Probability — Detailed Notes

Chapter 14 | CBSE Class 10 Mathematics (NCERT)

Study Diagrams (Visual Revision)

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Probability — Decision Map

Probability — Core Ideas

Probability — Formula Strip

14.1 Probability — A Theoretical Approach

Equally likely outcomes

A fair coin or fair die is unbiased: each face/side is as likely as any other.
Not every experiment has equally likely outcomes (e.g. bag with $4$ red and $1$ blue — “red” vs “blue” as colour-events are not equally likely, though each ball is).

Assumption in this chapter: experiments have equally likely outcomes.

Empirical vs theoretical

Empirical (Class IX): $P (E) = \frac{trials where E happened}{total trials}$ — needs many repetitions.
Theoretical (classical): under equally likely assumption, $P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes of the experiment}$ (Laplace’s definition.) Repetition is not required when the assumption holds.

Basic results

Elementary event: an event with exactly one outcome. Sum of probabilities of all elementary events of an experiment $= 1$ .
Complement: $E ˉ =$ “not $E$ ”.
$P (E ˉ) = 1 - P (E)$
Impossible event: $P = 0$ (e.g. getting $8$ on a die).
Sure (certain) event: $P = 1$ (e.g. number $< 7$ on a die).
For any event $E$ : $0 \leq P (E) \leq 1$

Better Method / Strategy
Prefer the complement when “at least”, “not”, or “different from” appears.
Always write $n (S)$ first (total outcomes), then count favourable carefully.

Worked ideas (NCERT-aligned)

Coin once: $P (H) = P (T) = \frac{1}{2}$ ; elementary events; sum $= 1$ .

Die: $P (number > 4) = \frac{2}{6} = \frac{1}{3}$ ; $P (number \leq 4) = \frac{4}{6} = \frac{2}{3}$ ; complements.

Cards: Well-shuffled $\Rightarrow$ equally likely.
$P (ace) = \frac{4}{52} = \frac{1}{13}$ ; $P (not ace) = 1 - \frac{1}{13} = \frac{12}{13}$ .

Complementary wins: If $P (S) = 0.62$ , then $P (R) = 1 - 0.62 = 0.38$ (exactly one of two must win).

Birthdays (non-leap): $P (different) = \frac{364}{365}$ ; $P (same) = \frac{1}{365}$ .

Bag / marbles: Count total objects of equal size; favourable $=$ count of desired colour.

Two coins: Outcomes $(H, H), (H, T), (T, H), (T, T)$ — four equally likely.
$P (at least one head) = \frac{3}{4} = 1 - \frac{1}{4}$ .

Two dice: $6 \times 6 = 36$ ordered pairs.
$P (sum = 8) = \frac{5}{36}$ ; $P (sum = 13) = 0$ ; $P (sum \leq 12) = 1$ .
Trap: sums $2, 3, \dots, 12$ are not $11$ equally likely outcomes.

Geometric probability (Examples 10–11 in NCERT): ratio of favourable length/area to total — marked not from examination point of view.

Playing cards — quick reference

Feature	Count
Total cards	$52$
Suits	$4$ (spades, hearts, diamonds, clubs) × $13$
Red / black	$26$ / $26$
Face cards (K, Q, J)	$12$
Aces	$4$

Useful card probabilities to keep ready

$P (ace) = \frac{4}{52} = \frac{1}{13}$
$P (face card) = \frac{12}{52} = \frac{3}{13}$
$P (red face card) = \frac{6}{52} = \frac{3}{26}$
$P (king of hearts) = \frac{1}{52}$
$P (not ace) = 1 - \frac{1}{13} = \frac{12}{13}$

Two-dice sum frequencies (for quick counting)

Sum	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Ways	$1$	$2$	$3$	$4$	$5$	$6$	$5$	$4$	$3$	$2$	$1$

Total ways $= 36$ . This table alone disproves “each sum has probability $\frac{1}{11}$ ”.

Common experiment checklist

Experiment	$∥ S ∥$ (typical)
One coin	$2$
Two coins	$4$
One die	$6$
Two dice	$36$
One card	$52$
Bag of $n$ identical-size balls	$n$

Better Method / Strategy — Building $S$
Use ordered outcomes when order matters (two dice, two distinct coins).
$(1, 4) \neq = (4, 1)$ . Never collapse to unordered sums unless probabilities are weighted.

Worked idea — Without replacement (board style)

Lot of $20$ bulbs with $4$ defective.
(i) $P (defective) = \frac{4}{20} = \frac{1}{5}$ .
(ii) If first is good and not replaced: $19$ left, $16$ good → $P (second good) = \frac{16}{19}$ .

Five diamond honour cards shuffled; queen drawn and set aside:
$P (second is ace) = \frac{1}{4}$ ; $P (second is queen) = 0$ .

Worked idea — “Same result” on three tosses

Coin tossed $3$ times: $8$ outcomes.
Win if HHH or TTT → $P (win) = \frac{2}{8} = \frac{1}{4}$ ; $P (lose) = \frac{3}{4}$ .

Worked idea — Arguments to judge (NCERT style)

“Two coins → three outcomes HH, TT, one-of-each, each $\frac{1}{3}$ ” → incorrect (mixed has two elementary outcomes).
“Die → odd or even, so $P (odd) = \frac{1}{2}$ ” → correct (three odd, three even faces).
“Sums on two dice are $11$ outcomes each $\frac{1}{11}$ ” → incorrect.

Equally likely or not? (quick judgment)

Situation	Equally likely?
Fair coin / fair die	Yes
True–false guess (if random)	Often modelled yes
Car starts or not	Not necessarily
Basketball shot make/miss	Not necessarily
Baby boy/girl (simplified model)	Often taken as yes in school problems

Empirical vs theoretical (NCERT note to reader)

Empirical = based on what did happen in trials.
Theoretical = predicts what will happen under assumptions.
As trials increase, empirical tends to approach theoretical for fair experiments.

NCERT summary checklist

Theoretical $P (E)$ definition (equally likely)
Sure event → $1$ ; impossible → $0$
$0 \leq P (E) \leq 1$
Elementary events; their probabilities sum to $1$
$P (E) + P (E ˉ) = 1$

Exam-Oriented Analysis

Question type	Typical marks	Key skill
Fill definitions / bounds	1	Exact NCERT wording
Bag / die / single card	1–2	Favourable count
Complement / “not E”	1–2	$1 - P (E)$
Two dice / two–three coins	2–3	Correct $∥ S ∥$
Without replacement	2–3	Update totals
Judge arguments	2	Equally likely test

Common mistakes

Treating unequally likely outcomes as equal (two coins → “HH, TT, HT” as three outcomes)
Using $11$ outcomes for sums on two dice
Favourable count errors on cards (face cards $= 12$ , not $16$ )
Writing probabilities outside $[0, 1]$
Forgetting complement when “not” is easier
Including ace as a face card
Forgetting order in two-dice pairs

Scoring tips

State $n (S)$ and list or count favourable outcomes
Reduce fractions; alternatively leave $\frac{a}{b}$
For two-part card questions, reuse $P (E ˉ) = 1 - P (E)$
In MCQ “which cannot be a probability”, eliminate negatives and values $> 1$
When justifying fairness of a coin toss, mention unbiased + equally likely