JEE Main 2026 (23 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Probability: 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 $\frac{m}{n}$ , $gcd (m, n) = 1$ , then $m + n$ is equal to .........

Q: How do I know the upper limit for $b$ is 90?

Look at the inequality $a \ge b + 10$. Since the maximum value $a$ can take is 100, the largest possible value for $b$ must satisfy $100 \ge b + 10$. Subtracting 10 from both sides gives $b \le 90$. If $b$ were 91, then $a$ would have to be at least 101, which is impossible in our set.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the problem as a grid of 100 by 100 possible pairs

(a, b)

. Since we are selecting without replacement, the diagonal where

a = b

is excluded. The condition

a - b \geq 10

defines a specific triangular region in this grid. Can you identify the boundaries of this region?

Hint 2

Hint 2: For each possible value of

b

, determine the range of valid values for

a

. If

b = 1

, what is the minimum value of

a

? If

b = 2

, what is the minimum value of

a

? Notice how the number of favorable outcomes for

a

changes as

b

increases.

Hint 3

Hint 3: You will find that for

b = 1

, there are 90 favorable values for

a

. For

b = 2

, there are 89. This creates a sum of the form

90 + 89 + 88 + \dots + 1

. Use the formula for the sum of the first

k

natural numbers,

\frac{k ( k + 1 )}{2}

, to calculate the total number of favorable outcomes.

Visualized Solution

Defining the Sample Space

Total numbers available: ${1, 2, 3, \dots, 100}$
We need to select two numbers, $a$ and $b$ .
Selection is done without replacement.

Total Possible Outcomes

Number of ways to select $a$ : $100$
Number of ways to select $b$ : $99$
Total ordered pairs $(a, b) = 100 \times 99 = 9900$

Analyzing the Condition $a - b \geq 10$

Given condition: $a - b \geq 10$
Rearranging the inequality: $a \geq b + 10$

Finding the Maximum Value of $b$

We know the maximum possible value for $a$ is $100$ .
Therefore, $b + 10 \leq 100$
This implies $b \leq 90$

Visualizing the Favorable Region

The condition $a \geq b + 10$ represents a specific region.
All valid pairs $(a, b)$ must lie in this region.

Counting Cases for $b = 1$

If $b = 1$ , then $a \geq 1 + 10 ⟹ a \geq 11$
Possible values for $a$ : ${11, 12, \dots, 100}$
Number of favorable cases $= 100 - 11 + 1 = 90$

Counting Cases for $b = 2$

If $b = 2$ , then $a \geq 2 + 10 ⟹ a \geq 12$
Possible values for $a$ : ${12, 13, \dots, 100}$
Number of favorable cases $= 100 - 12 + 1 = 89$

Counting Cases for $b = 90$

Continuing this pattern up to the maximum value of $b$ .
If $b = 90$ , then $a \geq 90 + 10 ⟹ a \geq 100$
Since $a \leq 100$ , the only possible value is $a = 100$ .
Number of favorable cases $= 1$

Summing the Favorable Outcomes

Total Favorable Cases $= 90 + 89 + 88 + \dots + 1$
This is the sum of the first $90$ natural numbers.
Formula: $\sum k = \frac{n ( n + 1 )}{2}$

Calculating the Sum

Substitute $n = 90$ into the formula.
Sum $= \frac{90 \times 91}{2}$
Sum $= 45 \times 91 = 4095$

Setting Up the Probability

Probability $P = \frac{Favorable Cases}{Total Cases}$
$P = \frac{4095}{9900}$

Simplifying the Probability Fraction

Divide numerator and denominator by their common factors.
Dividing by $45$ :
$4095 \div 45 = 91$
$9900 \div 45 = 220$
$P = \frac{91}{220}$

Identifying $m$ and $n$

The problem states $P = \frac{m}{n}$ where $gcd (m, n) = 1$ .
Our simplified fraction is $\frac{91}{220}$ .
Check coprimality: $91 = 7 \times 13$ , and $220 = 2^{2} \times 5 \times 11$ .
They share no common factors, so $gcd (91, 220) = 1$ .
Therefore, $m = 91$ and $n = 220$ .

Final Calculation: $m + n$

We need to find the value of $m + n$ .
$m + n = 91 + 220$
$m + n = 311$
Final Answer: $311$

00:00 / 00:00

The Sigma Insight: Classical Definition of Probability

Solution Diagram

Analyzing the Universe of Possibilities

To begin, we define the sample space for selecting two distinct numbers

a

and

b

from the set

{1, 2, 3, \dots, 100}

. Since the selection is made without replacement, the first number

a

has 100 possible choices, and the second number

b

has 99 possible choices.

The total number of ordered pairs

(a, b)

is given by:

100 \times 99 = 9900

This value represents the denominator of our probability calculation.

The Constraint

A Geometric Reality

We are tasked with finding the number of pairs that satisfy the condition

a - b \geq 10

, or equivalently,

a \geq b + 10

. We can determine the number of favorable outcomes by systematically testing values of

b

b = 1

, then

a \geq 11

. The possible values for

a

are

{11, 12, \dots, 100}

, which gives

100 - 11 + 1 = 90

choices.

b = 2

, then

a \geq 12

. The possible values for

a

are

{12, 13, \dots, 100}

, which gives

100 - 12 + 1 = 89

choices.

This pattern continues until

b = 90

, where

a \geq 100

. In this final case,

a

must be 100, providing exactly 1 choice.

The Elegance of Summation

The total number of favorable outcomes is the sum of the arithmetic progression:

90 + 89 + 88 + \dots + 1

Using the formula for the sum of the first

k

natural numbers,

S = \frac{k ( k + 1 )}{2}

, where

k = 90

, we calculate:

S = \frac{90 ( 90 + 1 )}{2} = 45 \times 91 = 4095

The Final Synthesis

The probability

P

is the ratio of favorable outcomes to the total number of outcomes:

P = \frac{4095}{9900}

To simplify this fraction, we divide both the numerator and the denominator by their greatest common divisor, which is 45:

\frac{4095 \div 45}{9900 \div 45} = \frac{91}{220}

Since

91 = 7 \times 13

and

220 = 2^{2} \times 5 \times 11

, the fraction is in its simplest form where

m = 91

and

n = 220

The final result,

m + n

, is:

91 + 220 = 311

Common Traps & FAQs

Wait, why is the total number of outcomes

100 \times 99

and not

100 \times 100

Great catch! The problem specifies that the numbers are selected 'without replacement'. This means once you pick the first number

a

, it is removed from the set, leaving only 99 choices for the second number

b

. If it were 'with replacement', then yes, it would be

100 \times 100

How do I know the upper limit for

b

is 90?

Look at the inequality

a \geq b + 10

. Since the maximum value

a

can take is 100, the largest possible value for

b

must satisfy

100 \geq b + 10

. Subtracting 10 from both sides gives

b \leq 90

. If

b

were 91, then

a

would have to be at least 101, which is impossible in our set.

Why can't we just use symmetry to solve this?

Symmetry is a powerful tool, but it can be tricky here. While the total number of pairs where

a > b

is equal to the number of pairs where

b > a

, the condition

a - b \geq 10

is not perfectly symmetric because of the 'gap' of 10. It is much safer and more reliable to count the favorable cases directly using the arithmetic progression we derived.

Animated Solution for Mathematics - Probability: 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 $\frac{m}{n}$ , $gcd (m, n) = 1$ , then $m + n$ is equal to .........

Enter Numerical Value:

Visualized Solution

Defining the Sample Space

Total Possible Outcomes

Analyzing the Condition $a - b \geq 10$

Finding the Maximum Value of $b$

Visualizing the Favorable Region

Counting Cases for $b = 1$

Counting Cases for $b = 2$

Counting Cases for $b = 90$

Summing the Favorable Outcomes

Calculating the Sum

Setting Up the Probability

Simplifying the Probability Fraction

Identifying $m$ and $n$

Final Calculation: $m + n$

The Sigma Insight: Classical Definition of Probability

Analyzing the Universe of Possibilities

The Constraint

The Elegance of Summation

The Final Synthesis

Similar Questions

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Enter Numerical Value:

Visualized Solution

Defining the Sample Space

Total Possible Outcomes

Analyzing the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a−b≥10

Finding the Maximum Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Visualizing the Favorable Region

Counting Cases for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=1

Counting Cases for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=2

Counting Cases for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=90

Summing the Favorable Outcomes

Calculating the Sum

Setting Up the Probability

Simplifying the Probability Fraction

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Classical Definition of Probability

Analyzing the Universe of Possibilities

The Constraint

The Elegance of Summation

The Final Synthesis

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