Introduction To Probability

Q: What is the classical definition of probability?

If an experiment has N equally likely, mutually exclusive and exhaustive outcomes and m of them favour event A, then P(A) = m/N. It works only when every outcome has the same chance; otherwise use weights or a probability tree, multiplying along branches.

Q: How do you convert odds into probability?

If the odds in favour of an event are m : n, the event succeeds in m of every m + n equally likely cases, so P = m/(m + n). Odds against m : n mean P = n/(m + n). For odds 2 : 5 in favour, the probability is 2/7, not 2/5.

Q: What is the formula for the probability of A or B or C?

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(C and A) + P(A and B and C). Pairwise overlaps are subtracted because they were counted twice, and the triple overlap is added back because it was removed too often.

Q: Why is a sum of 7 the most likely total with two dice?

Sum 7 can be made in 6 ordered ways, (1, 6) through (6, 1), more than any other total. In general a total s has 6 - |s - 7| ways, which peaks at s = 7. So P(sum 7) = 6/36 = 1/6, while totals 2 and 12 have only 1/36 each.

MathsProbabilityFor JEE aspirants

Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (sure). When a random experiment has equally likely outcomes, the classical probability of an event $A$ is $P (A) = \frac{n ( A )}{n ( S )}$ : favourable outcomes divided by all outcomes of the sample space $S$ . This introduction to probability builds the base for JEE Main and JEE Advanced: sample spaces, types of events, algebra of events, odds, the addition theorem for two and three events, and geometric probability, with fully solved examples.

On this page1Sample space2Events3Algebra of events4Classical probability5Addition theorem6Geometric probability7Strategy8Solved examples

Key Formulas - Quick Reference

★ Must learnClassical probability: $P (A) = \frac{n ( A )}{n ( S )}$ , with $0 \leq P (A) \leq 1$ , $P (ϕ) = 0$ , $P (S) = 1$
★ Must learnComplement: $P (A^{'}) = 1 - P (A)$
★ Must learnOdds in favour of $A$ are $m : n$ $\Rightarrow$ $P (A) = \frac{m}{m + n}$ and $P (A^{'}) = \frac{n}{m + n}$
★ Must learnAddition theorem: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ ; mutually exclusive: $P (A \cup B) = P (A) + P (B)$
$P (A \cap B^{'}) = P (A) - P (A \cap B)$ and $P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$
P(exactly one of $A, B$ ) $= P (A) + P (B) - 2 P (A \cap B)$
★ Must learn $P (A \cup B \cup C) = \sum P (A) - \sum P (A \cap B) + P (A \cap B \cap C)$
P(exactly two of $A, B, C$ ) $= \sum P (A \cap B) - 3 P (A \cap B \cap C)$ ; P(at least two) $= \sum P (A \cap B) - 2 P (A \cap B \cap C)$
P(exactly one of $A, B, C$ ) $= \sum P (A) - 2 \sum P (A \cap B) + 3 P (A \cap B \cap C)$
Bounds: $max {P (A), P (B)} \leq P (A \cup B) \leq min {1, P (A) + P (B)}$

1. Random Experiment and Sample Space

Random experiment: an experiment whose possible outcomes are all known in advance, but whose outcome in any one performance cannot be predicted until it is over. Tossing a coin, throwing a die and drawing a card from a pack of 52 are random experiments.

Sample space: the set of all possible outcomes of a random experiment, usually written $S$ . In set language, $S$ is the universal set. For one throw of a die, $S = {1, 2, 3, 4, 5, 6}$ ; for one toss of a coin, $S = {H, T}$ .

When two dice are rolled, every outcome is an ordered pair $(x, y)$ : $x$ on the first die and $y$ on the second. Keeping $(1, 2)$ and $(2, 1)$ separate is exactly what makes all $6 \times 6 = 36$ outcomes equally likely (Figure 1).

Figure 1: The 36 ordered pairs

(x, y)

for two dice. Doublets fill one diagonal and sum

7

fills the other, so each has probability

\frac{6}{36} = \frac{1}{6}

Exam Trick

For two dice, the number of ordered pairs with total $s$ is $6 - ∣ s - 7∣$ for $s = 2, 3, \dots, 12$ . The counts run $1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1$ . So sum $8$ has $5$ ways and sum $11$ has $2$ ways, with no listing needed.

Figure 2: Totals

2

12

occur in

1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1

ways (sum

36

). Total

s

has

6 - ∣ s - 7∣

ways, so

P (sum 8) = \frac{5}{36}

1.1 Multi-stage experiments: use a tree

When the second stage depends on the first (a head leads to another toss, a tail leads to a die), draw a tree. Each branch carries its own probability; the probability of an outcome is the product along its path (Figure 3).

Figure 3: Multiply along each branch:

P (H H) = \frac{1}{4}

but

P (T 1) = \frac{1}{12}

. The

8

outcomes are not equally likely, so

\frac{n ( A )}{n ( S )}

would give wrong answers here.

Caution: counting outcomes gives probability only when outcomes are equally likely. In Figure 3, $P (H H) = \frac{1}{4}$ but $P (T 1) = \frac{1}{12}$ , so writing $P (H H) = \frac{1}{8}$ is a classic trap.

Key idea

Count outcomes only when they are equally likely: two dice give 36 ordered pairs; unequal stages need a tree.

2. Events and Their Types

Event: any subset of the sample space. Getting a head, or getting a prime number on a die, are events. If $n (S) = n$ , exactly $2^{n}$ events (subsets) can be formed, including $ϕ$ and $S$ .

Type of event	Meaning	Example
Impossible event	The empty set $ϕ$ ; $P (ϕ) = 0$	Getting 7 on one die
Sure event	The whole sample space $S$ ; $P (S) = 1$	Getting a number less than 7 on one die
Simple event	Contains exactly one sample point	Getting HT when a coin is tossed twice
Compound event	Contains two or more sample points	Drawing a red suit: {heart, diamond}
Complement of $A$	All outcomes of $S$ not in $A$ ; written $A^{'}$ , $\overset{ˉ}{A}$ or $A^{c}$	Complement of "even" on a die is "odd"
Equally likely events	Events with the same chance of occurring	H and T for a fair coin
Mutually exclusive events	Cannot occur together: $A_{i} \cap A_{j} = ϕ$ for $i \neq = j$	Sum 4, sum 10 and sum 12 with two dice
Exhaustive events	At least one must occur: $E_{1} \cup E_{2} \cup \dots \cup E_{n} = S$	Heart, diamond, club, spade for a drawn card

For suits, take $S$ = {heart, spade, club, diamond}. Drawing a heart is the simple event {heart}; drawing a red card is the compound event {heart, diamond}. Equally likely needs a fair mechanism: for an unbiased die {1}, {2}, {3}, {4} are equally likely, but for a biased coin {H} and {T} are not.

Figure 4: Complement

A^{'}

, mutually exclusive events (

A \cap B = ϕ

) and a partition (

E_{1} \cup E_{2} \cup E_{3} = S

with no overlaps).

Pairwise versus overall. If $A_{i} \cap A_{j} = ϕ$ for every pair $i \neq = j$ , then $A_{1} \cap A_{2} \cap \dots \cap A_{n} = ϕ$ . The converse is false: three events can have an empty common part while two of them still overlap. Mutually exclusive always means pairwise disjoint.

2.1 Independent events (first look)

Events are independent if the occurrence of one does not change the chance of the other. With a pair of dice, "first die even" and "second die odd" are independent, because one die cannot influence the other. They are not mutually exclusive, since both can happen together. The full treatment, with $P (A \cap B) = P (A) P (B)$ , is in the Conditional Probability concept.

Mutually exclusive

A set idea: no common outcome, $A \cap B = ϕ$ .

Then $P (A \cup B) = P (A) + P (B)$ .

Example: 2 and 5 on one die.

Independent

A probability idea: $P (A \cap B) = P (A) P (B)$ .

One event gives no information about the other.

Example: first die even, second die odd.

If $A$ and $B$ are mutually exclusive with $P (A) > 0$ and $P (B) > 0$ , they are strongly dependent: once one happens, the other cannot.

Quick Recall: tap to check

An urn has 3 red and 2 blue balls. Write the sample space for drawing one ball.

Name the balls to keep outcomes equally likely:

S

= {R1, R2, R3, B1, B2}.

One die is thrown. Are "odd" and "even" mutually exclusive?

Yes,

{1, 3, 5} \cap {2, 4, 6} = ϕ

. They are also exhaustive, so they partition

S

Die thrown. Which pair is mutually exclusive: (a) odd, even (b) odd, number

\geq 4

(a) only. In (b) both contain 5.

Die thrown. Which system is exhaustive: (a) odd,

\leq 4

, the number 5 (b)

\leq 4

> 4

(b) only. System (a) misses 6 and system (c) misses 5.

3. Algebra of Events

Events are sets, so every statement in words has a set form. Keep this table handy while reading problems.

In words	Set notation
At least one of A, B occurs (A or B)	$A \cup B$ (also written $A + B$ )
Both A and B occur	$A \cap B$ (also written $A B$ )
A does not occur	$A^{'}$ , $\overset{ˉ}{A}$ or $A^{c}$
Neither A nor B occurs	$A^{'} \cap B^{'}$
A occurs but B does not	$A \cap B^{'}$
Exactly one of A, B occurs	$(A \cap B^{'}) \cup (A^{'} \cap B)$
Occurrence of A implies occurrence of B	$A \subseteq B$

De Morgan's laws: $(A \cup B)^{'} = A^{'} \cap B^{'}$ and $(A \cap B)^{'} = A^{'} \cup B^{'}$ .

Distributive laws: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ and $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

Figure 5: The four disjoint regions

A \cap B^{'}

A \cap B

A^{'} \cap B

and

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

build every two-event result. Adding

P (A)

and

P (B)

counts

A \cap B

twice.

3.1 Results that follow from the four regions

$P (A^{'}) = 1 - P (A)$ , since $A \cup A^{'} = S$ and $A$ , $A^{'}$ are mutually exclusive.
$P (A \cap B^{'}) = P (A) - P (A \cap B)$ , since $A \cap B^{'}$ and $A \cap B$ are disjoint and together make $A$ . Similarly $P (A^{'} \cap B) = P (B) - P (A \cap B)$ .
$P (A^{'} \cup B^{'}) = 1 - P (A \cap B)$ and $P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ (De Morgan).
P(exactly one of $A, B$ ) $= P (A \cap B^{'}) + P (A^{'} \cap B) = P (A) + P (B) - 2 P (A \cap B) = P (A \cup B) - P (A \cap B)$ .

Quick Recall: tap to check

Two coins are tossed,

A

= {HH, HT},

B

= {HT, TT}. Find

A \cup B

and

A \cap B

A \cup B

= {HH, HT, TT};

A \cap B

= {HT}.

A coin is tossed and a die thrown.

A

: head and an odd number;

B

: tail and an even number. Find

A \cup B

and

A \cap B

A \cup B

= {H1, H3, H5, T2, T4, T6};

A \cap B = ϕ

Write "neither

A

nor

B

" and "exactly one of

A

B

" in set form.

A^{'} \cap B^{'} = (A \cup B)^{'}

;

(A \cap B^{'}) \cup (A^{'} \cap B)

4. Classical Definition of Probability

If a random experiment can result in any one of $N$ equally likely, mutually exclusive and exhaustive outcomes, and exactly $m$ of them favour event $A$ , then

P (A) = \frac{m}{N} = \frac{number of favourable outcomes}{total number of outcomes} = \frac{n ( A )}{n ( S )}

Figure 6:

0 \leq P (A) \leq 1

. With one die,

P (7) = 0

P (6) = \frac{1}{6}

P (not 6) = \frac{5}{6}

and

P (less than 7) = 1

4.1 Odds in favour and odds against

If, out of $m + n$ equally likely cases, $m$ favour $A$ and $n$ do not, the odds in favour of $A$ are $m : n$ and the odds against $A$ are $n : m$ . Then $P (A) = \frac{m}{m + n}$ and $P (A^{'}) = \frac{n}{m + n} = 1 - P (A)$ .

Figure 7: Odds in favour

m : n

mean

m

good cases out of

m + n

, so

P (A) = \frac{m}{m + n}

. Odds

2 : 5

give

\frac{2}{7}

, not

\frac{2}{5}

Remark. In a finite sample space of equally likely outcomes, $P (A) = 1$ only for the sure event and $P (A) = 0$ only for the impossible event. In continuous (geometric) settings this link breaks: probability 0 does not mean impossible, and probability 1 does not mean certain.

Key idea

P (A) = \frac{n ( A )}{n ( S )}

needs equally likely outcomes; count numerator and denominator the same way (both ordered or both unordered).

5. Addition Theorem of Probability

For any two events $A$ and $B$ of the same experiment:

P (A \cup B) = P (A) + P (B) - P (A \cap B)

If $A$ and $B$ are mutually exclusive, no sample point lies in $A \cap B$ , so $P (A \cup B) = P (A) + P (B)$ .

Adding $P (A)$ and $P (B)$ counts the overlap $A \cap B$ twice (Figure 5), so it is subtracted once. For three events the same idea gives the inclusion-exclusion formula. Figure 8 shows why "exactly two" and "at least two" differ only in the multiple of $P (A \cap B \cap C)$ .

Figure 8: Seven regions of three events. Shade depth counts how many of

A

B

C

occur, which explains the

- 3 P (A \cap B \cap C)

(exactly two) and

- 2 P (A \cap B \cap C)

(at least two) terms.

$P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)$
P(at least two of $A, B, C$ ) $= P (A \cap B) + P (B \cap C) + P (C \cap A) - 2 P (A \cap B \cap C)$
P(exactly two of $A, B, C$ ) $= P (A \cap B) + P (B \cap C) + P (C \cap A) - 3 P (A \cap B \cap C)$
P(exactly one of $A, B, C$ ) $= \sum P (A) - 2 \sum P (A \cap B) + 3 P (A \cap B \cap C)$

General form for $n$ events (inclusion-exclusion):

P (r = 1 ⋃ n A_{r}) = r \sum P (A_{r}) - r_{1} < r_{2} \sum P (A_{r_{1}} \cap A_{r_{2}}) + r_{1} < r_{2} < r_{3} \sum P (A_{r_{1}} \cap A_{r_{2}} \cap A_{r_{3}})

- \dots + (- 1)^{n - 1} P (A_{1} \cap A_{2} \cap \dots \cap A_{n})

5.1 Bounds on P(A or B)

Since $A \subseteq A \cup B$ and $B \subseteq A \cup B$ , $P (A \cup B) \geq max {P (A), P (B)}$ . Since $P (A \cap B) \geq 0$ , $P (A \cup B) \leq P (A) + P (B)$ , and no probability exceeds 1. Also $P (A \cup B) \leq 1$ gives $P (A \cap B) \geq P (A) + P (B) - 1$ . Figure 9 shows the two extreme placements.

Figure 9: With

P (A) = 0.6

P (B) = 0.7

: the overlap is at least

0.6 + 0.7 - 1 = 0.3

and at most

min (0.6, 0.7) = 0.6

; the union runs from

0.7

1

max {P (A) + P (B) - 1, 0} \leq P (A \cap B) \leq min {P (A), P (B)}

JEE Advanced

Inclusion-exclusion in action: derangements. $n$ letters are put at random into $n$ addressed envelopes. Let $A_{i}$ be "letter $i$ is in its own envelope". Then $P (A_{i}) = \frac{1}{n}$ , $P (A_{i} \cap A_{j}) = \frac{( n - 2 )!}{n !}$ , and so on, so

P (no letter correct) = 1 - \frac{1}{1 !} + \frac{1}{2 !} - \frac{1}{3 !} + \dots + \frac{( - 1 ) ^{n}}{n !}

For $n = 4$ this is $\frac{9}{24} = \frac{3}{8}$ (Solved Example 33). As $n$ grows, the value tends to $\frac{1}{e} \approx 0.368$ , almost independent of $n$ .

Key idea

Add the pieces, subtract every overlap counted twice, add back what was removed too often: that is inclusion-exclusion.

Quick Recall: tap to check

A bag has 4 white, 3 red and 2 blue balls; one is drawn. Find P(white or red) and P(white and red).

\frac{7}{9}

(mutually exclusive, so add);

0

(one ball cannot be both).

Two fair dice are thrown. Find P(total

\leq 9

1 - P (10, 11, 12) = 1 - \frac{3 + 2 + 1}{36} = \frac{5}{6}

P (A) = 0.5

and

P (B) = 0.8

, what is the least possible value of

P (A \cap B)

0.5 + 0.8 - 1 = 0.3

6. Geometric Probability

Syllabus note: geometric (continuous) probability is not named in the JEE Main 2026 or JEE Advanced 2026 syllabus. It appeared in older JEE papers and builds strong intuition, so treat it as enrichment.

When outcomes form a continuum, counting is replaced by measuring length, area or volume. The following are taken as axioms:

If a point is taken at random on a segment $A B$ , the chance that it falls on a sub-segment $P Q$ is $\frac{P Q}{A B}$ : favourable length over total length.
If a point is taken at random in a region of area $S$ containing a region of area $σ$ , the chance that it falls in $σ$ is $\frac{σ}{S}$ : favourable area over total area. Volumes work the same way.

Figure 10: For continuous outcomes,

P = \frac{favourable length (or area)}{total length (or area)}

7. How to Attack a Probability Problem

Identify the experiment and write (or picture) the sample space $S$ .
Decide whether outcomes are equally likely. If not, draw a tree or assign weights.
Translate the words into set language: or $= \cup$ , and $= \cap$ , not $= A^{'}$ .
Look for a shortcut: complement for "at least one", addition theorem for "A or B", symmetry for dice totals.
Count favourable and total outcomes the same way, then check $0 \leq P \leq 1$ .

Figure 11: A decision path for basic probability questions. Most JEE errors come from skipping the first two checks: continuous outcomes and equally likely outcomes.

Figure 12: Revision map of the page. Each branch is one idea you must be able to state and use without looking.

8. Solved Examples

8.1 Sample spaces and events

Solved Example 1

Write the sample space of the experiment "a coin is tossed and a die is thrown".

Solution:

Every coin result pairs with every die result, giving $2 \times 6 = 12$ outcomes:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

Solved Example 2

A coin is tossed. If it shows a head, the coin is tossed again; otherwise a die is thrown. Write the sample space.

Solution:

S = {HH, HT, T1, T2, T3, T4, T5, T6}, which has 8 outcomes. The tree in Figure 3 shows how each outcome arises and why they are not equally likely.

Solved Example 3

Find the sample space for rolling a pair of dice once, and the number of elements in it.

Solution:

Let the first die be blue and the second green. If 1 shows on the blue die and 2 on the green die, record the ordered pair $(1, 2)$ . Every outcome is an ordered pair $(x, y)$ :

S = {(x, y) : x, y in {1, 2, 3, 4, 5, 6}}.

Figure 1 lists all of them: $n (S) = 6 \times 6 = 36$ . $(1, 2)$ and $(2, 1)$ are different outcomes; keeping them separate makes all 36 outcomes equally likely.

Solved Example 4

Write down all the events of the experiment "tossing a coin once".

Solution:

$S = {H, T}$ has 2 elements, so there are $2^{2} = 4$ events: $ϕ$ , {H}, {T} and {H, T}.

Solved Example 5

A die is thrown. Let

A

be "an odd number turns up" and

B

be "a number divisible by 3 turns up". Write the events (a)

A

B

(b)

A

and

B

Solution:

$A = {1, 3, 5}$ and $B = {3, 6}$ .

(a) $A$ or $B$ $= A \cup B = {1, 3, 5, 6}$

(b) $A$ and $B$ $= A \cap B = {3}$

Solved Example 6

Check whether the events are mutually exclusive: (a) {H} and {T} in one toss of a coin (b) {1, 2} and {2, 3} in one throw of a die.

Solution:

(a) ${H} \cap {T} = ϕ$ , so the events are mutually exclusive.

(b) ${1, 2} \cap {2, 3} = {2} \neq = ϕ$ , so the events are not mutually exclusive.

Solved Example 7

A die is thrown.

A

: "an even number turns up",

B

: "an odd prime turns up",

C

: "a number less than 4 turns up". Do

A

B

C

form an exhaustive system?

Solution:

$A = {2, 4, 6}$ , $B = {3, 5}$ , $C = {1, 2, 3}$ .

$A \cup B \cup C = {1, 2, 3, 4, 5, 6} = S$ , so the system is exhaustive. It is not mutually exclusive, since 2 lies in both $A$ and $C$ .

Solved Example 8

Three coins are tossed. Describe (i) two mutually exclusive events

A

B

(ii) three events that are mutually exclusive and exhaustive (iii) two events that are not mutually exclusive (iv) two events that are mutually exclusive but not exhaustive (v) three events that are mutually exclusive but not exhaustive.

Solution:

(i) $A$ : at least two heads; $B$ : at least two tails.

(ii) $A$ : at most one head; $B$ : exactly two heads; $C$ : exactly three heads.

(iii) $A$ : at most two tails; $B$ : exactly two heads.

(iv) $A$ : exactly one head; $B$ : exactly two heads.

(v) $A$ : exactly one tail; $B$ : exactly two tails; $C$ : exactly three tails.

Other correct choices are possible.

8.2 Classical probability

Solved Example 9

Two fair coins are tossed. What is the probability that at least one head occurs?

Solution:

$S = {H H, H T, T H, T T}$ . The coins are fair, so the outcomes are equally likely: if each has weight $w$ , then $4 w = 1$ , so $w = \frac{1}{4}$ .

$A$ = at least one head = {HT, TH, HH}, so $P (A) = \frac{3}{4}$ .

Check with the complement: $P (A) = 1 - P (T T) = 1 - \frac{1}{4} = \frac{3}{4}$ .

Solved Example 10

In a single throw of two fair dice, find the probability of (i) two 4s (ii) a doublet (iii) a five and a six (iv) a sum of 7.

Solution:

There are $6 \times 6 = 36$ equally likely outcomes.

(i) Only $(4, 4)$ is favourable: $P = \frac{1}{36}$ .

(ii) Doublets (1,1), (2,2), (3,3), (4,4), (5,5), (6,6): $P = \frac{6}{36} = \frac{1}{6}$ .

(iii) $(5, 6)$ and $(6, 5)$ : $P = \frac{2}{36} = \frac{1}{18}$ .

(iv) (1,6), (2,5), (3,4), (4,3), (5,2), (6,1): $P = \frac{6}{36} = \frac{1}{6}$ .

Solved Example 11

Two fair dice are thrown. Find the probability of getting a total of 8.

Solution:

$n (S) = 36$ , with $(1, 2)$ and $(2, 1)$ counted as separate points so that every outcome is equally likely.

Favourable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2). Required probability $= \frac{5}{36}$ , as the shortcut $6 - ∣8 - 7∣ = 5$ predicts.

Solved Example 12

Seven accidents occur in a week. What is the probability that all of them happen on the same day?

Solution:

Each accident can fall on any of the 7 days, so the total number of cases is $7^{7}$ . All seven on one day: choose that day in 7 ways.

P = \frac{7}{7 ^{7}} = \frac{1}{7 ^{6}}

Solved Example 13

A bag contains 5 white, 7 red and 4 black balls. A man draws 3 balls at random. Find the probability that all are white.

Solution:

Total balls $= 16$ , so $n (S) =^{16} C_{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$ .

All three white: $n (E) =^{5} C_{3} = 10$ .

Answer: $P (E) = \frac{10}{560} = \frac{1}{56}$ .

Solved Example 14

From 17 cards numbered 1, 2, 3, ..., 17, one card is drawn at random. Find the probability that its number is divisible by 3 or 7.

Solution:

Divisible by 3: 3, 6, 9, 12, 15. Divisible by 7: 7, 14. No number up to 17 is divisible by both, so the favourable count is $5 + 2 = 7$ .

Answer: $\frac{7}{17}$ .

Solved Example 15

Two cards are drawn at random from a pack of 52. Find the probability that (a) both are spades (b) one is a spade and one is a diamond.

Solution:

$n (S) =^{52} C_{2} = \frac{52 \times 51}{2} = 1326$ .

(a) $n (E_{1}) =^{13} C_{2} = 78$ , so $P (E_{1}) = \frac{78}{1326} = \frac{1}{17}$ .

(b) $n (E_{2}) =^{13} C_{1} \times^{13} C_{1} = 169$ , so $P (E_{2}) = \frac{169}{1326} = \frac{13}{102}$ .

Solved Example 16

A holds 3 tickets in a lottery with 3 prizes and 6 blanks. B holds 1 ticket in a lottery with 1 prize and 2 blanks. Compare their chances of success.

Solution:

For A: $n (S) =^{9} C_{3} = 84$ . A wins nothing only if all 3 tickets are blanks: $^{6} C_{3} = 20$ ways.

$P (A wins) = 1 - \frac{20}{84} = 1 - \frac{5}{21} = \frac{16}{21}$ .

For B: $P (B wins) = \frac{^{1} C _{1}}{^{3} C _{1}} = \frac{1}{3}$ .

Ratio of chances $= \frac{16/21}{1/3} = \frac{16}{7}$ , so A's chance is $\frac{16}{7}$ times B's.

Solved Example 17

A four-digit number is formed using the digits 0, 1, 2, 3, 4 without repetition. Find the probability that it is divisible by 4.

Solution:

Total: the first place cannot be 0, so there are $4 \times 4 \times 3 \times 2 = 96$ numbers, all equally likely.

Favourable: a number is divisible by 4 when its last two digits form a number divisible by 4. The possible endings are 04, 12, 20, 24, 32, 40.

Last two digits	Digits left for first two places	Ways (first place non-zero)
04	1, 2, 3	$3 \times 2 = 6$
12	0, 3, 4	$2 \times 2 = 4$
20	1, 3, 4	$3 \times 2 = 6$
24	0, 1, 3	$2 \times 2 = 4$
32	0, 1, 4	$2 \times 2 = 4$
40	1, 2, 3	$3 \times 2 = 6$

Answer: favourable $= 30$ , so $P = \frac{30}{96} = \frac{5}{16}$ .

Exam Trick

In digit problems, split the cases by "does the ending use 0?". If it does, the remaining digits are all non-zero and fill freely; if it does not, one remaining digit is 0 and cannot lead. This single split clears most JEE digit-probability questions.

Solved Example 18

A company ships items in boxes of 25. A sample of 3 items from each box is tested; if any defective is found, the whole box is screened, otherwise it is shipped. Find the probability that (a) a box with 3 defectives is shipped (b) a box with only one defective is sent back for screening.

Solution:

(a) The box is shipped only if all 3 sampled items come from the 22 good ones:

P = \frac{^{22} C _{3}}{^{25} C _{3}} = \frac{22 \cdot 21 \cdot 20}{25 \cdot 24 \cdot 23} = \frac{77}{115}

(b) The sample must include the single defective item, with the other 2 chosen from 24:

P = \frac{^{24} C _{2}}{^{25} C _{3}} = \frac{276}{2300} = \frac{3}{25}

Solved Example 19

Numbers

p

and

q

are chosen at random, with replacement, from {1, 2, ..., 10}. Find the probability that the roots of

x^{2} + p x + q = 0

are real and distinct.

Solution:

Real and distinct roots need $p^{2} > 4 q$ . Since $q \geq 1$ , we need $p \geq 3$ . Count values of $p$ for each $q$ :

$q$	$4 q$	Values of $p$	Count
1	4	3 to 10	8
2	8	3 to 10	8
3	12	4 to 10	7
4	16	5 to 10	6
5	20	5 to 10	6
6	24	5 to 10	6
7	28	6 to 10	5
8	32	6 to 10	5
9	36	7 to 10	4
10	40	7 to 10	4

Answer: favourable cases $= 59$ out of $10 \times 10 = 100$ , so $P = \frac{59}{100}$ .

Solved Example 20

In a single throw of 3 dice, find the probability of throwing (i) one, two and three (ii) a total of eleven (iii) a total less than eleven (iv) a total more than ten.

Solution:

Total cases $= 6^{3} = 216$ .

(i) The numbers 1, 2, 3 can appear on the three dice in $3! = 6$ ways: $P = \frac{6}{216} = \frac{1}{36}$ .

(ii) Ways to get 11 = coefficient of $x^{11}$ in $(x + x^{2} + \dots + x^{6})^{3}$ , which is 27. $P = \frac{27}{216} = \frac{1}{8}$ .

(iii) Favourable cases = sum of the coefficients of $x^{3}, x^{4}, \dots, x^{10}$ in $(x + \dots + x^{6})^{3}$ = coefficient of $x^{10}$ in $(x + \dots + x^{6})^{3} (1 + x + x^{2} + \dots)$ .

= coeff. of x^{10} in x^{3} (1 - x^{6})^{3} (1 - x)^{- 4} = coeff. of x^{7} in (1 - 3 x^{6}) (1 - x)^{- 4}

=^{10} C_{3} - 3 \cdot^{4} C_{3} = 120 - 12 = 108

So $P (total < 11) = \frac{108}{216} = \frac{1}{2}$ . Here we used: if $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots$ , then $f (x) (1 + x + x^{2} + \dots) = a_{0} + (a_{0} + a_{1}) x + (a_{0} + a_{1} + a_{2}) x^{2} + \dots$ .

(iv) "More than ten" is the complement of "less than eleven": $P = 1 - \frac{1}{2} = \frac{1}{2}$ .

Exam Trick

With 3 dice, replacing each face $k$ by $7 - k$ turns a total $s$ into $21 - s$ . So totals 3 to 10 and totals 11 to 18 are mirror images, giving $P (total \leq 10) = \frac{1}{2}$ with no algebra at all.

8.3 Addition theorem

Solved Example 21

A bag contains 4 white, 3 red and 4 green balls. A ball is drawn at random. Find the probability that it is white or green.

Solution:

Let $A$ : ball is white, $B$ : ball is green. These are mutually exclusive, so $P (A \cap B) = 0$ .

$P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{4}{11} + \frac{4}{11} - 0 = \frac{8}{11}$ .

Solved Example 22

A die is thrown.

A

: "an odd number",

B

: "a number divisible by 3",

C

: "a number

\leq 4

". Find the probability that exactly two of

A

B

C

occur.

Solution:

$A = {1, 3, 5}$ , $B = {3, 6}$ , $C = {1, 2, 3, 4}$ .

$A \cap B = {3}$ , $B \cap C = {3}$ , $A \cap C = {1, 3}$ , $A \cap B \cap C = {3}$ .

P (exactly two) = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} - 3 \times \frac{1}{6} = \frac{1}{6}

Check: only the outcome 1 lies in exactly two events ( $A$ and $C$ ).

Solved Example 23

For three events

A

B

C

: the probability that exactly one of

A

B

occurs is

1 - a

, exactly one of

B

C

1 - 2 a

, exactly one of

C

A

1 - a

, and all three occur together with probability

a^{2}

. Prove that the probability that at least one of

A

B

C

occurs is at least

\frac{1}{2}

Solution:

P(exactly one of $A, B$ ) $= P (A) + P (B) - 2 P (A \cap B)$ , and similarly for the other pairs. Adding the three given values:

(1 - a) + (1 - 2 a) + (1 - a) = 2 \sum P (A) - 2 \sum P (A \cap B)

so $\sum P (A) - \sum P (A \cap B) = \frac{3 - 4 a}{2}$ . Adding $P (A \cap B \cap C) = a^{2}$ (inclusion-exclusion):

P (A \cup B \cup C) = \frac{3 - 4 a}{2} + a^{2} = a^{2} - 2 a + \frac{3}{2}

Completing the square, $a^{2} - 2 a + \frac{3}{2} = (a - 1)^{2} + \frac{1}{2} \geq \frac{1}{2}$ . Hence $P (A \cup B \cup C) \geq \frac{1}{2}$ .

Solved Example 24

The odds in favour of three students P, Q and R standing first in an examination are

1 : 2

2 : 5

and

1 : 7

. Find the probability that one of them stands first.

Solution:

Let $A$ , $B$ , $C$ be the events that P, Q, R stand first: $P (A) = \frac{1}{3}$ , $P (B) = \frac{2}{7}$ , $P (C) = \frac{1}{8}$ .

Only one student can stand first, so the events are mutually exclusive:

P (A \cup B \cup C) = \frac{1}{3} + \frac{2}{7} + \frac{1}{8} = \frac{56 + 48 + 21}{168} = \frac{125}{168}

8.4 Geometric probability (enrichment)

Solved Example 25

A sphere is circumscribed about a cube. A point is chosen at random inside the sphere. Find the probability that it lies outside the cube.

Solution:

If the cube has edge $a$ , the sphere's diameter is the space diagonal $a 3$ , so its radius is $\frac{a 3}{2}$ .

V_{sphere} = \frac{4}{3} π (\frac{a 3}{2})^{3} = \frac{3 π a ^{3}}{2}

P = 1 - \frac{a ^{3}}{3 π a ^{3} /2} = 1 - \frac{2}{3 π} \approx 0.632

Solved Example 26

A line segment is divided at random into three parts. What is the probability that the parts can form a triangle?

Solution:

Let $A B = ℓ$ , and let $C$ , $D$ divide it with $A C = x$ , $C D = y$ , $D B = ℓ - x - y$ . Then $x > 0$ , $y > 0$ , $x + y < ℓ$ : the sample space is triangle $O P Q$ with legs $ℓ$ and area $\frac{ℓ ^{2}}{2}$ .

A triangle exists when each part is less than the sum of the other two:

$x + y > ℓ - x - y$ , i.e. $x + y > \frac{ℓ}{2}$
$x + (ℓ - x - y) > y$ , i.e. $y < \frac{ℓ}{2}$
$y + (ℓ - x - y) > x$ , i.e. $x < \frac{ℓ}{2}$

These cut out triangle $R S T$ (Figure 13), with legs $\frac{ℓ}{2}$ :

P = \frac{ar ( △ R S T )}{ar ( △ O P Q )} = \frac{\frac{1}{2} \cdot \frac{ℓ}{2} \cdot \frac{ℓ}{2}}{ℓ ^{2} /2} = \frac{1}{4}

Figure 13: Sample space

x + y < ℓ

(triangle

O P Q

, area

\frac{ℓ ^{2}}{2}

) and the favourable triangle

R S T

(area

\frac{ℓ ^{2}}{8}

), so

P = \frac{1}{4}

Solved Example 27

Two points are taken at random on a line segment of length

L

. Find the probability that the distance between them is at least

ℓ

, where

ℓ < L

Solution:

Let $A B$ be the segment and $C$ , $D$ the two points with $C$ nearer to $A$ . Put $A C = x$ and $C D = y$ . Then $x + y < L$ : the sample space is triangle $O P Q$ with legs $L$ and area $\frac{1}{2} L^{2}$ .

The event needs $y \geq ℓ$ together with $x + y < L$ : the triangle $R S Q$ with legs $L - ℓ$ and area $\frac{1}{2} (L - ℓ)^{2}$ (Figure 14).

P = \frac{\frac{1}{2} ( L - ℓ ) ^{2}}{\frac{1}{2} L ^{2}} = (\frac{L - ℓ}{L})^{2}

Figure 14: The gap

y \geq ℓ

cuts off triangle

R S Q

with legs

L - ℓ

, similar to

O P Q

, so

P = (\frac{L - ℓ}{L})^{2}

8.5 JEE-style problems

Solved Example 28

P (A \cup B) = \frac{3}{4}

P (A \cap B) = \frac{1}{4}

and

P (A^{'}) = \frac{2}{3}

, then

P (A^{'} \cap B)

equals
(A)

5/12

(B)

3/8

(C)

5/8

(D)

1/4

Solution:

$P (A) = 1 - \frac{2}{3} = \frac{1}{3}$ .
Addition theorem: $P (B) = P (A \cup B) - P (A) + P (A \cap B) = \frac{3}{4} - \frac{1}{3} + \frac{1}{4} = \frac{2}{3}$ .
$P (A^{'} \cap B) = P (B) - P (A \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$ .

Answer: (A) $\frac{5}{12}$ .

Solved Example 29

P (A) = 0.6

and

P (B) = 0.7

, then

P (A \cap B)

always lies in
(A)

[0.3, 0.6]

(B)

[0, 0.6]

(C)

[0.42, 0.6]

(D)

[0.3, 0.7]

Solution:

Lower bound: $P (A \cup B) \leq 1$ forces $P (A \cap B) \geq 0.6 + 0.7 - 1 = 0.3$ . Upper bound: $A \cap B \subseteq A$ gives $P (A \cap B) \leq 0.6$ . Both ends are reachable (Figure 9). Option (C) uses $0.42 = 0.6 \times 0.7$ , which is the value only if $A$ , $B$ are independent.

Answer: (A) $[0.3, 0.6]$ .

Solved Example 30

Three fair dice are thrown. Find the probability that the largest number shown is exactly 5.

Solution:

All three dice show at most 5 in $5^{3} = 125$ ways.
All three show at most 4 in $4^{3} = 64$ ways; these have maximum below 5.
Maximum exactly 5: $125 - 64 = 61$ ways out of $6^{3} = 216$ .

Answer: $\frac{61}{216}$ . In general, P(max $= k$ ) $= \frac{k ^{n} - ( k - 1 ) ^{n}}{6 ^{n}}$ for $n$ dice.

Solved Example 31

The letters of the word ASSASSIN are arranged at random. Find the probability that no two S's are together.

Solution:

Total arrangements: $\frac{8 !}{4 ! 2 !} = 840$ (4 S's, 2 A's).
Arrange the other letters A, A, I, N first: $\frac{4 !}{2 !} = 12$ ways. They create 5 gaps.
Put the 4 S's in 4 different gaps: $^{5} C_{4} = 5$ ways. Favourable $= 12 \times 5 = 60$ .

Answer: $\frac{60}{840} = \frac{1}{14}$ .

Solved Example 32

A number is chosen at random from 1 to 100. Find the probability that it is divisible by 2, 3 or 5.

Solution:

Let $A$ , $B$ , $C$ be divisibility by 2, 3, 5: $n (A) = 50$ , $n (B) = 33$ , $n (C) = 20$ , $n (A \cap B) = 16$ , $n (B \cap C) = 6$ , $n (C \cap A) = 10$ , $n (A \cap B \cap C) = 3$ .

n (A \cup B \cup C) = 50 + 33 + 20 - 16 - 6 - 10 + 3 = 74

Answer: $\frac{74}{100} = \frac{37}{50}$ .

Solved Example 33

Four letters are placed at random in four addressed envelopes, one in each. Find the probability that no letter goes into its own envelope.

Solution:

Let $A_{i}$ : letter $i$ is in its own envelope. By inclusion-exclusion,

P (at least one correct) =^{4} C_{1} \cdot \frac{3 !}{4 !} -^{4} C_{2} \cdot \frac{2 !}{4 !} +^{4} C_{3} \cdot \frac{1 !}{4 !} -^{4} C_{4} \cdot \frac{0 !}{4 !}

= 1 - \frac{1}{2} + \frac{1}{6} - \frac{1}{24} = \frac{15}{24}

Answer: $1 - \frac{15}{24} = \frac{9}{24} = \frac{3}{8}$ . Check by listing: 9 of the $4! = 24$ arrangements are derangements.

Practice Questions

A coin is tossed twice; if the second toss gives a head, a die is thrown. Write the sample space.Answer: HT, TT, HH1 to HH6, TH1 to TH6 (14 outcomes)
Two dice are rolled. $A$ : sum $> 8$ ; $B$ : 2 on either die; $C$ : sum $\geq 7$ and a multiple of 3. Find $A \cap B$ , $B \cap C$ , $A \cap C$ and state which pairs are mutually exclusive.Answer: $A \cap B = ϕ$ , $B \cap C = ϕ$ , $A \cap C$ = {(3,6), (6,3), (4,5), (5,4), (6,6)}; A, B yes; B, C yes; A, C no
Die thrown. $A$ : odd, $B$ : divisible by 3, $C$ : number $\leq 4$ . Find P(at least two of $A, B, C$ ).Answer: $\frac{1}{3}$
For the same $A$ , $B$ , $C$ , find P(exactly one occurs).Answer: $\frac{2}{3}$
A card is drawn from a pack of 52. Find P(king or heart).Answer: $\frac{4 + 13 - 1}{52} = \frac{4}{13}$
The letters of the word SUCCESS are arranged at random. Find the probability that all the S's are together.Answer: $\frac{5 ! /2 !}{7 ! / ( 3 ! 2 !)} = \frac{60}{420} = \frac{1}{7}$
A segment of length $a$ is cut at a random point. Find the probability that neither part exceeds $b$ , where $a > b$ and $2 b > a$ (geometric).Answer: $\frac{2 b - a}{a}$
A 10 m cloth is divided at random among three brothers. Find the probability that nobody gets more than 4 m (geometric).Answer: $\frac{1}{25}$

Common Mistakes to Avoid

Watch out

Treating $(1, 2)$ and $(2, 1)$ as one outcome for two dice. The 36 ordered pairs are equally likely; unordered pairs are not.
Using $\frac{n ( A )}{n ( S )}$ when outcomes are not equally likely, as in a coin followed by a die. Multiply along tree branches instead.
Converting odds $m : n$ to $P = \frac{m}{n}$ . The correct value is $\frac{m}{m + n}$ .
Forgetting to subtract $P (A \cap B)$ in "A or B" questions, which double counts outcomes such as the king of hearts in "king or heart".
Mixing ordered and unordered counting: if the denominator uses $^{n} C_{r}$ , the numerator must also count unordered selections.
Assuming mutually exclusive events are independent. Two events with non-zero probabilities cannot be both.
Using $- 2 P (A \cap B \cap C)$ for "exactly two"; that coefficient belongs to "at least two". Exactly two uses $- 3$ .
Stopping at $\sum P (A) - \sum P (A \cap B)$ for three events and forgetting to add back $P (A \cap B \cap C)$ .

Frequently Asked Questions

What is the classical definition of probability?

If an experiment has $N$ equally likely, mutually exclusive and exhaustive outcomes and $m$ of them favour event $A$ , then $P (A) = \frac{m}{N}$ . It works only when every outcome has the same chance; otherwise use weights or a probability tree, multiplying along branches.

What is the difference between mutually exclusive and exhaustive events?

Mutually exclusive events cannot happen together, so they have no common outcome. Exhaustive events together cover the whole sample space, so at least one must happen. Events that are both form a partition of $S$ , and their probabilities add up to exactly 1.

How do you convert odds into probability?

If the odds in favour of an event are $m : n$ , the event succeeds in $m$ of every $m + n$ equally likely cases, so $P = \frac{m}{m + n}$ . Odds against $m : n$ mean $P = \frac{n}{m + n}$ . For odds $2 : 5$ in favour, the probability is $\frac{2}{7}$ , not $\frac{2}{5}$ .

What is the formula for the probability of A or B or C?

$P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)$ . Pairwise overlaps are subtracted because they were counted twice, and the triple overlap is added back because it was removed too often.

Why is a sum of 7 the most likely total with two dice?

Sum 7 can be made in 6 ordered ways, $(1, 6)$ through $(6, 1)$ , more than any other total. In general a total $s$ has $6 - ∣ s - 7∣$ ways, which peaks at $s = 7$ . So $P (sum 7) = \frac{6}{36} = \frac{1}{6}$ , while totals 2 and 12 have only $\frac{1}{36}$ each.

Is an event with probability zero always impossible?

In a finite sample space of equally likely outcomes, yes: only the empty event has probability 0. With continuous outcomes it is not: the chance that a random point on a segment lands exactly at its midpoint is 0, yet it can happen. Probability measures likelihood, not possibility.

Which basic probability topics are asked in JEE Main?

The JEE Main syllabus lists probability of an event, the addition and multiplication theorems, Bayes' theorem and the probability distribution of a random variate. Most introductory questions combine classical probability with permutations and combinations, such as selecting balls, cards or digits, so counting skill matters as much as formulas.

Is geometric probability part of the JEE Advanced syllabus?

The JEE Advanced 2026 syllabus lists random experiments, types of events, addition and multiplication rules, conditional probability, independence, total probability, Bayes' theorem and counting-based probability. Geometric (length or area based) probability is not named, so treat it as enrichment rather than a core topic.

Previous year questions on Introduction To Probability

28 questions from past papers, each with a step-by-step solution.

Show all 28 questions

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