What is theoretical probability?

Theoretical probability P(E) is the number of outcomes favourable to event E divided by the total number of equally likely outcomes of the experiment — no repeated trials needed, just counting.

What does it mean for outcomes to be equally likely?

Equally likely outcomes are ones where none is more probable than another, such as a fair coin's head/tail or a fair die's six faces. Not every experiment has this property — for example, drawing a ball from a bag with unequal colour counts does not make the colours equally likely.

What is an elementary event, and what do all elementary event probabilities sum to?

An elementary event has exactly one outcome from the experiment. The probabilities of all elementary events in an experiment always add up to exactly 1, since together they account for every possible outcome.

What are complementary events in probability?

The complement of an event E, written as 'not E', is the event that E does not happen. P(E) + P(not E) always equals 1, so P(not E) = 1 − P(E) — a useful shortcut when 'not E' is easier to reason about than E directly.

What is the range of possible probability values?

Every probability satisfies 0 ≤ P(E) ≤ 1. A probability of 0 means the event is impossible, and a probability of 1 means the event is certain (sure) to happen.

Chapter 14 · CBSE Class 10 Maths

Probability

Two ways to learn — a crisp formula reference, or a story that makes complementary events feel obvious.

Equally likely outcomesTheoretical probabilityElementary eventsComplementary events

Also useful:Chapter 13 — Statistics NCERT Solutions — Class 10 Maths 1-on-1 CBSE Class 10 Tuition

CBSEClass 10MathematicsChapter 14⏱ 7 min crisp · 10 min story

Four scenes that build the ideas from scratch, step by step.

How this works

Four scenes, one thread: probability is just counting favourable outcomes out of all equally likely ones — and once you know an event's probability, its exact opposite (the complement) comes for free by subtracting from 1. Read the story, then tap "The Maths" for the formal result.

Equally Likely, or Not?

Toss a fair coin. Head and tail are equally likely — neither has any reason to come up more often. Now imagine a bag with 4 red balls and just 1 blue ball. Are "drawing red" and "drawing blue" equally likely too?

Student

No — there are way more red balls, so red should come up more often. But drawing ANY one specific ball (say, one particular red ball among the 4) is still equally likely compared to any other specific ball.

Exactly — not every experiment automatically has equally likely outcomes at the level you care about. Coins and dice happen to work out nicely; a red/blue ball count does not, unless you're comparing individual balls rather than colours.

Theoretical probability P(E) = (favourable outcomes) / (total equally likely outcomes)

Coin: P(head) = 1/2 (1 favourable outcome, 2 total)

The Elementary Building Blocks

A bag has one red, one blue, and one yellow ball — each equally likely to be drawn. Since each event (drawing red, drawing blue, drawing yellow) has exactly ONE favourable outcome, each is called an elementary event.

Teacher

Add up the probabilities of ALL the elementary events in this bag. What do you notice?

P(red) = 1/3, P(blue) = 1/3, P(yellow) = 1/3

Sum = 1/3 + 1/3 + 1/3 = 1

This isn't a coincidence — since every possible outcome is accounted for exactly once, the elementary event probabilities must always add up to exactly 1, no matter the experiment.

This also works for non-elementary events that together cover everything: on a die, "greater than 4" (outcomes 5,6 → P=2/6) and "4 or less" (outcomes 1,2,3,4 → P=4/6) together sum to 2/6+4/6=1, since between them they cover every possible roll.

Not E

On a die, "greater than 4" and "4 or less" aren't just two separate events that happen to sum to 1 — one is literally EVERYTHING the other isn't. That relationship has a name: complementary events.

Teacher

If P(E) + P(not E) always equals 1, what's the fastest way to find P(not E) once you already know P(E)?

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

This shortcut turns awkward "not X" questions into quick subtraction. Drawing a card from a 52-card deck: P(ace) = 4/52 = 1/13. Rather than counting all 48 non-ace cards directly, just subtract.

P(not an ace) = 1 − 1/13 = 12/13

Two special cases worth knowing by name: an impossible event (like rolling an 8 on a standard die) has probability exactly 0. A sure/certain event (like rolling a number less than 7) has probability exactly 1. Every other probability sits strictly between them: 0 ≤ P(E) ≤ 1.

The Surprising Birthday

Two friends, Savita and Hamida. What's the probability they share the same birthday (ignoring leap years, so 365 possible days)? Counting every way two birthdays could match sounds complicated — Hamida's birthday could coincide with Savita's on any of 365 specific days, in principle.

Student

Instead, what if I find the probability their birthdays are DIFFERENT first — that sounds much easier to count directly?

Savita's birthday: any of 365 days

Hamida's birthday different from Savita's: 364 favourable days out of 365

P(different birthdays) = 364/365

"Same birthday" is exactly the complement of "different birthdays" — so the harder-sounding probability falls out immediately by subtraction.

P(same birthday) = 1 − P(different birthdays) = 1 − 364/365 = 1/365

The Big Picture

Every idea in this chapter branches from one formula: count favourable outcomes over all equally likely ones. Elementary events always sum to 1, and the complement rule (P(not E) = 1 − P(E)) turns hard counting problems into easy subtraction.

Basic probability formula?

P(E) = favourable / total (equally likely)

Range of any probability?

0 ≤ P(E) ≤ 1

Sum of all elementary events?

Always exactly 1

Complement rule?

P(not E) = 1 − P(E)

'Not X' is hard to count directly?

Find P(X) first, subtract from 1

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