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JEE Mains · Maths · STD 12 - 13. probability

Let the sum of two positive integers be \(24\) . If the probability, that their product is not less than \(\frac{3}{4}\) times their greatest positive product, is \(\frac{\mathrm{m}}{\mathrm{n}}\), where \(\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\), then \(\mathrm{n}-\mathrm{m}\) equals :

  1. A \(9\)
  2. B \(11\)
  3. C \(8\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(10\)

Step-by-step Solution

Detailed explanation

\( x+y=24, x, y \in N \) \( A M>G M \Rightarrow x y \leq 144 \) \( x y \geq 108\) Favorable pairs of \((\mathrm{x}, \mathrm{y})\) are \( (13,11),(12,12),(14,10),(15,9),(16,8), \) \( (17,7),(18,6),(6,18),(7,17),(8,16),(9,15), \) \( (10,14),(11,13)\) i.e. \(13\) cases Total…