Probability: Class 11 Maths · Chapter 14

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

75 min
2 simulations
43 practice questions
20 MCQs

What you will learn in Probability

Describe events as subsets of a sample space and decide from an outcome whether an event has occurred
Classify events as impossible, sure, simple or compound, and build 'not A', 'A or B', 'A and B' and 'A but not B' with set operations
Test whether given events are mutually exclusive, exhaustive, or both
State the three axioms of probability, check whether an assignment of probabilities to outcomes is valid, and find the probability of an event by adding outcome probabilities
Derive P(E) = n(E)/n(S) for equally likely outcomes and use counting, permutations and combinations to apply it
Prove the addition rule for 'A or B' and the complement rule for 'not A', and apply them, including the special case of mutually exclusive events
Solve board- and entrance-level problems on coins, dice, cards, committees and random orders, including the addition rule for three events

Original companion text aligned to NCERT Mathematics Class 11, Chapter 14, pp. 289–313. Not affiliated with or endorsed by NCERT.

14.1 Event

Before a cricket match the two captains walk out for the toss, and from then on the commentators talk about chances: of rain, of a big total, of a collapse. This chapter makes such talk exact in two moves. First we describe every 'happening' we might care about as a set of outcomes, so that the algebra of sets from Chapter 1 does the bookkeeping. Then we state three short rules, the axioms, that any probability must obey, and derive from them every formula you will use.

Stay with the coin and the die. Suppose all you care about is whether you get a head together with an even number. Three outcomes give that: H2, H4 and H6. Collect them into the set $E = {H 2, H 4, H 6},$ a subset of $S$ . Any yes-or-no question about the result picks out a subset in the same way, as Fig. 14.1 shows, and any subset answers some such question.

2/5'A head and an even number' is $E = {H 2, H 4, H 6} .$

Fig. 14.1An event is just a subset of the sample space.NCERT p. 289

Twelve tiles in two rows show the outcomes of tossing a coin and rolling a die: H1 to H6 on top, T1 to T6 below. In the first step no tile is lit and the label says S has 12 outcomes. In the second step H2, H4 and H6 light up for the event head with an even number. In the third step H5, H6, T5 and T6 light up for a number more than 4. In the fourth step nothing lights up, because no outcome shows a 7, so the event is the empty set. In the last step the result H4 is marked with a tick inside the outlined event E, so E has occurred.

Table 14.1Some events of the coin-and-die experiment

The event in words	Its subset of $S$
a tail with a 6	${T 6}$
an odd number, either face	${H 1, H 3, H 5, T 1, T 3, T 5}$
a head with a number below 3	${H 1, H 2}$
a number less than 7	$S$ (all 12 outcomes)
a tail with a number above 6	$ϕ$ (no outcome)

Definition

Event

An event (घटना) is a set of outcomes of an experiment, that is, a subset of its sample space $S$ . Every subset qualifies, including $ϕ$ and $S$ itself.

ExampleFor a coin tossed with a die, ${H 2, H 4, H 6}$ is the event 'a head and an even number'.

Because every subset counts, a sample space with $n$ outcomes has $2^{n}$ events, one for each subset in its power set (Chapter 1). Even the small coin-and-die experiment has $2^{12} = 4096$ events, which is why we describe events by rules rather than by listing them all.

14.1.1 Occurrence of an event

Run the experiment once. Say the coin shows heads and the die shows 4, so the outcome is H4. This outcome is an element of $E = {H 2, H 4, H 6},$ and we say that $E$ has occurred (last step of Fig. 14.1). Had the result been T4, the outcome would lie outside $E$ and $E$ would not have occurred.

So occurring is just membership. A trial ends with one outcome $ω$ ; every event containing $ω$ has occurred in that trial, and every event that leaves $ω$ out has not. One outcome makes many events occur at once: H4 also makes 'a head', 'an even number' and 'a number greater than 3' occur, because it belongs to all three subsets.

14.1.2 Types of events

Events are sorted by how many sample points they contain. Take a spinner with eight equal sectors numbered 1 to 8, so $S = {1, 2, \dots, 8},$ and look at four events (Fig. 14.2).

Impossible event. 'The pointer stops on a number greater than 8' contains no outcome of $S$ , so as a set it is $ϕ$ . The empty set is called the impossible event: no trial can make it occur.
Sure event. 'The pointer stops on a number less than 9' contains every outcome, so it is $S$ itself, the sure event: every trial makes it occur.
Simple event. 'The pointer stops on 5' is ${5}$ . An event with exactly one sample point is a simple (or elementary) event. A sample space with $n$ outcomes has exactly $n$ simple events; the spinner has eight, ${1}$ to ${8}$ .
Compound event. 'The pointer stops on a perfect square' is ${1, 4}$ . An event with more than one sample point is a compound event.

4/4'A perfect square' has two outcomes: ${1, 4}$ .

Fig. 14.2The number of sample points decides the type of an event.NCERT p. 290

A spinner disc divided into eight equal sectors numbered 1 to 8, with a text panel on the right. In the first step no sector is lit and the panel reads more than 8, the empty set, impossible event. In the second step all eight sectors are lit and the panel reads less than 9, the whole sample space, sure event. In the third step only sector 5 is lit and the panel reads stops on 5, the set containing 5, simple event. In the fourth step sectors 1 and 4 are lit and the panel reads a perfect square, the set containing 1 and 4, compound event.

Definition

Impossible and sure events

The empty set $ϕ$ is the impossible event and the whole sample space $S$ is the sure event.

Definition

Simple and compound events

An event with exactly one sample point is simple (elementary); an event with more than one sample point is compound.

ExampleTossing three coins, 'three tails' $= {T T T}$ is simple, while 'all three faces alike' $= {H H H, T T T}$ is compound.

Quick check

1 of 3

Three coins are tossed. Which of these events is simple?

exactly two headsall three faces the samethree tailsat least one tail

Show answer

Answer (C) three tails

'Three tails' is

{T T T},

a single sample point. 'All three faces the same' is

{H H H, T T T},

which has two points, so it is compound.

14.1.3 Algebra of events

Chapter 1 combined sets by complement, union, intersection and difference. Events are sets, so the same operations build new events from old ones, and each operation has a plain-English reading. Let $A$ and $B$ be events of an experiment with sample space $S$ .

For a running example, choose a month of the year at random, so $S$ is the set of 12 months. Let $A$ be 'the month has 31 days' and $B$ be 'the month's name begins with J'. Then $A$ = {Jan, Mar, May, Jul, Aug, Oct, Dec} and $B$ = {Jan, Jun, Jul}.

'Not $A$ '. Gather the outcomes of $S$ that lie outside $A$ . They form the complementary event $A^{'} = S - A$ , which happens in precisely those trials where $A$ fails. Here $A^{'}$ = {Feb, Apr, Jun, Sep, Nov}, the months shorter than 31 days.
' $A$ or $B$ '. The union $A \cup B$ holds the outcomes in $A$ , in $B$ or in both, so 'A or B' occurs when at least one of them occurs. Here $A \cup B$ = {Jan, Mar, May, Jun, Jul, Aug, Oct, Dec}.
' $A$ and $B$ '. The intersection $A \cap B$ , the outcomes common to the two events, occurs when both occur. Here it is {Jan, Jul}.
' $A$ but not $B$ '. The difference $A - B = A \cap B^{'}$ occurs when $A$ occurs and $B$ does not. Here it is {Mar, May, Aug, Oct, Dec}.

3/5 $A \cup B$ : every month with at least one tag.

Fig. 14.3One sample space, four new events built from A and B.NCERT p. 291

Twelve tiles, Jan to Dec, in three rows of four. Tiles of months with 31 days carry a small tag A at the top left, and months whose names begin with J carry a tag B at the top right. In the first step only the tags show. In the second step the five tiles without tag A are filled, for not A. In the third step every tile with tag A or tag B is filled, eight months, for A or B. In the fourth step only Jan and Jul, which carry both tags, are filled, for A and B. In the fifth step the five tiles with tag A but without tag B are filled, for A but not B.

Table 14.2Event language and set language

In words	As a set
not $A$	$A^{'}$ , or $S - A$
$A$ or $B$ (at least one)	$A \cup B$
$A$ and $B$ (both)	$A \cap B$
$A$ but not $B$ (only $A$ )	$A - B = A \cap B^{'}$
neither $A$ nor $B$	$A^{'} \cap B^{'} = (A \cup B)^{'}$
exactly one of $A$ , $B$	$(A - B) \cup (B - A)$

The 'neither' row is De Morgan's law in disguise: 'not A and not B' is the same as 'not (A or B)'. Keep this table at hand; most worded problems are solved by translating each phrase into one of its rows.

Example 14.1NCERT Example 1 · p. 292

Divisors and primes

A number is picked at random from 1 to 12. Let $A$ be 'the number divides 12' and $B$ be 'the number is prime'. Write as sets the events (i) $A$ and $B$ , (ii) $A$ or $B$ , (iii) not $A$ , (iv) $A$ but not $B$ , (v) neither $A$ nor $B$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which set is the event 'exactly one of

A

and

B

occurs'?

A \cup B

A^{'} \cap B^{'}

(A \cap B)^{'}

(A - B) \cup (B - A)

Show answer

Answer (D) $(A - B) \cup (B - A)$

Exactly one means

A

without

B

, or

B

without

A

A \cup B

also contains the outcomes where both occur, and

(A \cap B)^{'}

also contains those where neither occurs.

14.1.4 Mutually exclusive events

Draw one card from a well-shuffled pack of 52. Let $A$ be 'a heart' and $B$ be 'a black card'. Hearts are red, so no card is in both events: whichever of them occurs, the other cannot. Now let $C$ be 'a king'. The king of hearts lies in $A$ and in $C$ , so these two events can happen together (Fig. 14.4).

3/3Hearts and kings share the king of hearts: $A \cap C \neq ϕ .$

Fig. 14.4Disjoint events cannot happen together; overlapping events can.NCERT p. 292

The 52 cards are drawn as a grid with one row per suit, spades, hearts, diamonds and clubs, and one column per rank from ace to king. In the first step nothing is marked. In the second step the hearts row is filled and the two black rows, spades and clubs, are hatched; the filled and hatched cells never share a cell, and the label says no common card. In the third step the hearts row is filled and the king column is outlined; the king of hearts is the one cell in both, marked with a star, and the label says they share one card.

Definition

Mutually exclusive events

Events $A$ and $B$ are mutually exclusive (परस्पर अपवर्जी) when they have no outcome in common, $A \cap B = ϕ,$ so that no single trial can make both of them happen.

ExampleIn one draw, 'a heart' and 'a black card' are mutually exclusive; 'a heart' and 'a king' are not.

Example 14.2NCERT Example 2 · p. 293

Which pairs cannot happen together?

Two dice are rolled. Consider the events $A$ : 'the product of the numbers is odd', $B$ : 'the sum is odd', $C$ : 'both dice show the same number' and $D$ : 'the product is greater than 30'. Which of the six pairs of events can never occur together?

Try it yourself first. Then reveal the solution one step at a time.

Checking events on 36 outcomes by hand is slow. The lab below draws the two-dice sample space as a grid, so you can see at a glance whether two events overlap and what their union, intersection and complements look like.

InteractiveTwo-dice event lab

An event lab on the 36 outcomes of rolling two dice. Two drop-down menus choose events A and B from twelve presets, such as the sum is even, a double, or the first die shows 6. Seven buttons pick the event to show: A, B, not A, A or B, A and B, A but not B, and neither. A six by six grid, with the first die down the side and the second across the top, fills the cells of the chosen event, and every cell carries small A and B tags when it belongs to those events. A panel gives the number of outcomes and the probability as a fraction of 36, checks the addition rule, and says whether A and B are mutually exclusive and whether together they are exhaustive. Nothing moves until the learner acts.

Fig. 14.5Choose events

A

and

B

and an operation. The grid fills the outcomes of that event; the panel counts them and says whether

A

and

B

are mutually exclusive or exhaustive.

14.1.5 Exhaustive events

Back to the eight-sector spinner. Let $A$ be 'a number at most 3' $= {1, 2, 3}$ , $B$ be 'an even number' $= {2, 4, 6, 8}$ and $C$ be 'a number at least 5' $= {5, 6, 7, 8} .$ Every number from 1 to 8 lies in at least one of them, so $A \cup B \cup C = S$ : whatever the spinner shows, at least one of the three events occurs. Events with this property are called exhaustive.

Definition

Exhaustive events

A family of events $E_{1}, E_{2}, \dots, E_{n}$ is exhaustive (निःशेष) when together they use up the sample space: $E_{1} \cup E_{2} \cup \dots \cup E_{n} = S .$ Then no trial can end with all of them missing.

The events $A$ , $B$ , $C$ above overlap, for $2 \in A \cap B$ and $6 \in B \cap C,$ so they are exhaustive but not mutually exclusive. Now split the spinner differently: $P$ = 'at most 3', $Q$ = '4 or 5' and $R$ = 'at least 6'. These cover $S$ and no two of them share an outcome, so exactly one of them occurs at every spin (Fig. 14.6). Such a family is called mutually exclusive and exhaustive; it partitions the sample space.

Mutually exclusive and exhaustive

\begin{aligned} E_{1} \cup \dots \cup E_{n} = S, \\ E_{i} \cap E_{j} = ϕ (i \neq j) \end{aligned}

(14.1)

where

$E_{1}, \dots, E_{n}$: the events of the family
$i \neq j$: any two different events of the family

2/3Every number is covered, but 2, 6 and 8 are covered twice.

Fig. 14.6Exhaustive: every column has a dot. A partition: every column has exactly one.NCERT p. 293

Eight numbered cells, 1 to 8, run along the top. Below are three rows, one per event, with a dot under each number that belongs to the event. In the first step the rows are A at most 3, B even and C at least 5. In the second step a count row appears under the dots: every number has at least one dot, so the events are exhaustive, but 2, 6 and 8 have two dots each, marked as overlaps. In the third step the rows change to P at most 3, Q 4 or 5, and R at least 6, and every number now has exactly one dot, which makes a partition.

Example 14.3NCERT Example 3 · p. 294

Guessing a quiz

A student guesses the answers to three true-or-false questions, so each answer is right (R) or wrong (W). Let $A$ be 'all three right', $B$ be 'exactly two right' and $C$ be 'at most one right'. Do $A$ , $B$ and $C$ form a mutually exclusive and exhaustive family?

Try it yourself first. Then reveal the solution one step at a time.

Sorting outcomes by the value of some count, as here, always gives a partition. Splitting the months into 31-day months, 30-day months and February is another.

Quick check

1 of 3

A die is rolled. Which family is mutually exclusive and exhaustive?

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

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

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

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

Show answer

Answer (B) ${1}, {2, 3}, {4, 5, 6}$

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

share no outcome and together give

S

. The first and last families overlap (at 2 and at 3); the third misses 5, so it is not exhaustive.

14.2 Axiomatic Approach to Probability

Words such as probably, likely and hardly ever rank chances without measuring them. Probability theory puts a number on each event. You already know two ways to choose that number. The classical way divides the count of favourable outcomes by the count of equally likely outcomes. The experimental way uses the fraction of trials in which the event happened.

The axiomatic approach, set out by the Russian mathematician A. N. Kolmogorov in 1933, works one level higher. It does not tell you which numbers to use. It lists three rules that any sensible assignment must obey, and everything else is proved from them.

AxiomAxioms of probability

Suppose an experiment has sample space $S$ . A probability (प्रायिकता) on $S$ is a rule $P$ that attaches to each event $E \subseteq S$ a real number $P (E)$ with three properties. (i) Non-negativity: $P (E) \geq 0$ for every event $E$ . (ii) Normalisation: the sure event gets $P (S) = 1$ . (iii) Additivity: whenever $E \cap F = ϕ,$ $P (E \cup F) = P (E) + P (F) .$

So the inputs of $P$ are events, the members of the power set of $S$ , and its outputs are real numbers. The axioms do not say that these numbers stay below 1; that is a consequence, proved below together with another one used all the time.

DerivationTwo quick consequences of the axioms

The impossible event has probability 0. For any event $E$ , the sets $E$ and $ϕ$ are disjoint and $E \cup ϕ = E .$ Axiom (iii) gives $P (E) = P (E) + P (ϕ),$ so $P (ϕ) = 0$ .

No probability exceeds 1. $E$ and $E^{'}$ are disjoint with $E \cup E^{'} = S$ , so axioms (iii) and (ii) give $P (E) + P (E^{'}) = 1.$ As $P (E^{'}) \geq 0$ by axiom (i), $P (E) \leq 1$ .

When the sample space is finite, $S = {ω_{1}, ω_{2}, \dots, ω_{n}},$ the axioms boil down to a simple recipe. Write $P (ω_{i})$ for $P ({ω_{i}}),$ the probability of the elementary event ${ω_{i}}$ . Then $0 \leq P (ω_{i}) \leq 1$ for each $i$ , and

Outcome probabilities add to 1

\sum_{i = 1}^{n} P (ω_{i}) = 1

(14.2)

Probability of an event

P (A) = \sum_{ω_{i} \in A} P (ω_{i})

(14.3)

where

$A$: any event
$ω_{i} \in A$: the sum runs over the outcomes in $A$ only

Both follow from axiom (iii): the simple events are pairwise disjoint, so applying (iii) again and again adds their probabilities. Over all of $S$ the total is $P (S) = 1$ , which is (14.2); over the outcomes of $A$ it is $P (A)$ , which is (14.3). Conversely, any numbers in $[0, 1]$ that add to 1 can be attached to the outcomes, and (14.3) then defines a $P$ satisfying all three axioms.

The axioms therefore do not fix the numbers. Drop a drawing pin on the floor: it lands point up ( $U$ ) or tipped over on its side ( $D$ ). Nothing about its shape makes the two equally likely. For every $p$ with $0 \leq p \leq 1$ , the choice $P (U) = p$ , $P (D) = 1 - p$ obeys the axioms, so there are infinitely many valid assignments. Which one describes a real pin is a question for experiment. Even for a coin, $P (H) = 0.6$ , $P (T) = 0.4$ is mathematically valid; it simply describes a bent coin rather than a fair one.

SimulationA spinner with unequal sectors

A spinner lab with three sectors of unequal size: green covers half the disc, blue a third and red a sixth. Buttons spin it once, 10, 100 or 1000 times. A bar chart shows the relative frequency of each colour next to its theoretical probability, and a readout tracks the relative frequency of green. Nothing moves until the learner presses a spin button.

Fig. 14.7The sectors cover

\frac{1}{2}

\frac{1}{3}

and

\frac{1}{6}

of the disc. Press a spin button and compare each bar with its theoretical value.

Try this

Before spinning: write the assignment $P (Green), P (Blue), P (Red)$ for this disc and check it against the axioms.
Spin 1000 times. Is the relative frequency of green close to $\frac{1}{2}$ ? Add the three relative frequencies: what total do you always get?
After only 10 spins the frequencies also add to 1 and are never negative, yet they may be far from the sector sizes. Which axioms do they obey anyway?

Example 14.4NCERT Example 4 · p. 297

Valid or not?

A sample space is $S = {ω_{1}, ω_{2}, ω_{3}, ω_{4}, ω_{5}} .$ Which of these assignments of probabilities to the outcomes are valid?

Table 14.3Five proposed assignments

Row	$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$
(a)	0.1	0.3	0.2	0.25	0.15
(b)	0.5	0	0.5	0	0
(c)	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{16}$	$\frac{1}{16}$
(d)	0.3	0.3	0.3	0.3	−0.2
(e)	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$

Try it yourself first. Then reveal the solution one step at a time.

14.2.1 Probability of an event

Once every outcome has its probability, (14.3) gives the probability of any event: add up the probabilities of its outcomes. The outcomes need not be equally likely.

Take a die loaded so that a 6 turns up with probability $\frac{1}{4}$ , while the other five faces share the remaining $\frac{3}{4}$ equally, $\frac{3}{20} = 0.15$ each. For the event 'an even number' we add the pieces for 2, 4 and 6: $0.15 + 0.15 + 0.25 = 0.55.$ For 'at least 5' we add $0.15 + 0.25 = 0.40.$ Fig. 14.8 pictures the total probability 1 as a strip cut into these pieces.

3/4P(even) $= 0.15 + 0.15 + 0.25 = 0.55.$

Fig. 14.8Share the total 1 among the outcomes; an event gets the sum of its shares.NCERT p. 298

A long horizontal strip stands for total probability 1, with 0 at its left end and 1 at its right. In the first step it is a single piece labelled P of S equals 1. In the second step it is cut into six pieces labelled 1 to 6: five pieces of 0.15 and a longer last piece of 0.25. In the third step the pieces for 2, 4 and 6 rise above the strip and the sum 0.15 plus 0.15 plus 0.25 equals 0.55 appears, the probability of an even number. In the fourth step the pieces for 5 and 6 rise instead, giving 0.40 for at least 5.

ExampleNCERT p. 298

Two matches, unequal chances

An analyst assigns these probabilities to the results of a team's next two matches, where W is a win and L a loss: $P (W W) = 0.36,$ $P (W L) = 0.24,$ $P (L W) = 0.24,$ $P (L L) = 0.16.$ Is the assignment valid? Find the probability that the team (i) wins at least once, (ii) gets the same result in both matches.

Try it yourself first. Then reveal the solution one step at a time.

14.2.2 Probabilities of equally likely outcomes

Very often symmetry (a fair coin, a balanced die, a well-shuffled pack) makes all outcomes equally likely. Suppose $S = {ω_{1}, \dots, ω_{n}}$ and $P (ω_{i}) = p$ for every $i$ . Then (14.2) reads $n p = 1$ , so $p = \frac{1}{n} .$ If an event $E$ has $m$ outcomes, (14.3) adds $m$ copies of $\frac{1}{n}$ :

Equally likely outcomes

P (E) = \frac{n (E)}{n (S)} = \frac{m}{n}

(14.4)

where

$n (E) = m$: the number of outcomes favourable to $E$
$n (S) = n$: the total number of outcomes

This is the Class 10 formula, now derived from the axioms, and the derivation shows its limits: it holds only when the outcomes are equally likely. For the loaded die of Section 14.2.1, $P (6) = \frac{1}{4},$ not $\frac{1}{6}$ . For a fair coin tossed with a fair die, all 12 outcomes are equally likely, so $P (head and even) = \frac{3}{12} = \frac{1}{4} .$

Quick check

1 of 3

For

S = {a, b, c},

which assignment is not valid?

0.7, 0.4, - 0.1

0.5, 0.5, 0

\frac{1}{3}, \frac{1}{3}, \frac{1}{3}

0.1, 0.2, 0.7

Show answer

Answer (A) $0.7, 0.4, - 0.1$

0.7 + 0.4 - 0.1 = 1,

but a negative value breaks axiom (i). The other three lists are non-negative and add to 1.

14.2.3 Probability of the event 'A or B'

Pick one token at random from twenty tokens numbered 1 to 20. Let $A$ be 'an even number' and $B$ be 'a multiple of 3'. Then $n (A) = 10$ and $n (B) = 6$ , yet $A \cup B$ has 13 tokens, not 16. The tokens 6, 12 and 18 belong to both events, and the sum $n (A) + n (B)$ counts each of them twice (Fig. 14.9). Subtracting the overlap once repairs the count.

\begin{aligned} P (A \cup B) & = \frac{10 + 6 - 3}{20} \\ = \frac{13}{20} \end{aligned}

3/4Tiles 6, 12 and 18 were counted twice.

Fig. 14.9Adding the two counts counts the overlap twice; subtract it once.NCERT Fig. 14.1 · p. 301

Twenty tiles numbered 1 to 20 sit in four rows of five, with a running count below. In the first step the ten even tiles are filled and tagged A, and the count reads n of A equals 10. In the second step the six multiples of 3 get a dashed border and a tag B, and the count reads 10 plus 6 equals 16. In the third step tiles 6, 12 and 18, which carry both tags, get a thick ring and the label counted twice. In the fourth step the tiles outside A or B fade, the rings change to counted once, and the count reads 16 minus 3 equals 13, the size of A or B.

Theorem 14.1Addition rule for two events

For any two events $A$ and $B$ , the probability that at least one of them occurs is $P (A) + P (B)$ minus the probability $P (A \cap B)$ that both occur. In symbols this is (14.5) below.

From the axioms. Cut $A \cup B$ and $B$ into disjoint pieces: $A \cup B = A \cup (B - A)$ and $B = (A \cap B) \cup (B - A) .$ In each union the two pieces share no outcome, so axiom (iii) applies to both.

Write $x$ for $P (B - A)$ . The first union gives $P (A \cup B) = P (A) + x,$ and the second gives $P (B) = P (A \cap B) + x .$

Subtract the second equation from the first. The unknown $x$ cancels, and moving $P (B)$ to the right-hand side gives exactly the rule.

From sample points. For a finite $S$ , write $P (A) + P (B)$ as a sum of $P (ω)$ by (14.3). Points of $A - B$ and of $B - A$ appear once, but points of $A \cap B$ appear twice, once in each sum. Taking one copy of $P (A \cap B)$ away leaves each point of $A \cup B$ counted once, which is $P (A \cup B)$ .

Addition rule

\begin{aligned} P (A \cup B) \\ = P (A) + P (B) \\ - P (A \cap B) \end{aligned}

(14.5)

When $A$ and $B$ are mutually exclusive, $A \cap B = ϕ$ and $P (ϕ) = 0$ , so the rule shrinks to $P (A \cup B) = P (A) + P (B),$ which is axiom (iii). In words: probabilities of 'or' simply add only when the events cannot overlap. Since $P (A \cap B) \geq 0,$ every pair of events also satisfies $P (A \cup B) \leq P (A) + P (B) .$

Try it in the two-dice event lab: choose two overlapping events and the operation 'A or B'. The panel adds $n (A)$ and $n (B)$ , takes away $n (A \cap B)$ and lands exactly on $n (A \cup B)$ .

Quick check

1 of 3

P (A) = 0.45

P (B) = 0.3

and

P (A \cap B) = 0.12.

Find

P (A \cup B)

Show answer

Answer 0.63

0.45 + 0.3 - 0.12 = 0.63.

14.2.4 Probability of event 'not A'

Draw one token from twenty-five tokens numbered 1 to 25 and let $A$ be 'a perfect square'. Then $A = {1, 4, 9, 16, 25}$ and $P (A) = \frac{5}{25} = \frac{1}{5} .$ The other 20 tokens make up $A^{'}$ , so $P (A^{'}) = \frac{20}{25} = \frac{4}{5},$ which is exactly $1 - P (A)$ (Fig. 14.10). This is no accident.

3/3 $A$ and $A^{'}$ fill $S$ : $0.2 + 0.8 = 1$ .

Fig. 14.10An event and its complement split the sample space, so their probabilities add to 1.NCERT p. 301

Twenty-five numbered tiles in a five by five grid with a panel on the right. In the first step the five perfect squares, 1, 4, 9, 16 and 25, are filled with a solid border and the panel reads A, 5 of 25, P of A equals 0.2. In the second step the other twenty tiles are hatched with a dashed border and the panel reads not A, 20 of 25, 0.8. In the third step both kinds are shown together, every tile is in exactly one of them, and the panel reads 0.2 plus 0.8 equals 1.

Theorem 14.2Complement rule

For every event $A$ , $P (not A) = P (A^{'}) = 1 - P (A) .$

Complement rule

P (A^{'}) = 1 - P (A)

(14.6)

The rule pays off whenever 'not $A$ ' is easier to count than $A$ , typically in at least one questions. Roll a die twice: 'no six' has $5 \times 5 = 25$ of the 36 outcomes, so P(at least one six) $= 1 - \frac{25}{36} = \frac{11}{36} .$

SimulationDrawing from a full pack

A card-drawing lab with a standard pack of 52. Buttons draw one card, 10, 100 or 1000 cards, each time from a full reshuffled pack. A bar chart shows how often each suit has turned up beside its theoretical share of one quarter, and a readout compares the relative frequency of a face card with the theoretical value 3 over 13. Nothing moves until the learner presses a draw button.

Fig. 14.11Each draw is one card from a full, freshly shuffled pack. The readout tracks 'a face card'; its complement is every other draw.

Try this

Before drawing: what are P(a face card) and P(not a face card)?
Draw 1000 cards. Subtract the face-card frequency from 1 and compare it with your value of P(not a face card).
Do the four suit bars settle near $\frac{1}{4}$ each? Which two bars together estimate P(a red card)?

Example 14.5NCERT Example 5 · p. 302

One card from a pack

A card is taken at random from a well-shuffled pack of 52, all cards equally likely. Find the probability that it is (i) a face card, (ii) not a king, (iii) a red face card, (iv) neither a heart nor a face card.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.6NCERT Example 6 · p. 303

A jar of laddoos

A jar holds 15 laddoos of the same size: 6 besan, 5 motichoor and 4 coconut. One is taken out without looking. Find the probability that it is (i) besan, (ii) coconut, (iii) not motichoor, (iv) besan or coconut.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.7NCERT Example 7 · p. 304

Two applicants

Meera and Kabir apply for the same summer internship. The probability that Meera is selected is 0.3, that Kabir is selected is 0.25, and that both are selected is 0.1. Find the probability that (a) neither is selected, (b) at least one of them is not selected, (c) exactly one of them is selected.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.8NCERT Example 8 · p. 305

Choosing a team

A quiz team of 3 is chosen at random from 4 girls and 3 boys. Find the probability that the team has (a) no boy, (b) exactly one boy, (c) at least two boys.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

P (A) = \frac{3}{8},

what is

P (not A)

Show answer

Answer 0.625

1 - \frac{3}{8} = \frac{5}{8} = 0.625.

Miscellaneous Examples

The last examples mix this chapter with counting from Chapter 6. The plan is always the same: describe an equally likely sample space, count it with permutations or combinations, count the event, and divide. Before counting, look for a symmetry that may give the answer outright.

3/3A is before B in one order of each pair: $\frac{3}{6} = \frac{1}{2} .$

Fig. 14.12Swapping A and B pairs the orders off, so A comes first in exactly half of them.NCERT p. 308

Six cards show the six orders of the letters A, B and C, arranged in three rows of two. In the first step the six orders are simply listed: ABC, ACB and CAB on the left, BAC, BCA and CBA on the right. In the second step a double arrow labelled swap A and B joins the two cards in each row, showing that each right card is its left partner with A and B exchanged. In the third step the left card of each pair, the one where A comes before B, is outlined and ticked, and the label reads 3 of 6, so P equals one half.

Example 14.9NCERT Example 9 · p. 308

A photo line

Five friends A, B, C, D and E stand in a row for a photo in a random order, all orders equally likely. Find the probability that (i) A stands somewhere to the left of B, (ii) A, B and C stand in that left-to-right order, not necessarily side by side, (iii) A and B take the two end places, (iv) A and B stand next to each other.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.10NCERT Example 10 · p. 309

Aces in a hand

A hand of 5 cards is dealt from a well-shuffled pack of 52. Find the probability that it contains (i) no ace, (ii) exactly three aces, (iii) at least three aces.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.11NCERT Example 11 · p. 309

Addition rule for three events

Prove the addition rule for three events, (14.7) below: for any events $A$ , $B$ , $C$ , the probability of $A \cup B \cup C$ is the sum of the three single probabilities, minus the probabilities of the three pairwise intersections, plus $P (A \cap B \cap C) .$

Try it yourself first. Then reveal the solution one step at a time.

Addition rule for three events

\begin{aligned} P (A \cup B \cup C) \\ = P (A) + P (B) \\ + P (C) \\ - P (A \cap B) \\ - P (B \cap C) \\ - P (A \cap C) \\ + P (A \cap B \cap C) \end{aligned}

(14.7)

Example 14.12NCERT Example 12 · p. 310

Three medals

Six runners P, Q, R, S, T and U run a 100 m race with no ties, all finishing orders equally likely. Find the probability that (a) P, Q and R win gold, silver and bronze respectively, (b) P, Q and R win the three medals in some order.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Four friends, including Asha and Ben, sit in a row of four chairs at random. Find the probability that Asha and Ben sit next to each other.

Show answer

Answer 0.5

Glue them into a block:

3! \times 2 = 12

of the

4! = 24

orders, so

\frac{1}{2}

Practice · 15 questions

Practice questions with answers (NCERT “Miscellaneous Exercise on Chapter 14”)

1
NumericalMediumNCERT Miscellaneous Exercise, Q1
A carton holds 6 mango, 9 orange and 15 litchi juice packs. Four packs are taken out at random. Find the probability that
1. all four are orange;
  Show answer
  Answer 0.004597701149425287
  $\frac{^{9} C_{4}}{^{30} C_{4}} = \frac{126}{27 405} = \frac{2}{435} \approx 0.0046.$
2. at least one is litchi.
  Show answer
  Answer 0.9501915708812261
  Complement of 'no litchi': $1 - \frac{^{15} C_{4}}{^{30} C_{4}} = 1 - \frac{1365}{27 405} = \frac{248}{261} \approx 0.950.$
NumericalMediumNCERT Miscellaneous Exercise, Q2
Five cards are drawn from a well-shuffled pack of 52. Find the probability of getting 2 hearts and 3 clubs.
Show answer
Answer 0.008583433373349339
$\frac{^{13} C_{2} \times^{13} C_{3}}{^{52} C_{5}} = \frac{78 \times 286}{2 598 960} = \frac{143}{16 660} \approx 0.0086.$
3
NumericalEasyNCERT Miscellaneous Exercise, Q3
A die has the number 4 on three faces, 5 on two faces and 6 on one face. It is rolled once. Find
1. P(5);
  Show answer
  Answer 0.3333333333333333
  $\frac{2}{6} = \frac{1}{3} .$
2. P(4 or 6);
  Show answer
  Answer 0.6666666666666666
  $\frac{3 + 1}{6} = \frac{2}{3} .$
3. P(not 4).
  Show answer
  Answer 0.5
  $1 - \frac{3}{6} = \frac{1}{2} .$
4
NumericalHardNCERT Miscellaneous Exercise, Q4
In a school fete raffle, 5000 tickets are sold and 20 of them win identical prizes. Find, to four decimal places, the probability of winning nothing if you buy
1. one ticket;
  Show answer
  Answer 0.996
  $\frac{4980}{5000} = 0.996.$
2. two tickets;
  Show answer
  Answer 0.992
  $\frac{^{4980} C_{2}}{^{5000} C_{2}} = \frac{4980 \times 4979}{5000 \times 4999} \approx 0.9920.$
3. five tickets.
  Show answer
  Answer 0.9802
  $\frac{^{4980} C_{5}}{^{5000} C_{5}} \approx 0.9802.$
5
NumericalHardNCERT Miscellaneous Exercise, Q5
120 students, you and a friend among them, are split at random into a section of 50 and a section of 70. Find the probability that
1. you two are in the same section;
  Show answer
  Answer 0.5098039215686274
  $\frac{50}{120} \cdot \frac{49}{119} + \frac{70}{120} \cdot \frac{69}{119} = \frac{7280}{14 280} = \frac{26}{51} \approx 0.510.$
2. you are in different sections.
  Show answer
  Answer 0.49019607843137253
  Complement of (a): $\frac{25}{51} \approx 0.490.$
NumericalHardNCERT Miscellaneous Exercise, Q6
Four letters are written to four different people and four envelopes are addressed. The letters are put into the envelopes at random, one in each. What is the chance that one or more of the letters reach the right envelope?
Show solution
Call the letters 1 to 4. The arrangements with no letter in its own place are 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321: nine of them. So P(at least one correct) $= 1 - \frac{9}{24} = \frac{15}{24} = \frac{5}{8} .$
7
NumericalMediumNCERT Miscellaneous Exercise, Q7
$A$ and $B$ are events with $P (A) = 0.62$ , $P (B) = 0.48$ and $P (A \cap B) = 0.27.$ Find
1. $P (A \cup B)$ ;
  Show answer
  Answer 0.83
  $0.62 + 0.48 - 0.27 = 0.83.$
2. $P (A^{'} \cap B^{'})$ ;
  Show answer
  Answer 0.17
  $1 - P (A \cup B) = 0.17.$
3. $P (A \cap B^{'})$ ;
  Show answer
  Answer 0.35
  $0.62 - 0.27 = 0.35.$
4. $P (B \cap A^{'})$ .
  Show answer
  Answer 0.21
  $0.48 - 0.27 = 0.21.$
NumericalMediumNCERT Miscellaneous Exercise, Q8
Six volunteers run a school blood-donation camp. One of them is chosen at random to speak to the press. Find the probability that the speaker is female or under 25.
Table 14.4The volunteers
Name Gender Age (years)
Tanvi F 22
Arjun M 31
Farhan M 24
Gurpreet F 35
Lalit M 28
Neha F 19
Show answer
Answer 0.6666666666666666
Female: 3; under 25: 3 (Tanvi, Farhan, Neha); both: Tanvi and Neha. So $\frac{3 + 3 - 2}{6} = \frac{2}{3} .$
9
NumericalHardNCERT Miscellaneous Exercise, Q9
Four-digit numbers greater than 6000 are formed at random from the digits 0, 2, 5, 6, 9. Find the probability that the number is divisible by 5 when
1. digits may repeat;
  Show answer
  Answer 0.39759036144578314
  All: $2 \times 5^{3} - 1 = 249$ (removing 6000). Ending in 0 or 5: $2 \times 5^{2} \times 2 - 1 = 99.$ P $= \frac{99}{249} = \frac{33}{83} \approx 0.398.$
2. digits may not repeat.
  Show answer
  Answer 0.5
  All: $2 \times 4 \times 3 \times 2 = 48.$ Ending in 0 or 5: $2 \times 2 \times (3 \times 2) = 24.$ P $= \frac{1}{2}$ .
NumericalEasyNCERT Miscellaneous Exercise, Q10
A bicycle lock has three wheels, each with the digits 0 to 9, and it opens for one sequence of three different digits. What is the probability that a random guess of three different digits opens it?
Show answer
Answer 0.001388888888888889
$^{10} P_{3} = 10 \times 9 \times 8 = 720$ sequences, one correct: $\frac{1}{720} \approx 0.0014.$
Quick answerMedium
Assertion (A): For any two events, $P (A \cup B) \leq P (A) + P (B) .$ Reason (R): $P (A \cap B) \geq 0$ for every pair of events.
Both are true and R explains ABoth are true but R does not explain AA is true but R is falseA is false but R is true
Show answer
Answer (A) Both are true and R explains A
By (14.5), $P (A \cup B)$ is $P (A) + P (B)$ with the non-negative number $P (A \cap B)$ taken away, and taking away a non-negative number cannot increase a sum.
Quick answerMedium
Events $A$ , $B$ , $C$ are mutually exclusive and exhaustive with $P (A) = 2 P (B) = 3 P (C) .$ Then $P (A)$ is
$\frac{1}{2}$ $\frac{3}{11}$ $\frac{2}{3}$ $\frac{6}{11}$
Show answer
Answer (D) $\frac{6}{11}$
$P (A) + \frac{P (A)}{2} + \frac{P (A)}{3} = 1,$ so $\frac{11}{6} P (A) = 1$ and $P (A) = \frac{6}{11} .$
Quick answerMedium
Two fair dice are rolled. Find the probability that the sum is 8 or the dice show a double.
Show answer
Answer 0.2777777777777778
Sum 8: 5 outcomes; doubles: 6; both: $(4, 4)$ . So $\frac{5 + 6 - 1}{36} = \frac{10}{36} = \frac{5}{18} \approx 0.278.$
Quick answerHard
$P (A) = 0.4$ and $P (B) = 0.5$ . Which interval contains every possible value of $P (A \cap B)$ , and nothing more?
$[0.1, 0.4]$ $[0, 0.5]$ $[0.2, 0.4]$ $[0, 0.4]$
Show answer
Answer (D) $[0, 0.4]$
$P (A \cap B) \leq P (A) = 0.4$ (reached when $A \subseteq B$ ). From $P (A \cup B) \leq 1$ we only get $P (A \cap B) \geq - 0.1,$ so the lower limit is 0 (reached when $A$ , $B$ are mutually exclusive).
Quick answerHard
Four cards are drawn from a well-shuffled pack of 52. Find the probability that all four are of different suits.
Show answer
Answer 0.10549819927971188
$\frac{13^{4}}{^{52} C_{4}} = \frac{28 561}{270 725} \approx 0.1055.$

Name	Gender	Age (years)
Tanvi	F	22
Arjun	M	31
Farhan	M	24
Gurpreet	F	35
Lalit	M	28
Neha	F	19

Recap

Probability: key points and summary

An event is any subset of the sample space; it occurs when the outcome belongs to it (Section 14.1).
$ϕ$ is the impossible event and $S$ the sure event; simple events have one sample point, compound events more (Section 14.1.2).
'Not A' is $A^{'} = S - A$ , 'A or B' is $A \cup B$ , 'A and B' is $A \cap B$ , 'A but not B' is $A - B = A \cap B^{'}$ (Section 14.1.3).
$A$ and $B$ are mutually exclusive when $A \cap B = ϕ$ (Section 14.1.4).
Events are exhaustive when their union is $S$ ; if they are also pairwise disjoint they partition $S$ ((14.1)).
Axioms: $P (E) \geq 0$ , $P (S) = 1$ , and $P (E \cup F) = P (E) + P (F)$ for mutually exclusive $E$ , $F$ ; so $P (ϕ) = 0$ (Section 14.2).
For finite $S$ , outcome probabilities lie in $[0, 1]$ and add to 1 ((14.2)).
The probability of an event is the sum of the probabilities of its outcomes ((14.3)).
With $n$ equally likely outcomes, $P (E) = \frac{n (E)}{n (S)}$ ((14.4)).
For any two events, add $P (A)$ and $P (B)$ , then subtract $P (A \cap B)$ so the overlap counts once; for mutually exclusive events nothing is subtracted (Theorem 14.1).
The complement rule $P (A^{'}) = 1 - P (A)$ is the quick route to 'at least one' (Theorem 14.2).
For three events, add the singles, subtract the pairs and add back the triple ((14.7)).

Timeline

Historical note: from gambling tables to axioms

c. 2500 BCE
Clay dice at Mohenjo-daro
Cubical dice of baked clay have been dug up at Harappan sites, so games of chance in India are very old.
1560s
Cardano on games of chance
The Italian physician Gerolamo Cardano (1501–1576) wrote Liber de Ludo Aleae, the first book on the mathematics of dice. It was printed only in 1663.
1654
Pascal and Fermat
A gambler's questions about dice and an interrupted game reached Blaise Pascal, whose letters with Pierre de Fermat solved them and founded the subject.
1657
Huygens's textbook
Christiaan Huygens (1629–1695) published the first printed treatise on probability, built on the idea of expected value.
1713
Bernoulli's law
Jacob Bernoulli's Ars Conjectandi, published after his death, proved that relative frequencies settle near probabilities over many trials.
1718–1812
De Moivre and Laplace
Abraham de Moivre's Doctrine of Chances and Pierre-Simon Laplace's 1812 treatise made probability a large, systematic theory.
1800s
The Russian school
P. L. Chebyshev (1821–1894) and his student A. A. Markov (1856–1922) proved general limit theorems.
1933
Kolmogorov's axioms
A. N. Kolmogorov (1903–1987) treated probability as a function on sets obeying a few axioms, the approach of this chapter.

Flash cards · 6

Rules to remember

1 / 6

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 14.1, Q1	Practice 14.1 · Q1
NCERT Exercise 14.1, Q2	Practice 14.1 · Q2
NCERT Exercise 14.1, Q3	Practice 14.1 · Q3
NCERT Exercise 14.1, Q4	Practice 14.1 · Q4
NCERT Exercise 14.1, Q5	Practice 14.1 · Q5
NCERT Exercise 14.1, Q6	Practice 14.1 · Q6
NCERT Exercise 14.1, Q7	Practice 14.1 · Q7
NCERT Exercise 14.2, Q1	Practice 14.2 · Q1
NCERT Exercise 14.2, Q2	Practice 14.2 · Q2
NCERT Exercise 14.2, Q3	Practice 14.2 · Q3
NCERT Exercise 14.2, Q4	Practice 14.2 · Q4
NCERT Exercise 14.2, Q5	Practice 14.2 · Q5
NCERT Exercise 14.2, Q6	Practice 14.2 · Q6
NCERT Exercise 14.2, Q7	Practice 14.2 · Q7
NCERT Exercise 14.2, Q8	Practice 14.2 · Q8
NCERT Exercise 14.2, Q9	Practice 14.2 · Q9
NCERT Exercise 14.2, Q10	Practice 14.2 · Q10
NCERT Exercise 14.2, Q11	Practice 14.2 · Q11
NCERT Exercise 14.2, Q12	Practice 14.2 · Q12
NCERT Exercise 14.2, Q13	Practice 14.2 · Q13
NCERT Exercise 14.2, Q14	Practice 14.2 · Q14
NCERT Exercise 14.2, Q15	Practice 14.2 · Q15
NCERT Exercise 14.2, Q16	Practice 14.2 · Q16
NCERT Exercise 14.2, Q17	Practice 14.2 · Q17
NCERT Exercise 14.2, Q18	Practice 14.2 · Q18
NCERT Exercise 14.2, Q19	Practice 14.2 · Q19
NCERT Exercise 14.2, Q20	Practice 14.2 · Q20
NCERT Exercise 14.2, Q21	Practice 14.2 · Q21
NCERT Miscellaneous Exercise, Q1	Miscellaneous Exercise on Chapter 14 · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous Exercise on Chapter 14 · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous Exercise on Chapter 14 · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous Exercise on Chapter 14 · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous Exercise on Chapter 14 · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous Exercise on Chapter 14 · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous Exercise on Chapter 14 · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous Exercise on Chapter 14 · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous Exercise on Chapter 14 · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous Exercise on Chapter 14 · Q10

Quick check: MCQs on Probability

14.1.2 Types of events3 questions
14.1.3 Algebra of events3 questions
14.1.5 Exhaustive events3 questions
14.2.2 Probabilities of equally likely outcomes3 questions
14.2.3 Probability of the event 'A or B'3 questions
14.2.4 Probability of event 'not A'3 questions
Miscellaneous Examples2 questions

Glossary · 12

Key terms (English–Hindi)

complementary eventपूरक घटनाpoorak ghatna: For an event $A$ , the event $A^{'} = S - A$ of all outcomes not in $A$ ; it occurs exactly when $A$ does not.First seen in the chapter →
compound eventमिश्र घटनाmishr ghatna: An event with more than one sample point.First seen in the chapter →

equally likely outcomesसम-संभाव्य परिणामsam-sambhavya parinaam: Outcomes that all have the same probability, $\frac{1}{n}$ when there are $n$ of them.First seen in the chapter →
eventघटनाghatna: Any subset of the sample space. It occurs when the outcome of the experiment belongs to it.
exhaustive eventsनिःशेष घटनाएँnihshesh ghatnaayen: Events whose union is the whole sample space, so at least one of them occurs at every trial.First seen in the chapter →

impossible eventअसंभव घटनाasambhav ghatna: The empty set $ϕ$ viewed as an event: no outcome makes it occur.First seen in the chapter →

mutually exclusive eventsपरस्पर अपवर्जी घटनाएँparaspar apavarjee ghatnaayen: Events that cannot occur together: $A \cap B = ϕ$ .

probabilityप्रायिकताpraayikta: A function $P$ that gives each event a number $P (E)$ with $P (E) \geq 0$ , $P (S) = 1$ and $P (E \cup F) = P (E) + P (F)$ for mutually exclusive $E$ , $F$ .First seen in the chapter →

random experimentयादृच्छिक परीक्षणyadrichchhik pareekshan: An experiment with more than one possible outcome, all known in advance, whose actual outcome cannot be predicted.First seen in the chapter →

sample spaceप्रतिदर्श समष्टिpratidarsh samashti: The set $S$ of all possible outcomes of a random experiment; each element is a sample point.First seen in the chapter →
simple eventसरल घटनाsaral ghatna: An event with exactly one sample point, such as ${5}$ ; also called an elementary event.First seen in the chapter →
sure eventनिश्चित घटनाnishchit ghatna: The whole sample space $S$ viewed as an event: it occurs at every trial.First seen in the chapter →

Probability: Class 11 Maths · Chapter 14

What you will learn in Probability

14.1 Event

14.1.1 Occurrence of an event

14.1.2 Types of events

14.1.3 Algebra of events

14.1.4 Mutually exclusive events

14.1.5 Exhaustive events

Practice 14.1: questions with answers

14.2 Axiomatic Approach to Probability

14.2.1 Probability of an event

14.2.2 Probabilities of equally likely outcomes

14.2.3 Probability of the event 'A or B'

14.2.4 Probability of event 'not A'

Practice 14.2: questions with answers

Miscellaneous Examples

Practice questions with answers (NCERT “Miscellaneous Exercise on Chapter 14”)

Probability: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Probability

Key terms (English–Hindi)

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