Probability Formulas for JEE Main and Advanced

MathsClass 11 and 12, ProbabilityFor JEE Main and Advanced

Probability formulas for JEE Main and Advanced on one page: the basic rules and the addition rule, conditional probability and independence, total probability and Bayes' theorem, mean and variance of a random variable and the binomial distribution, each with the condition under which it holds. The table shows which formulas decided the most questions in recent JEE papers, and section 6 solves ten of them step by step. Probability is a chapter in both Class 11 and Class 12.

Which formulas JEE uses most

We read all 53 questions on probability in our 48 past papers (JEE Main 2025-2026, JEE Advanced 2022-2026) and noted the formulas each solution needs; one question can need several. The formulas used most:

Formula	Questions
$P (E) = \frac{n ( E )}{n ( S )}$ for equally likely outcomes	1834%
Random variable: $\sum p_{i} = 1$ , $E (X) = \sum x_{i} p_{i}$ , $Var (X) = E (X^{2}) - [E (X)]^{2}$	1223%
Bayes' theorem: $P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum _{j} P ( E _{j} ) P ( A ∣ E _{j} )}$	917%
$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ and $P (A \cap B) = P (A) P (B ∣ A)$	815%
Total probability: $P (A) = \sum_{i} P (E_{i}) P (A ∣ E_{i})$	713%
Independent events: $P (A \cap B) = P (A) P (B)$ ; repeated independent trials	713%
$P (E^{'}) = 1 - P (E)$ ; at least one $= 1 -$ none	611%
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , and the three-event form	36%

Show the other 1

Formula	Questions
Binomial: $P (X = r) =^{n} C_{r} p^{r} q^{n - r}$ , mean $n p$ , variance $n pq$	24%

1. Basic rules

Probability	Equals	Note
$P (E)$	$\frac{n ( E )}{n ( S )}$	Outcomes equally likely; $0 \leq P (E) \leq 1$
$P (E^{'})$	$1 - P (E)$	Use it for at least one: $1 - P (none)$
$P (A \cup B)$	$P (A) + P (B) - P (A \cap B)$	$= P (A) + P (B)$ when $A$ , $B$ are mutually exclusive
$P (A \cup B \cup C)$	$\sum P (A) - \sum P (A \cap B) + P (A \cap B \cap C)$	Three sums: the singles, the three pairs, then the triple
$P (A \cap B^{'})$	$P (A) - P (A \cap B)$	$A$ but not $B$
$P (A^{'} \cap B^{'})$	$1 - P (A \cup B)$	Neither (De Morgan)
$P (A^{'} \cup B^{'})$	$1 - P (A \cap B)$	Not both (De Morgan)
$P (exactly one of A, B)$	$P (A) + P (B) - 2 P (A \cap B)$

2. Conditional probability and independence

Conditional probability

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) \neq = 0

Multiplication rule: $P (A \cap B) = P (A) P (B ∣ A)$ , and for three events $P (A \cap B \cap C) = P (A) P (B ∣ A) P (C ∣ A \cap B)$ . (The events you condition on must have non-zero probability.) Also $P (A^{'} ∣ B) = 1 - P (A ∣ B)$ .

Independent events

P (A \cap B) = P (A) P (B)

Equivalently, when $P (B) \neq = 0$ , $P (A ∣ B) = P (A)$ . If $A$ , $B$ are independent, so are $A^{'}$ and $B$ , $A$ and $B^{'}$ , $A^{'}$ and $B^{'}$ .

At least one of mutually independent events $A_{1}, \dots, A_{n}$ :
$P (at least one) = 1 - i = 1 \prod n (1 - P (A_{i}))$
Three events are mutually independent exactly when all four product rules hold: the three pairs and $P (A \cap B \cap C) = P (A) P (B) P (C)$ . Pairwise independence is not enough.
Mutually exclusive is not independent: if $A \cap B = \emptyset$ with $P (A), P (B) > 0$ , then $P (A \cap B) = 0 \neq = P (A) P (B)$ .
Repeated independent trials: with success probability $p$ on each trial and $q = 1 - p$ , the first success comes on trial $k$ with probability $q^{k - 1} p$ . When two players take turns and the first to succeed wins, the player who starts wins with probability $\frac{p _{1}}{1 - q _{1} q _{2}}$ , where $p_{1}$ , $p_{2}$ are the players' chances on one turn and $q_{1} = 1 - p_{1}$ , $q_{2} = 1 - p_{2}$ .

3. Total probability and Bayes' theorem

$E_{1}, E_{2}, \dots, E_{n}$ split the sample space: they are mutually exclusive, cover every outcome and each has $P (E_{i}) > 0$ . $A$ is any event (with $P (A) \neq = 0$ for Bayes' theorem).

Total probability

P (A) = i = 1 \sum n P (E_{i}) P (A ∣ E_{i})

Use it when the event can happen through several routes: a bag is chosen first, a ball is moved from one bag to another, a coin is picked from a mixed lot.

Bayes' theorem

P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{\sum _{j} P ( E _{j} ) P ( A ∣ E _{j} )}

Use it when the result $A$ is known and the question asks which route it came from. The denominator is $P (A)$ from the total-probability rule.

4. Random variables and the binomial distribution

	Formula	Note
Probabilities add to 1	$\sum p_{i} = 1$	Find an unknown $k$ in a table first
Mean	$μ = E (X) = \sum x_{i} p_{i}$
Variance	$σ^{2} = E (X^{2}) - μ^{2} = \sum x_{i}^{2} p_{i} - μ^{2}$	$σ$ is the standard deviation
Linear change	$E (a X + b) = a E (X) + b$ , $Var (a X + b) = a^{2} Var (X)$	Adding $b$ does not change the variance

Binomial distribution

P (X = r) =^{n} C_{r} p^{r} q^{n - r}, q = 1 - p

$n$ independent trials, each with two outcomes and the same success probability $p$ ; $X$ counts the successes, $r = 0, 1, \dots, n$ . Mean $n p$ , variance $n pq$ , and $P (X \geq 1) = 1 - q^{n}$ .

Drawing with replacement gives a binomial distribution. Drawing without replacement does not: count with $^{n} C_{r}$ instead (as in PYQ 5 below).

5. Exam traps

Watch out

Mutually exclusive ( $A \cap B = \emptyset$ ) and independent ( $P (A \cap B) = P (A) P (B)$ ) are different ideas; two events with positive probability cannot be both.
$P (A ∣ B) \neq = P (B ∣ A)$ in general; Bayes' theorem connects them.
$P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ , not $1 - P (A \cap B)$ (De Morgan).
In Bayes questions the denominator is the total probability of what was observed, over every route.
Without replacement the draws are not independent: use $^{n} C_{r}$ counts, not the binomial formula.
Find the unknown constant in a probability table from $\sum p_{i} = 1$ (and $p_{i} \geq 0$ ) before anything else.
$Var (X) = E (X^{2}) - [E (X)]^{2}$ : $E (X^{2}) = \sum x_{i}^{2} p_{i}$ is the mean of the squares, not the square of the mean.
For three events, pairwise independence does not give $P (A \cap B \cap C) = P (A) P (B) P (C)$ .

6. Past-paper questions, solved

Ten questions from our past papers, chosen to cover the formulas used most. The question and the answer key are as set in the paper; the working is ours.

PYQ 1: Bayes' theorem

JEE Main 2026 Apr 8 Shift 2, Mathematics Q6

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\frac{2}{5}$ , $\frac{1}{5}$ and $\frac{2}{5}$ . The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5}$ , $\frac{1}{3}$ and $\frac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :

(A) $\frac{11}{37}$
(B) $\frac{12}{37}$
(C) $\frac{13}{37}$
(D) $\frac{14}{37}$

Solution:

Routes: bus, scooter, car with $P = \frac{2}{5}, \frac{1}{5}, \frac{2}{5}$ ; late on each with $\frac{1}{5}, \frac{1}{3}, \frac{1}{4}$ .
Total probability of being late:
$P (L) = \frac{2}{25} + \frac{1}{15} + \frac{1}{10} = \frac{12 + 10 + 15}{150} = \frac{37}{150}$
Bayes:
$P (bus ∣ L) = \frac{2/25}{37/150} = \frac{12}{37}$

Answer key (B) $\frac{12}{37}$

PYQ 2: Bayes' theorem

JEE Advanced 2024 Paper 1, Mathematics Section 1 Q2

A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\frac{1}{2}$ . Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$ . Then the probability that the student knows the answer of a randomly chosen question is

(A) $\frac{1}{12}$
(B) $\frac{1}{7}$
(C) $\frac{5}{7}$
(D) $\frac{5}{12}$

Solution:

Let $k = P (knows)$ . A known answer is always correct; a guess is correct with probability $\frac{1}{2}$ , so $P (correct) = k + \frac{1 - k}{2}$ .
Bayes, with $G$ for guessed and $C$ for correct:
$P (G ∣ C) = \frac{( 1 - k ) /2}{k + ( 1 - k ) /2} = \frac{1}{6}$
Cross-multiply: $3 (1 - k) = k + \frac{1 - k}{2}$ , so $6 - 6 k = k + 1$ and $k = \frac{5}{7}$ .

Answer key (C) $\frac{5}{7}$

PYQ 3: Total probability

JEE Main 2026 Jan 23 Shift 2, Mathematics Q11

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\frac{p}{q}$ , $g cd (p, q) = 1$ , then $p + q$ is equal to

(A) 22
(B) 23
(C) 24
(D) 21

Solution:

The ball moved from bag B is white with probability $\frac{6}{10}$ (bag A then has $10$ white out of $18$ ) or black with probability $\frac{4}{10}$ ( $9$ white out of $18$ ).
$P (white) = \frac{6}{10} \cdot \frac{10}{18} + \frac{4}{10} \cdot \frac{9}{18} = \frac{96}{180} = \frac{8}{15}$
So $p + q = 8 + 15 = 23$ .

Answer key (B) 23

PYQ 4: Binomial distribution

JEE Main 2026 Jan 24 Shift 1, Mathematics Q17

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is :-

(A) $\frac{7}{1 0 ^{7}}$
(B) $\frac{81}{1 0 ^{8}}$
(C) $\frac{67}{1 0 ^{8}}$
(D) $\frac{73}{1 0 ^{8}}$

Solution:

With replacement, the $8$ draws are independent with $p = \frac{10}{100} = \frac{1}{10}$ (defective) and $q = \frac{9}{10}$ .
$P (X \geq 7) = P (7) + P (8) =^{8} C_{7} (\frac{1}{10})^{7} \frac{9}{10} + (\frac{1}{10})^{8} = \frac{72 + 1}{1 0 ^{8}}$
So the probability is $\frac{73}{1 0 ^{8}}$ .

Answer key (D) $\frac{73}{1 0 ^{8}}$

PYQ 5: Mean and variance

JEE Main 2025 Jan 28 Shift 1, Mathematics Q19

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denotes the number of defective oranges, then the variance of $x$ is :

(A) $\frac{28}{75}$
(B) $\frac{14}{25}$
(C) $\frac{26}{75}$
(D) $\frac{18}{25}$

Solution:

Two oranges from $10$ ( $3$ defective), without replacement; $X$ is the number of defective ones:
$P (0) P (1) P (2) = \frac{^{7} C _{2}}{^{10} C _{2}} = \frac{21}{45} = \frac{7 \cdot 3}{45} = \frac{21}{45} = \frac{^{3} C _{2}}{45} = \frac{3}{45}$
$E (X) E (X^{2}) = \frac{21 + 2 \cdot 3}{45} = \frac{3}{5} = \frac{21 + 4 \cdot 3}{45} = \frac{11}{15}$
$Var (X) = \frac{11}{15} - \frac{9}{25} = \frac{28}{75}$

Answer key (A) $\frac{28}{75}$

PYQ 6: Conditional probability, addition rule, De Morgan

JEE Main 2025 Jan 22 Shift 2, Mathematics Q17

If $A$ and $B$ are two events such that $P (A \cap B) = 0.1$ , and $P (A ∣ B)$ and $P (B ∣ A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P ( A ˉ \cup B ˉ )}{P ( A ˉ \cap B ˉ )}$ is :

(A) $\frac{5}{3}$
(B) $\frac{4}{3}$
(C) $\frac{9}{4}$
(D) $\frac{7}{4}$

Solution:

$12 x^{2} - 7 x + 1 = (3 x - 1) (4 x - 1)$ , so $P (A ∣ B)$ and $P (B ∣ A)$ are $\frac{1}{3}$ and $\frac{1}{4}$ in some order.
$P (B) = \frac{P ( A \cap B )}{P ( A ∣ B )}$ and $P (A) = \frac{P ( A \cap B )}{P ( B ∣ A )}$ : in either order ${P (A), P (B)} = {0.3, 0.4}$ , so $P (A \cup B) = 0.3 + 0.4 - 0.1 = 0.6$ .
De Morgan: $P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = 1 - P (A \cap B) = 0.9$ and $P (\overset{ˉ}{A} \cap \overset{ˉ}{B}) = 1 - P (A \cup B) = 0.4$ give the ratio $\frac{0.9}{0.4}$ , which is $\frac{9}{4}$ .

Answer key (C) $\frac{9}{4}$

PYQ 7: Independent trials in turns

JEE Main 2025 Jan 24 Shift 1, Mathematics Q11

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

(A) $\frac{9}{17}$
(B) $\frac{9}{19}$
(C) $\frac{8}{17}$
(D) $\frac{8}{19}$

Solution:

On one throw of two dice: a sum of $5$ in $4$ ways, so $p_{A} = \frac{1}{9}$ ; a sum of $8$ in $5$ ways, so $p_{B} = \frac{5}{36}$ .
A wins on throw $1$ , or after one full round with no success, or after two, and so on:
$P (A) = p_{A} + q_{A} q_{B} p_{A} + (q_{A} q_{B})^{2} p_{A} + \dots = \frac{p _{A}}{1 - q _{A} q _{B}}$
$q_{A} q_{B} P (A) = \frac{8}{9} \cdot \frac{31}{36} = \frac{62}{81} = \frac{1/9}{19/81} = \frac{9}{19}$

Answer key (B) $\frac{9}{19}$

PYQ 8: Addition rule for three events

JEE Advanced 2022 Paper 1, Mathematics Section 1 Q3

In a study about a pandemic, data of 900 persons was collected. It was found that

190 persons had symptom of fever,
220 persons had symptom of cough,
220 persons had symptom of breathing problem,
330 persons had symptom of fever or cough or both,
350 persons had symptom of cough or breathing problem or both,
340 persons had symptom of fever or breathing problem or both,
30 persons had all three symptoms (fever, cough and breathing problem).

If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is _____________.

Solution:

From $n (F \cup C) = n (F) + n (C) - n (F \cap C)$ : $n (F \cap C) = 80$ ; likewise $n (C \cap B) = 90$ and $n (F \cap B) = 70$ .
Add the three sets ( $190 + 220 + 220 = 630$ ), take away the three pairs ( $80 + 90 + 70 = 240$ ) and add back the people with all three:
$n (F \cup C \cup B) = 630 - 240 + 30 = 420$
so $900 - 420 = 480$ people have no symptom.
Only fever $= 190 - 80 - 70 + 30 = 70$ , only cough $= 80$ , only breathing $= 90$ : exactly one symptom for $240$ people.
At most one symptom: $480 + 240 = 720$ , and $P = \frac{720}{900} = 0.8$ .

Answer key 0.8

PYQ 9: Independence and conditional probability

JEE Advanced 2026 Paper 1, Mathematics Section 2 Q1

Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_{1}$ be the event that the ball chosen belonged to Box I and let $E_{2}$ be the event that the ball chosen belonged to Box II. Let $F_{1}$ be the event that the ball chosen is red and let $F_{2}$ be the event that the ball chosen is green.

Then which of the following statements is (are) TRUE ?

(A) The events $E_{1}$ and $F_{1}$ are independent
(B) The events $E_{2}$ and $F_{2}$ are dependent
(C) The conditional probability $P (F_{1} ∣ E_{1})$ is equal to the conditional probability $P (F_{1} ∣ E_{2})$
(D) The conditional probability $P (F_{1} ∣ E_{1})$ is greater than the conditional probability $P (F_{2} ∣ E_{2})$

Solution:

All $35$ balls together: $14$ red, $21$ green; Box I gave $15$ ( $6$ red), Box II $20$ ( $8$ red). So $P (E_{1}) = \frac{3}{7}$ , $P (F_{1}) = \frac{2}{5}$ and $P (E_{1} \cap F_{1}) = \frac{6}{35}$ .
(A) $\frac{3}{7} \cdot \frac{2}{5} = \frac{6}{35} = P (E_{1} \cap F_{1})$ : independent, true. (B) $P (E_{2} \cap F_{2}) = \frac{12}{35} = \frac{4}{7} \cdot \frac{3}{5}$ : independent too, so false.
(C) and (D):
$P (F_{1} ∣ E_{1}) P (F_{1} ∣ E_{2}) P (F_{2} ∣ E_{2}) = \frac{6}{15} = \frac{2}{5} = \frac{8}{20} = \frac{2}{5} = \frac{12}{20} = \frac{3}{5}$
So (C) is true, and (D) is false, since $\frac{2}{5} < \frac{3}{5}$ .

Answer key (A) The events $E_{1}$ and $F_{1}$ are independent, (C) The conditional probability $P (F_{1} ∣ E_{1})$ is equal to the conditional probability $P (F_{1} ∣ E_{2})$

PYQ 10: Complement

JEE Main 2026 Jan 22 Shift 1, Mathematics Q4

Two distinct numbers a and b are selected at random from 1, 2, 3, …, 50. The probability, that their product ab is divisible by 3, is

(A) $\frac{561}{1225}$
(B) $\frac{664}{1225}$
(C) $\frac{272}{1225}$
(D) $\frac{8}{25}$

Solution:

$ab$ is divisible by $3$ unless neither number is. $1$ to $50$ has $16$ multiples of $3$ and $34$ other numbers.
$P = 1 - \frac{^{34} C _{2}}{^{50} C _{2}} = 1 - \frac{561}{1225} = \frac{664}{1225}$

Answer key (B) $\frac{664}{1225}$

Every formula on this page is checked with a computer algebra system, and every worked answer is recomputed and matched with the answer key. Past papers: JEE Main 2025-2026, JEE Advanced 2022-2026; the questions were read by hand (lists checked 2026-10-15).

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