CBSE Class 10 Mathematics Probability Notes

What You Get in These Notes

✅  Plain-English explanations of every concept — no jargon without a definition

✅  All formulas derived from scratch so you understand them, not just memorise them

✅  Worked examples for every concept, matching actual CBSE board exam question styles

✅  Common mistakes section — the exact errors that cost students marks every year

✅  Complete formula summary table for last-minute revision before the exam

✅  Practice questions sorted by 1-mark, 3-mark, and 5-mark categories

✅  Solved examples covering coins, dice, cards, and coloured balls — all CBSE favourites

✅  Printable PDF version available — same content, clean layout, ready to annotate

Key Topics in This Chapter

• Basic terminology: experiment, outcome, event, sample space

• Theoretical probability — definition and formula

• Complementary events and their probability

• Impossible events (P = 0) and certain events (P = 1)

• Probability problems on coins, dice, playing cards, and coloured balls

• Experimental vs Theoretical probability — comparison and connection

Term

Definition

Example

Random Experiment

An experiment whose outcome cannot be predicted with certainty in advance

Tossing a coin, rolling a die

Outcome

A single possible result of a random experiment

Getting Heads when tossing a coin

Sample Space (S)

The set of ALL possible outcomes of an experiment

S = {H, T} for a coin toss

Event

A specific outcome or a set of outcomes from the sample space

Getting an even number on a die

Favourable Outcomes

Outcomes in the sample space that satisfy the event

Even numbers: {2, 4, 6}

Equally Likely Outcomes

Outcomes that have the same chance of occurring

All faces of a fair die

Probability of an Event

A number between 0 and 1 measuring likelihood

P(Head) = 1/2

Complementary Event

Event that occurs when the original event does NOT occur

Not getting a Head = getting a Tail

Experiment

Sample Space

Total Outcomes

Tossing 1 coin

{H, T}

2

Tossing 2 coins

{HH, HT, TH, TT}

4

Tossing 3 coins

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

8

Rolling 1 die

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

6

Rolling 2 dice

{(1,1),(1,2),...,(6,6)} — ordered pairs

36

Drawing 1 card from a deck

All 52 cards

52

Drawing from coloured balls

All balls (each treated as distinct)

Total no. of balls

Important: Always Count, Never Assume

Never assume the sample space without writing it out first — especially for two coins or two dice.

Two coins: many students write {HH, HT, TT} — WRONG. The correct space is {HH, HT, TH, TT}.

HT and TH are different outcomes because the coins are distinct (coin 1 and coin 2).

Two dice: sample space has 6 × 6 = 36 ordered pairs. Always use ordered pairs (a, b).

P(E)  =  Number of outcomes favourable to event E

         ─────────────────────────────────────────

         Total number of equally likely outcomes

P(E)  =  n(E)  /  n(S)

where  n(E) = number of favourable outcomes

       n(S) = total number of outcomes in sample space S

0  ≤  P(E)  ≤  1

P(E) = 0  →  Impossible event  (can never happen)

P(E) = 1  →  Certain event     (will always happen)

0 < P(E) < 1  →  Event may or may not happen

P(E) Value

Type of Event

Real Example

P = 0

Impossible event

Rolling a 7 on a standard die

P = 1/6

Unlikely event

Rolling a specific number (e.g., 4) on a die

P = 1/2

Equally likely

Getting Heads on a fair coin

P = 5/6

Likely event

NOT rolling a specific number on a die

P = 1

Certain event

Getting a number ≤ 6 on a standard die

Worked Example 3.1 — Basic Probability (Die)

Problem: A fair die is rolled once. Find the probability of:

  (a) Getting a number greater than 4

  (b) Getting an even number

  (c) Getting a number divisible by 3

Sample Space S = {1, 2, 3, 4, 5, 6}  →  n(S) = 6

(a) Numbers greater than 4: {5, 6}  →  n(E) = 2

    P(E) = 2/6 = 1/3

(b) Even numbers: {2, 4, 6}  →  n(E) = 3

    P(E) = 3/6 = 1/2

(c) Numbers divisible by 3: {3, 6}  →  n(E) = 2

    P(E) = 2/6 = 1/3

Worked Example 3.2 — Basic Probability (Coins)

Problem: Two fair coins are tossed simultaneously. Find the probability of:

  (a) Getting exactly one Head

  (b) Getting at least one Tail

  (c) Getting no Heads

Sample Space S = {HH, HT, TH, TT}  →  n(S) = 4

(a) Exactly one Head: {HT, TH}  →  n(E) = 2  →  P = 2/4 = 1/2

(b) At least one Tail: {HT, TH, TT}  →  n(E) = 3  →  P = 3/4

(c) No Heads (= both Tails): {TT}  →  n(E) = 1  →  P = 1/4

P(E')  =  1  −  P(E)

Equivalently:   P(E)  +  P(E')  =  1

where  E' = complementary event of E  (read as 'E prime' or 'not E')

Also:  P(E) + P(not E) = 1  →  always true for any event E

Worked Example 4.1 — Complementary Events

Problem: A bag contains 5 red, 3 blue, and 2 green balls. One ball is drawn randomly.

Find: (a) P(red)  (b) P(not red)  (c) P(not green)

Total balls = 5 + 3 + 2 = 10

(a) P(red) = 5/10 = 1/2

(b) P(not red) = 1 − P(red) = 1 − 1/2 = 1/2

    Verify: not red = blue or green = (3+2)/10 = 5/10 = 1/2 ✓

(c) P(green) = 2/10 = 1/5

    P(not green) = 1 − 1/5 = 4/5

Worked Example 4.2 — Using Complement to Simplify

Problem: A die is rolled. Find the probability of getting a number that is NOT a perfect square.

It is easier to find: P(perfect square) first.

Perfect squares on a die: {1, 4}  →  n(E) = 2  →  P(perfect square) = 2/6 = 1/3

P(not a perfect square) = 1 − 1/3 = 2/3

When to Use the Complement Method

Use P(E') = 1 − P(E) whenever the words 'at least', 'at most', 'not', or 'none' appear.

Example: 'at least one Head from 3 coins' → find P(no heads) = 1/8 → answer = 1 − 1/8 = 7/8

This avoids listing many favourable outcomes and reduces chances of error.

Category

Details

Count

Total cards

One full deck

52

Suits

Hearts (♥), Diamonds (♦), Clubs (♣), Spades (♠)

4 suits

Cards per suit

Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King

13 each

Red cards

Hearts + Diamonds

26

Black cards

Clubs + Spades

26

Face cards (picture)

Jack, Queen, King in each suit

12 (3×4)

Non-face number cards

Ace, 2 through 10 in each suit

40

Aces

One per suit

4

Kings

One per suit

4

Queens

One per suit

4

Jacks

One per suit

4

Red face cards

J, Q, K of Hearts and Diamonds

6

Black aces

Ace of Clubs + Ace of Spades

2

Key Card Facts to Memorise

Total = 52.  Red = 26.  Black = 26.  Face cards = 12.  Aces = 4.

Ace is NOT a face card — face cards are only Jack, Queen, King.

There is NO Joker in the standard CBSE deck of 52 cards.

Each suit has exactly: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K — that is 13 cards.

Red suits = Hearts and Diamonds.  Black suits = Clubs and Spades.

Worked Example 5.1 — One Card Drawn

Problem: A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of:

  (a) A king   (b) A red card   (c) A face card   (d) The ace of spades

  (e) A card that is NOT a face card

n(S) = 52

(a) Kings: 4 (one per suit)  →  P = 4/52 = 1/13

(b) Red cards: 26            →  P = 26/52 = 1/2

(c) Face cards: 12           →  P = 12/52 = 3/13

(d) Ace of spades: 1         →  P = 1/52

(e) Not a face card: 52−12 = 40  →  P = 40/52 = 10/13

    OR: P(not face) = 1 − 3/13 = 10/13 ✓

Worked Example 5.2 — Cards with Multiple Conditions

Problem: From a deck of 52 cards, find the probability of drawing a card that is:

  (a) A red face card   (b) A black king   (c) A queen or a jack

n(S) = 52

(a) Red face cards = J, Q, K of Hearts + J, Q, K of Diamonds = 6

    P = 6/52 = 3/26

(b) Black kings = King of Clubs + King of Spades = 2

    P = 2/52 = 1/26

(c) Queens (4) + Jacks (4) = 8

    P = 8/52 = 2/13

Worked Example 6.1 — Two Dice

Problem: Two dice are rolled simultaneously. Find the probability of:

  (a) Getting a sum of 7

  (b) Getting a sum of at least 10

  (c) Getting the same number on both dice (doublets)

  (d) Getting a sum less than 4

n(S) = 36

(a) Sum = 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)  →  n = 6  →  P = 6/36 = 1/6

(b) Sum ≥ 10: (4,6),(5,5),(6,4),(5,6),(6,5),(6,6)  →  n = 6  →  P = 6/36 = 1/6

(c) Doublets: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)  →  n = 6  →  P = 6/36 = 1/6

(d) Sum < 4: (1,1),(1,2),(2,1)  →  n = 3  →  P = 3/36 = 1/12

Sum

Favourable Pairs

Count

Probability

2

(1,1)

1

1/36

3

(1,2),(2,1)

2

2/36 = 1/18

4

(1,3),(2,2),(3,1)

3

3/36 = 1/12

5

(1,4),(2,3),(3,2),(4,1)

4

4/36 = 1/9

6

(1,5),(2,4),(3,3),(4,2),(5,1)

5

5/36

7

(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)

6

6/36 = 1/6

8

(2,6),(3,5),(4,4),(5,3),(6,2)

5

5/36

9

(3,6),(4,5),(5,4),(6,3)

4

4/36 = 1/9

10

(4,6),(5,5),(6,4)

3

3/36 = 1/12

11

(5,6),(6,5)

2

2/36 = 1/18

12

(6,6)

1

1/36

Worked Example 6.2 — Three Coins

Problem: Three fair coins are tossed. Find the probability of:

  (a) Exactly 2 Heads   (b) At least 1 Head   (c) At most 1 Tail

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}  →  n(S) = 8

(a) Exactly 2 Heads: {HHT, HTH, THH}  →  n = 3  →  P = 3/8

(b) At least 1 Head: use complement.

    P(no heads) = P(TTT) = 1/8

    P(at least 1 Head) = 1 − 1/8 = 7/8

(c) At most 1 Tail = 0 Tails or 1 Tail: {HHH, HHT, HTH, THH}  →  n = 4  →  P = 4/8 = 1/2

    OR: P(at most 1 Tail) = P(2+ Heads)  →  same as ≥ 2 Heads = 4/8 = 1/2 ✓

Worked Example 7.1 — Balls in a Bag

Problem: A bag contains 3 red, 5 black, and 2 white balls. One ball is drawn at random. Find:

  (a) P(black)   (b) P(not white)   (c) P(red or black)   (d) P(green)

Total balls = 3 + 5 + 2 = 10  →  n(S) = 10

(a) P(black) = 5/10 = 1/2

(b) P(not white) = 1 − P(white) = 1 − 2/10 = 8/10 = 4/5

    OR: red + black = 3 + 5 = 8  →  P = 8/10 = 4/5 ✓

(c) P(red or black) = (3+5)/10 = 8/10 = 4/5

(d) P(green) = 0/10 = 0  [no green balls — impossible event]

Worked Example 7.2 — Number Slips

Problem: Cards numbered 1 to 25 are placed in a box. One card is picked at random. Find:

  (a) P(a prime number)   (b) P(divisible by 5)   (c) P(a perfect square)

  (d) P(a multiple of 3 or 7)

n(S) = 25

(a) Primes from 1–25: {2,3,5,7,11,13,17,19,23}  →  n = 9  →  P = 9/25

(b) Multiples of 5: {5,10,15,20,25}  →  n = 5  →  P = 5/25 = 1/5

(c) Perfect squares: {1,4,9,16,25}  →  n = 5  →  P = 5/25 = 1/5

(d) Multiples of 3: {3,6,9,12,15,18,21,24}  →  8 numbers

    Multiples of 7: {7,14,21}  →  3 numbers

    Common (multiples of 21): {21}  →  1 number

    Total = 8 + 3 − 1 = 10  →  P = 10/25 = 2/5

Worked Example 7.3 — Finding Unknown Quantity

Problem: A bag has 5 red and 8 blue balls. How many red balls should be added so that the

probability of drawing a red ball becomes 5/8?

Let x = number of red balls to be added.

New red = 5 + x,  New total = 13 + x

P(red) = (5 + x) / (13 + x) = 5/8

8(5 + x) = 5(13 + x)

40 + 8x = 65 + 5x

3x = 25

x = 25/3  — not a whole number, so check problem for typos or recalculate.

Note: In exam questions, x will always be a positive integer. Set up equation carefully.

Feature

Experimental (Empirical)

Theoretical (Classical)

Definition

Based on actual observed data from experiments

Based on mathematical reasoning and equally likely outcomes

Formula

P(E) = Number of times E occurred / Total trials

P(E) = n(E) / n(S)

Requires

Conducting an actual experiment

Only logical analysis of outcomes

Result

Can vary with each set of trials

Fixed — always the same for the same experiment

Approaches

Theoretical value as trials → ∞

The fixed 'true' probability

Example

Tossing a coin 100 times, getting 48 Heads → P ≈ 0.48

Theoretical P(Head) = 0.5

Used when

Experiment can be conducted; outcomes not equally likely

Outcomes are equally likely; experiment is hypothetical

The Law of Large Numbers — Key Concept

As the number of trials in an experiment increases, the experimental probability

gets closer and closer to the theoretical probability.

Example: Flip a fair coin 10 times → might get 4 Heads (P = 0.4, not 0.5)

         Flip 10,000 times → will get very close to 5,000 Heads (P ≈ 0.5)

This is why theoretical probability predicts what will happen 'in the long run'.

Example of Impossible Event

Why It Is Impossible

Rolling a 7 on a standard die

A die only has faces 1 to 6

Drawing a green card from a standard deck

No green cards exist in a standard 52-card deck

Getting 3 Heads from 2 coin tosses

Maximum Heads from 2 coins is 2

Choosing a negative number from {1,2,3,4,5}

All numbers are positive

Example of Certain Event

Why It Is Certain

Getting a number ≤ 6 on a die

All outcomes (1,2,3,4,5,6) satisfy this

Drawing a red or black card from a standard deck

All 52 cards are either red or black

Getting H or T when a coin is tossed

There are no other outcomes possible

Choosing an odd or even number from {1,2,3,4,5}

Every number is either odd or even

Frequently Asked 1-Mark Question Type

Q: 'Is probability of an event always between 0 and 1?' → Yes: 0 ≤ P(E) ≤ 1

Q: 'Can P(E) be negative?' → No, never.

Q: 'Can P(E) be greater than 1?' → No, never.

Q: 'P(E) + P(not E) = ?' → Always equals 1.

Q: 'P(E) = 0 means...?' → The event is impossible.

Q: 'P(E) = 1 means...?' → The event is certain.

Mistake

What Goes Wrong

Correct Approach

Wrong sample space for 2 coins

Writing {HH, HT, TT} — misses TH

Full space: {HH, HT, TH, TT} — 4 outcomes

Ace counted as face card

Adding Ace to 12 face cards → 16

Face cards = only Jack, Queen, King = 12

Not simplifying probability

Leaving answer as 6/36 without simplifying

Always reduce fraction to lowest terms: 6/36 = 1/6

P(E) > 1

Dividing wrong way: n(S)/n(E)

Always: P = n(E) / n(S), not the other way

Forgetting ordered pairs for 2 dice

Treating (1,2) and (2,1) as same

They are DIFFERENT outcomes — both must be counted

Wrong complement calculation

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

P(not E) = 1 − P(E)  [subtract, not add]

P(impossible) = -1 or undefined

Trying to calculate instead of stating

P(impossible event) = 0 — state directly

Treating balls of same colour as identical

Not counting all balls in total

Total = ALL balls. Every ball is equally likely.

Concept

Formula / Fact

Notes

Theoretical Probability

P(E) = n(E) / n(S)

n(E) = favourable, n(S) = total

Range of Probability

0 ≤ P(E) ≤ 1

Always between 0 and 1

Impossible Event

P(E) = 0

No favourable outcomes

Certain Event

P(E) = 1

All outcomes are favourable

Complementary Event

P(E') = 1 − P(E)

E' = 'not E'

Sum Rule

P(E) + P(E') = 1

Always true

1 Coin Sample Space

{H, T}  →  n(S) = 2

2 Coins Sample Space

{HH, HT, TH, TT}  →  n(S) = 4

HT ≠ TH

3 Coins Sample Space

n(S) = 8  (2³)

1 Die Sample Space

{1,2,3,4,5,6}  →  n(S) = 6

2 Dice Sample Space

n(S) = 36  (6×6)

Ordered pairs (a,b)

Standard Deck

n(S) = 52

26 red, 26 black

Face Cards

J, Q, K in 4 suits = 12

Ace is NOT a face card

Aces in a deck

4 (one per suit)

Red/Black face cards

6 red, 6 black

Board Exam Strategy for Probability

1. Always write the sample space first — never skip this step even if it seems obvious.

2. For 2 coins: write all 4 outcomes. For 2 dice: remember there are 36 ordered pairs.

3. Face cards = 12 (J, Q, K only). Ace is NOT a face card. Memorise the deck structure.

4. Use the complement method for 'at least', 'at most', or 'not' questions — it saves time.

5. Always simplify fractions completely in your final answer.

6. P(E) + P(E') = 1 is a useful check — always verify your answer with this.

7. Show the formula, substitution, and simplified answer for full step marks.

8. Probability is one of the most scoring chapters — zero careless errors means full marks.