Mathematics · Probability

JEE Main 2026 — 23 January, Morning Shift — Question 24

From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that $\mathrm{a}-\mathrm{b} \geq 10$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$ , then $\mathrm{m}+\mathrm{n}$ is equal to $\_\_\_\_$ .

Answer: 311

Numerical answer — enter this value.

Step-by-step solution

$a-b \geq 10$

Total cases $=100 \times 99$

Fav. Cases $=1+2+3+\ldots 90$

Req. $\operatorname{Prob}=\frac{1+2+\ldots+90}{100 \times 99}$

$\frac{\mathrm{m}}{\mathrm{n}}=\frac{90\left(\frac{91}{2}\right)}{100(99)}=\frac{91}{220}$

$\mathrm{m}+\mathrm{n}=311$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Probability
Topic: Problems based on P & C

Let the mean and variance of 8 numbers -10,-7,-1, x, y, 9,2,16 be 7/2 and 293/4 , respectively. Then the mean of 4 numbers x, y, x+ y+1, x…

Answer: 311

Step-by-step solution

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