Chapter 15 — Introduction to Probability | Class 9 CBSE Maths 2026-27

Section 15.1

Probability Concepts & Scale

Probability measures how likely an event is to happen. It always lies between 0 (impossible) and 1 (certain).

0Impossible

0.25Unlikely

0.5Even chance

0.75Likely

1Certain

The probability scale — from 0 (impossible) to 1 (certain). 0.5 means equally likely to happen or not happen.

📋 Key Vocabulary

Term	Definition	Example
Experiment	An activity with uncertain outcomes	Rolling a die
Outcome	A single result of the experiment	Getting a 4
Sample Space (S)	Set of ALL possible outcomes	S = {1,2,3,4,5,6}
Event (E)	A subset of the sample space	E = {2,4,6} (even numbers)
Favourable outcomes	Outcomes that satisfy the event	3 outcomes for E above
Random experiment	Each outcome is equally likely	Fair coin, fair die

📐 Probability Formula

P(E) = Number of favourable outcomes / Total number of outcomes

Also written: P(E) = n(E) / n(S) | Always: 0 ≤ P(E) ≤ 1
P(impossible event) = 0 | P(certain event) = 1 | P(E) + P(not E) = 1

Section 15.2

Empirical Probability

Empirical (experimental) probability is based on actual experiments and observations — not theory. It approaches theoretical probability as the number of trials increases.

Formula

Empirical Probability

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

The more trials you do, the closer the empirical probability gets to the true (theoretical) probability. This is the Law of Large Numbers.

Example 1 · 2M

A coin was tossed 200 times. Heads came up 110 times. Find empirical P(Heads).

1

P(Heads) = 110/200 = 11/20 = 0.55

2

Theoretical P(Heads) = 0.5. The empirical value 0.55 is close — with more trials it would get closer to 0.5.

P(Heads) = 11/20 = 0.55

Example 2 · 2M

Die rolled 300 times. Number 6 appeared 45 times. Find empirical P(6). Compare with theoretical.

1

Empirical P(6) = 45/300 = 3/20 = 0.15

2

Theoretical P(6) = 1/6 ≈ 0.167. Empirical (0.15) is close but not exactly equal.

Empirical P(6) = 3/20 = 0.15

Section 15.3 — Most Important

Theoretical Probability

Theoretical probability assumes all outcomes are equally likely. Based on the definition of the experiment, not on conducting it.

🪙 Coin

Sample space: S = {H, T}
n(S) = 2

P(Head) = 1/2
P(Tail) = 1/2

🎲 Die (Fair, 6 faces)

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

P(any number) = 1/6
P(even) = 3/6 = 1/2
P(prime) = {2,3,5} = 3/6 = 1/2

Example 3 · 3M

A card is drawn from a well-shuffled deck of 52 cards. Find P(King), P(Red card), P(Heart), P(Face card).

1

P(King) = 4/52 = 1/13 (4 kings in deck)

2

P(Red card) = 26/52 = 1/2 (26 red cards)

3

P(Heart) = 13/52 = 1/4 (13 hearts)

4

P(Face card) = 12/52 = 3/13 (J,Q,K × 4 suits = 12)

🃏 Standard Deck of 52 Cards — Quick Reference

Suit	Colour	Cards	Total
♠ Spades	Black	A,2,3,4,5,6,7,8,9,10,J,Q,K	13
♣ Clubs	Black	A,2,3,4,5,6,7,8,9,10,J,Q,K	13
♥ Hearts	Red	A,2,3,4,5,6,7,8,9,10,J,Q,K	13
♦ Diamonds	Red	A,2,3,4,5,6,7,8,9,10,J,Q,K	13
Total	26 black, 26 red	4 Aces, 4 Kings, 4 Queens, 4 Jacks	52

Section 15.4

Sample Space & Events

Understanding sample spaces systematically avoids missing outcomes.

Example 4 · 3M

Two coins tossed simultaneously. Write sample space and find P(exactly 1 Head), P(at least 1 Head).

1

S = {HH, HT, TH, TT}. n(S) = 4

2

Exactly 1 Head: {HT, TH}. P = 2/4 = 1/2

3

At least 1 Head: {HH, HT, TH}. P = 3/4

P(exactly 1 Head) = 1/2, P(at least 1 Head) = 3/4

Example 5 · 3M

Two dice rolled. Write n(S) and find P(sum = 7), P(doublet), P(sum > 10).

1

n(S) = 6×6 = 36

2

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

3

Doublet: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) → 6 outcomes. P = 6/36 = 1/6

4

Sum > 10: (5,6),(6,5),(6,6) → 3 outcomes. P = 3/36 = 1/12

P(sum=7) = 1/6, P(doublet) = 1/6, P(sum>10) = 1/12

Section 15.5

Tree Diagrams

Tree diagrams visually list all outcomes of multi-step experiments — very useful for two-stage or multi-stage probability.

Tree diagram for two coin tosses — multiply probabilities along each branch to get outcome probability

NCERT Exercise · Solved

Exercise 15.1 & Mixed

Q1 · 2M

A bag contains 3 red, 5 blue, 2 green balls. Find P(Red), P(Blue), P(Green), P(not Red).

1

Total = 10. P(Red)=3/10, P(Blue)=5/10=1/2, P(Green)=2/10=1/5

2

P(not Red) = 1−3/10 = 7/10

P(Red)=3/10, P(Blue)=1/2, P(Green)=1/5, P(not Red)=7/10

Q2 · 3M

Cards numbered 1–20 are placed in a box. One card is drawn at random. Find P(prime number), P(divisible by 5), P(perfect square).

1

n(S)=20. Primes: {2,3,5,7,11,13,17,19}=8. P(prime)=8/20=2/5

2

Div by 5: {5,10,15,20}=4. P(div 5)=4/20=1/5

3

Perfect squares: {1,4,9,16}=4. P(sq)=4/20=1/5

P(prime)=2/5, P(div by 5)=1/5, P(perfect square)=1/5

Q3 · 2M

P(winning a game) = 0.3. What is P(losing)?

1

P(losing) = 1 − P(winning) = 1 − 0.3 = 0.7

P(losing) = 0.7

Smart Study

8 Essential Tips — Probability

1

P always between 0 and 1

0 ≤ P(E) ≤ 1 always. P=0 means impossible, P=1 means certain. If you get P>1 or P<0, you made an error.

2

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

Complementary events sum to 1. If P(rain) = 0.3, then P(no rain) = 0.7. Use this to find P(not E) = 1 − P(E).

3

List sample space carefully

For two dice: n(S)=36. For two coins: n(S)=4. Count all equally likely outcomes — don't miss any.

4

Cards deck: 52 total

4 suits × 13 cards. 26 red (hearts+diamonds), 26 black. 4 Aces, 4 Kings, 4 Queens, 4 Jacks, 16 face cards total (J+Q+K × 4 suits).

5

Empirical ≠ Theoretical exactly

Empirical probability from experiments rarely equals the theoretical value exactly. As trials → ∞, empirical → theoretical (Law of Large Numbers).

6

Tree diagrams: multiply branches

For compound events, probability of a path = product of probabilities on each branch. Sum of all end-probabilities = 1.

7

"At least one" = 1 − P(none)

P(at least 1 Head) = 1 − P(no Heads) = 1 − P(all Tails). This complement approach is faster than listing all favourable outcomes.

8

Probability is always a fraction/decimal

Write final answer as a simplified fraction or decimal. CBSE expects simplified: 6/36 = 1/6, 2/4 = 1/2.

Quick Reference

Chapter 15 — Key Facts

Situation	n(S)	Common Events
1 coin	2	P(H)=1/2, P(T)=1/2
2 coins	4	P(2H)=1/4, P(1H)=1/2, P(0H)=1/4
1 die	6	P(6)=1/6, P(even)=1/2, P(prime)=1/2
2 dice	36	P(sum=7)=1/6, P(doublet)=1/6
52 cards	52	P(ace)=1/13, P(red)=1/2, P(heart)=1/4, P(face)=3/13

CBSE Practice — FINAL Chapter

50 Practice Questions

MCQ · 1M · 2M · 3M · 5M · Case-Based — all with solutions. You've made it to the last chapter! 🎉

Introduction toProbability