From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then…

Year	2026	2025	2024	Total
Questions	9	17	9	35

More Probability Previous-Year Questions — Page 2

Q24 jee_main_2026_24_january_evening Sets and Probability

Let

S

be a set of 5 elements and

P(S)

denote the power set of

S

. Let

E

be an event of choosing an ordered pair

(A, B)

from the set

P(S) \times P(S)

such that

A B =

. If the probability of the event

E

3p2q

, where

p, q in N

, then

p + q

is equal to

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

Probability P(E) = Number of favorable outcomesTotal number of outcomes

Core Logic

Let $S = a, b, c, d, e$ $S = \{a, b, c, d, e\}$. Since $S$ $S$ has 5 elements, $P(S)$ $P(S)$ contains $2⁵ = 32$ $2^5 = 32$ subsets. The total number of ordered pairs $(A, B)$ $(A, B)$ that can be formed from $P(S) \times P(S)$ $P(S) \times P(S)$ is:

32 \times 32 = (2⁵)² = 2¹⁰ = 4⁵

Alternatively, consider element-wise mapping. For each of the 5 elements in $S$ $S$, it has 4 choices with respect to sets $A$ $A$ and $B$ $B$:

Status in A	Status in B
Present ($\checkmark$)	Present ($\checkmark$)
Present ($\checkmark$)	Absent (x)
Absent (x)	Present ($\checkmark$)
Absent (x)	Absent (x)

Step 1: Satisfying the Condition

For $A B =$ $A \cap B = \emptyset$, no element can be present in both $A$ $A$ and $B$ $B$ simultaneously. This rules out the choice where an element is ($\checkmark$) in $A$ $A$ and ($\checkmark$) in $B$ $B$.

Thus, each of the 5 elements has exactly 3 valid choices to ensure disjointness. Favorable cases = $3⁵$ $3^5$.

Step 2: Calculating Probability

Probability $P = FavorableTotal$ :

P = (3⁵)/(4⁵) = (3⁵)/((2²)⁵) = 3⁵2¹⁰

Comparing this with $(3^p)/(2^q)$ $\frac{3^p}{2^q}$:

p = 5, q = 10

p + q = 5 + 10 = 15

Pattern Recognition

Set operations mapping down to element-wise Boolean states (In/Out) transform combinatorial subset problems directly into base-state exponentiation problems ( $3ⁿ$ $3^n$ vs $4ⁿ$ $4^n$). Disjoint sets exclude exactly one state: (In, In).

Chapter Mix

Class 12 Maths: Probability Class 11 Maths: Sets

Q9 jee_main_2026_28_january_morning Bayes Theorem

A bag contains 10 balls out of which

k

are red and

(10 - k)

are black, where

0 \leq k \leq 10

. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is :

A. $(7)/(11)$ $\frac{7}{11}$
B. $(7)/(55)$ $\frac{7}{55}$
C. $(7)/(110)$ $\frac{7}{110}$
D. $(14)/(55)$ $\frac{14}{55}$

Solution

Related Formula

P(Eᵢ | A) = (P(A|Eᵢ)P(Eᵢ))/(Σ P(A|Eⱼ)P(Eⱼ))

Core Logic

Let event $A$ $A$ be drawing 3 black balls. Let $Ek$ $E_k$ be the event that the bag initially has $k$ $k$ red balls and $(10-k)$ $(10-k)$ black balls. Assuming all compositions (values of $k$ $k$ from 0 to 10) are equally likely initially, $P(Ek) = (1)/(11)$ $P(E_k) = \frac{1}{11}$ for all $k$ $k$. We want to find $P(E₁ | A)$ $P(E_1 | A)$.

Step 1: Set up the Conditional Probabilities

P(A | Ek) = 10-kC₃¹⁰C₃

This is only non-zero when $10-k \geq 3 k \leq 7$ $10-k \geq 3 \implies k \leq 7$.

Using Bayes' theorem:

P(E₁ | A) = P(A | E₁) P(E₁)Σk=0⁷ P(A | Ek) P(Ek)

Since $P(Ek)$ $P(E_k)$ is constant, it cancels out:

P(E₁ | A) = ⁹C₃ / ¹⁰C₃Σk=0⁷ (10-kC₃ / ¹⁰C₃) = ⁹C₃Σk=0⁷ 10-kC₃

Step 2: Series Summation

The denominator is:

Σk=0⁷ 10-kC₃ = ¹⁰C₃ + ⁹C₃ + ⁸C₃ + + ³C₃

Using the identity $Σr=mⁿ rCm = ⁿ⁺¹Cm+1$ (Hockey-stick identity):

³C₃ + ⁴C₃ + + ¹⁰C₃ = ¹¹C₄

Step 3: Final Probability Calculation

P(E₁ | A) = ⁹C₃¹¹C₄ = ((9 \times 8 \times 7)/(3 \times 2 \times 1))/((11 \times 10 \times 9 \times 8)/(4 \times 3 \times 2 \times 1)) = (84)/(330) = (14)/(55)

Pattern Recognition

When picking items and calculating inverse probabilities over all uniform possible states, the denominator transforms into a simple combinations sum evaluated via the hockey-stick identity: $Σ rCk = r+1Ck+1$ $\sum ^{r}C_k = ^{r+1}C_{k+1}$.

Chapter Mix

Class 12 Mathematics: Probability Class 11 Mathematics: Permutations and Combinations

Q4 jee_main_2026_28_january_evening Random Variables and Expectation

The probability distribution of a random variable

X

is given below :

$X$ $X$	$4k$ $4k$	$(30)/(7)k$ $\frac{30}{7}k$	$(32)/(7)k$ $\frac{32}{7}k$	$(34)/(7)k$ $\frac{34}{7}k$	$(36)/(7)k$ $\frac{36}{7}k$	$(38)/(7)k$ $\frac{38}{7}k$	$(40)/(7)k$ $\frac{40}{7}k$	$6k$ $6k$
$P(X)$ $P(X)$	$(2)/(15)$ $\frac{2}{15}$	$(1)/(15)$ $\frac{1}{15}$	$(2)/(15)$ $\frac{2}{15}$	$(1)/(5)$ $\frac{1}{5}$	$(1)/(15)$ $\frac{1}{15}$	$(2)/(15)$ $\frac{2}{15}$	$(1)/(5)$ $\frac{1}{5}$	$(1)/(15)$ $\frac{1}{15}$

E(X) = (263)/(15)

, then

P(X < 20)

is equal to:

A. $(3)/(5)$ $\frac{3}{5}$
B. $(8)/(15)$ $\frac{8}{15}$
C. $(11)/(15)$ $\frac{11}{15}$
D. $(14)/(15)$ $\frac{14}{15}$

Solution

Related Formula

E(X) = Σ Xᵢ P(Xᵢ)

Core Logic

Calculate the expected value:

E(X) = (4k)(2)/(15) + ((30k)/(7))(1)/(15) + ((32k)/(7))(2)/(15) + ((34k)/(7))(1)/(5) + ((36k)/(7))(1)/(15) + ((38k)/(7))(2)/(15) + ((40k)/(7))(1)/(5) + (6k)(1)/(15)

Factor out $k$ $k$ and sum the products to solve for $k$ $k$ using the given condition $E(X) = (263)/(15)$ $E(X) = \frac{263}{15}$.

Execution

Summing the terms:

E(X) = (k)/(15 \times 7) [ (28 \times 2) + 30(1) + 32(2) + 34(3) + 36(1) + 38(2) + 40(3) + 42(1) ]

E(X) = (k)/(105) [ 56 + 30 + 64 + 102 + 36 + 76 + 120 + 42 ] = (526k)/(105)

Given $E(X) = (263)/(15)$ $E(X) = \frac{263}{15}$:

(526k)/(105) = (263)/(15) \Rightarrow (526k)/(7) = 263 \Rightarrow 2k = 7 \Rightarrow k = (7)/(2)

Substitute $k$ $k$ to find the actual values of $X$ $X$: $4k = 14$ $4k = 14$, $(30)/(7)k = 15$ $\frac{30}{7}k = 15$, $(32)/(7)k = 16$ $\frac{32}{7}k = 16$, $(34)/(7)k = 17$ $\frac{34}{7}k = 17$, $(36)/(7)k = 18$ $\frac{36}{7}k = 18$, $(38)/(7)k = 19$ $\frac{38}{7}k = 19$, $(40)/(7)k = 20$ $\frac{40}{7}k = 20$, $6k = 21$ $6k = 21$.

We need $P(X < 20)$ $P(X < 20)$:

P(X < 20) = ΣX=14¹⁹ P(X)

= (2)/(15) + (1)/(15) + (2)/(15) + (3)/(15) + (1)/(15) + (2)/(15) = (11)/(15)

Pattern Recognition

When $E(X)$ $E(X)$ is provided with a parameterized state variable $k$ $k$, summing the discrete expected value directly unlocks the exact sample space scaling.

Chapter Mix

Class 12 Maths: Probability

Q60 jee_main_2025_02_april_evening Bayes' Theorem

Given three identical bags each containing 10 balls, whose colours are as follows:

array|l|l|l|l| & Red & Blue & Green Bag I & 3 & 2 & 5 Bag II & 4 & 3 & 3 Bag III & 5 & 1 & 4 array

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is

p

and if the ball is Green, the probability that it is from bag III is

q

, then the value of

((1)/(p) + (1)/(q))

is:

A. 6
B. 9
C. 7
D. 8

Solution

Related Formula

Bayes' Theorem: P(E₁|A) = P(E₁) P(A|E₁)Σi=1ⁿ P(Eᵢ) P(A|Eᵢ)

Core Logic

This is a conditional probability problem. We apply Bayes' Theorem twice: first for the Red ball, then for the Green ball.

Step 1: Solve for p (Red ball)

Let $E₁, E₂, E₃$ $E_1, E_2, E_3$ be the events of choosing Bag I, Bag II, and Bag III respectively. Since bags are identical, $P(E₁) = P(E₂) = P(E₃) = (1)/(3)$ .

The probabilities of drawing a Red ball from each bag are:

$P(R|E₁) = (3)/(10)$ $P(R|E_1) = \frac{3}{10}$
$P(R|E₂) = (4)/(10)$ $P(R|E_2) = \frac{4}{10}$
$P(R|E₃) = (5)/(10)$ $P(R|E_3) = \frac{5}{10}$

Applying Bayes' Theorem:

p = P(E₁|R) = (P(E₁) P(R|E₁))/(P(E₁)P(R|E₁) + P(E₂)P(R|E₂) + P(E₃)P(R|E₃))

p = ((3)/(10))/((3)/(10) + (4)/(10) + (5)/(10)) = (3)/(12) = (1)/(4)

Thus, $(1)/(p) = 4$ $\frac{1}{p} = 4$.

Step 2: Solve for q (Green ball)

The probabilities of drawing a Green ball from each bag are:

$P(G|E₁) = (5)/(10)$ $P(G|E_1) = \frac{5}{10}$
$P(G|E₂) = (3)/(10)$ $P(G|E_2) = \frac{3}{10}$
$P(G|E₃) = (4)/(10)$ $P(G|E_3) = \frac{4}{10}$

Applying Bayes' Theorem:

q = P(E₃|G) = (P(E₃) P(G|E₃))/(P(E₁)P(G|E₁) + P(E₂)P(G|E₂) + P(E₃)P(G|E₃))

q = ((4)/(10))/((5)/(10) + (3)/(10) + (4)/(10)) = (4)/(12) = (1)/(3)

Thus, $(1)/(q) = 3$ $\frac{1}{q} = 3$.

Step 3: Calculate the requested value

Sum the inverse values:

(1)/(p) + (1)/(q) = 4 + 3 = 7

Pattern Recognition

Simplification of Bayes' denominator: Since all prior events have identical probability $P(Eᵢ) = 1/k$ $P(E_i) = 1/k$, they cancel out of the Bayes' fraction entirely, allowing you to work directly with the raw ball counts.

Chapter Mix

Class 12 Mathematics: Probability

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Binomial Distribution

Step 2: Sum the Probabilities

Pattern Recognition

Chapter Mix

Reference Study Guides

More Probability Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Satisfying the Condition

Step 2: Calculating Probability

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up the Conditional Probabilities

Step 2: Series Summation

Step 3: Final Probability Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for p (Red ball)

Step 2: Solve for q (Green ball)

Step 3: Calculate the requested value

Pattern Recognition

Chapter Mix

More Probability Questions — jee_main_2026_24_january_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

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