Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn…

Year	2026	2025	2024	Total
Questions	9	17	9	35

Q10 jee_main_2026_21_jan_evening Probability Distribution

A random variable X takes values 0, 1, 2, 3 with probabilities

(2a + 1)/(30), (8a - 1)/(30), (4a + 1)/(30)

, b respectively, where

a, b in R

. Let

μ

and

σ

respectively be the mean and standard deviation of X such that

σ² + μ² = 2

. Then

(a)/(b)

is equal to:

A. $30$ $30$
B. $3$ $3$
C. $60$ $60$
D. $12$ $12$

Solution

Related Formula

Sum of probabilities: Σ P(X=xᵢ) = 1

Variance formula: σ² = E(X²) - μ² E(X²) = σ² + μ² = 2

E(X²) = Σ xᵢ² P(xᵢ)

Core Logic

Given Probability Distribution:

x	0	1	2	3
p(x)	$(2a+1)/(30)$ $\frac{2a+1}{30}$	$(8a-1)/(30)$ $\frac{8a-1}{30}$	$(4a+1)/(30)$ $\frac{4a+1}{30}$	$b$ $b$

Step 1: Set up variance equation

We know $σ² + μ² = Σ xᵢ² P(xᵢ) = 2$ .

0²((2a+1)/(30)) + 1²((8a-1)/(30)) + 2²((4a+1)/(30)) + 3²(b) = 2

(8a-1)/(30) + (16a+4)/(30) + 9b = 2

(24a+3)/(30) + 9b = 2

24a + 270b + 3 = 60 24a + 270b = 57

Dividing by 3:

8a + 90b = 19 (1)

Step 2: Total Probability Equation

Sum of all probabilities equals 1:

(2a+1)/(30) + (8a-1)/(30) + (4a+1)/(30) + b = 1

(14a+1)/(30) + b = 1

14a + 30b + 1 = 30 14a + 30b = 29 (2)

Step 3: Solve the Linear System

From (2), multiply by 3: $42a + 90b = 87$ $42a + 90b = 87$. Subtract (1) from this new equation:

(42a + 90b) - (8a + 90b) = 87 - 19

34a = 68 a = 2

Substitute $a = 2$ $a = 2$ back into (1):

8(2) + 90b = 19 16 + 90b = 19 90b = 3 b = (1)/(30)

We need $(a)/(b)$ $\frac{a}{b}$:

(a)/(b) = (2)/(1/30) = 60

Pattern Recognition

Notice that $σ² + μ²$ $\sigma^2 + \mu^2$ is simply the second moment $E(X²)$ $E(X^2)$. Avoid calculating $μ$ $\mu$ independently. Create a simultaneous system using $E(X²)=c$ $E(X^2)=c$ and $Σ p=1$ $\sum p=1$.

Chapter Mix

Class 12 Maths: Probability

Q4 jee_main_2026_22_january_morning Divisibility In Probability

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

Solution

Related Formula

P(Event) = 1 - P(Complement of Event)

Core Logic

For the product $ab$ $ab$ to be divisible by 3, at least one of the numbers $a$ $a$ or $b$ $b$ must be a multiple of 3.

It is easier to find the probability of the complement event: neither $a$ $a$ nor $b$ $b$ is a multiple of 3.

Step 1: Counting Favorable vs Total Outcomes

Total numbers = 50. Number of multiples of 3 in the set $1, 2, \dots, 50$ $\{1, 2, \ldots, 50\}$ is $(50)/(3) = 16$ $\lfloor \frac{50}{3} \rfloor = 16$.

Numbers that are NOT multiples of 3 = $50 - 16 = 34$ $50 - 16 = 34$.

The probability that both chosen numbers are NOT multiples of 3 is the number of ways to choose 2 numbers from the 34, divided by the total ways to choose 2 numbers from 50:

P(Not divisible by 3) = ³⁴C₂⁵⁰C₂

³⁴C₂ = (34 \times 33)/(2) = 561

⁵⁰C₂ = (50 \times 49)/(2) = 1225

Step 2: Final Probability

Required probability $P(ab is divisible by 3) = 1 - P(Not divisible by 3)$

P = 1 - (561)/(1225) = (1225 - 561)/(1225) = (664)/(1225)

Pattern Recognition

Whenever a probability question asks for 'at least one' condition (like a product being divisible by a prime), always use the complement rule: $1 - P(none)$ $1 - P(\text{none})$. It transforms a complex multi-case problem into a single combination calculation.

Chapter Mix

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

Q7 jee_main_2026_22_january_morning Probability Distribution

If a random variable

x

has the probability distribution

x	0	1	2	3	4	5	6	7
p(x)	0	2k	k	3k	$2k²$ $2k^2$	2k	$k²+k$ $k^2+k$	$7k²$ $7k^2$

then

P(3 < x\leq 6)

is equal to

A. $0.34$ $0.34$
B. $0.22$ $0.22$
C. $0.64$ $0.64$
D. $0.33$ $0.33$

Solution

Related Formula

Σ P(xᵢ) = 1

Core Logic

For a valid probability distribution, the sum of all probabilities must equal 1.

0 + 2k + k + 3k + 2k² + 2k + (k² + k) + 7k² = 1

Combine like terms:

10k² + 9k - 1 = 0

(10k - 1)(k + 1) = 0

Since $k$ $k$ cannot be negative (probabilities must be non-negative), we get $k = (1)/(10)$ $k = \frac{1}{10}$.

Step 1: Calculate the Desired Probability

We need to find $P(3 < x \leq 6)$ $\mathrm{P}(3 < x \leq 6)$. This includes the probabilities for $x = 4, 5, 6$ $x = 4, 5, 6$.

P(3 < x \leq 6) = P(x=4) + P(x=5) + P(x=6)

P(3 < x \leq 6) = 2k² + 2k + (k² + k) = 3k² + 3k

Substitute $k = (1)/(10)$ $k = \frac{1}{10}$:

= 3((1)/(100)) + 3((1)/(10)) = 0.03 + 0.3 = 0.33

Pattern Recognition

Sum of probabilities always equals 1. This generates a standard quadratic in $k$ $k$. Always discard the negative root since $P(xᵢ) \geq 0$ $P(x_i) \geq 0$ for all $i$ $i$.

Chapter Mix

Class 12 Maths: Probability

Q24 jee_main_2026_23_january_morning Random Selection

From the first

100

natural numbers, two numbers first

a

and then

b

are selected randomly without replacement. If the probability that

a - b \geq 10

is

(m)/(n)

,

(m, n) = 1

, then

m + n

is equal to _____.

Numerical Answer. Answer: 311 to 311

Solution

Core Logic

Total ways to select $a$ $a$ and $b$ $b$ from $100$ $100$ natural numbers (ordered and without replacement) is $100 \times 99$ $100 \times 99$. We require $a - b \geq 10$ $a - b \geq 10$, which implies $a \geq b + 10$ $a \geq b + 10$.

Step 1: Count Favorable Outcomes

Iterate over possible values of $a$ $a$: If $a = 100$ $a = 100$, $b$ $b$ can be anything from $1$ $1$ to $90$ $90$ ( $90$ $90$ cases). If $a = 99$ $a = 99$, $b$ $b$ can be anything from $1$ $1$ to $89$ $89$ ( $89$ $89$ cases). ... If $a = 11$ $a = 11$, $b$ $b$ can only be $1$ $1$ ( $1$ $1$ case). Total favorable cases $= 1 + 2 + 3 + + 90$ $= 1 + 2 + 3 + \dots + 90$. Using sum of first $n$ $n$ natural numbers:

Nfav = (90 \times 91)/(2)

Step 2: Calculate Probability

Probability $P = NfavNtotal = ((90 \times 91)/(2))/(100 \times 99)$

P = (90 \times 91)/(2 \times 100 \times 99) = (91)/(2 \times 10 \times 11) = (91)/(220)

Step 3: Final Answer

Here $m = 91$ $m = 91$ and $n = 220$ $n = 220$. They are coprime (gcd is 1).

m + n = 91 + 220 = 311

Pattern Recognition

Ordering constraints like $a-b\geq k$ $a-b\geq k$ on a discrete uniform sample space always collapse into a simple arithmetic progression sum. Start explicitly from the highest/lowest valid bounds to quickly identify the series length.

Chapter Mix

Class 11 Maths: Probability Class 11 Maths: Sequences and Series

Q11 jee_main_2026_23_january_evening Total Probability Theorem

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

(p)/(q)

,

(p, q) = 1

, then

p + q

is equal to

A. $22$ $22$
B. $23$ $23$
C. $24$ $24$
D. $21$ $21$

Solution

Related Formula

Total Probability Theorem:

P(E) = P(E|A₁)P(A₁) + P(E|A₂)P(A₂)

Core Logic

Total Probability Theorem diagram for Q11 - JEE Main 2026 Evening

Let Event

WB

be drawing a white ball from Bag B and mixing it in Bag A. Let Event

BB

be drawing a black ball from Bag B and mixing it in Bag A.

Probability of drawing White from Bag B:

P(WB) = (6)/(10) = (3)/(5)

Probability of drawing Black from Bag B:

P(BB) = (4)/(10) = (2)/(5)

Step 1: Calculating Final Probability

If a white ball is transferred, Bag A now has 10 White and 8 Black balls (18 total). Probability of drawing White from A given $WB$ $W_B$:

P(WA | WB) = (10)/(18)

If a black ball is transferred, Bag A now has 9 White and 9 Black balls (18 total). Probability of drawing White from A given $BB$ $B_B$:

P(WA | BB) = (9)/(18)

Total probability of drawing a white ball from Bag A:

P(WA) = P(WA | WB)P(WB) + P(WA | BB)P(BB)

P(WA) = ((10)/(18))((3)/(5)) + ((9)/(18))((2)/(5))

P(WA) = (30)/(90) + (18)/(90) = (48)/(90) = (8)/(15)

Step 2: Final Calculation

We have $P = (p)/(q) = (8)/(15)$ $P = \frac{p}{q} = \frac{8}{15}$. Since $(8, 15) = 1$ $\gcd(8, 15) = 1$, $p = 8$ $p = 8$ and $q = 15$ $q = 15$.

p + q = 8 + 15 = 23

Pattern Recognition

Standard transfer problem. Always set up the exhaustive branches of the initial transfer event and map the new probability state of the target bag.

Chapter Mix

Class 12 Maths: Probability

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is 29/45, then n is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Set up Equation and Solve

Pattern Recognition

Chapter Mix

Reference Study Guides

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Step 2: Total Probability Equation

Step 3: Solve the Linear System

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Step 1: Counting Favorable vs Total Outcomes

Step 2: Final Probability

Pattern Recognition

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Step 1: Calculate the Desired Probability

Pattern Recognition

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Solution

Core Logic

Step 1: Count Favorable Outcomes

Step 2: Calculate Probability

Step 3: Final Answer

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Solution

Related Formula

Core Logic

Step 1: Calculating Final Probability

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

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