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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Probability

From the first 100 natural numbers, two numbers first a and then b...

From the first 100 natural numbers, two numbers first aa and then bb are selected randomly without replacement. If the probability that a−b⩾10a-b \geqslant 10 is mn,gcd⁡(m,n)=1\frac{m}{n}, \operatorname{gcd}(m, n)=1, then m+nm+n is equal to ____\_\_\_\_.

Correct answer

0

Step-by-step solution

Total ordered selections =100×99=9900= 100 \times 99 = 9900.
Favorable (a−b≥10a - b \geq 10): for each bb from 1 to 90, aa ranges from b+10b+10 to 100, giving 91−b91 - b choices.
Count =∑b=190(91−b)=∑k=190k=90×912=4095= \sum_{b=1}^{90}(91-b) = \sum_{k=1}^{90}k = \frac{90 \times 91}{2} = 4095.
Probability =40959900=91220= \frac{4095}{9900} = \frac{91}{220} (after dividing by 45).
gcd⁡(91,220)=1\gcd(91, 220) = 1, so m=91,n=220m = 91, n = 220.
m+n=311m + n = 311.
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