JEE Mains · Maths · STD 11 - 14. probability
From the first 100 natural numbers, two numbers a and b are selected randomly without replacement. If the probability that \( a-b \ge 10 \) is \( \frac{m}{n} \), with \(gcd(m, n)=1\), then \( m+n \) is equal to:
- A 310
- B 311
- C 312
- D 313
Answer & Solution
Correct Answer
(B) 311
Step-by-step Solution
Detailed explanation
\(a-b \geq 10\) Total cases = \( 100 \times 99 \). Fav. Cases \(=1+2+3+\ldots 90\) Req. Prob \(=\frac{1+2+\ldots+90}{100 \times 99}\) \(\frac{ m }{ n }=\frac{90\left(\frac{91}{2}\right)}{100(99)}=\frac{91}{220}\) \(m + n =311\)
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