one square is chosen at random, find the probability that it is shaded.
Solution
Main diagonal
5 squares.
Other diagonal
5 squares.
Overlap
The centre square lies on both diagonals, so subtract it once.
Shaded squares
5+5−1=9.
Probability
925
Last-Minute Review
7. 60-Second Probability Revision
Random experiment: exact outcome uncertain beforehand.
Outcome: one possible result.
Sample space S: all possible outcomes.
Event E: outcomes satisfying a condition.
Formula: P(E)=n(E)/n(S) for equally likely outcomes.
Complement: P(not E)=1−P(E).
Range: 0≤P(E)≤1.
Impossible / certain: probabilities 0 / 1.
Two coins: HH, HT, TH, TT.
Two dice: 36 ordered pairs.
Deck: 52 total, 26 red, 26 black, 12 face, 4 aces.
Words: distinguish exactly, at least, at most and neither.
Prime trap: 1 is not prime.
Card trap: ace is not a face card.
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