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JEE Mains · Maths · STD 11 - 14. probability

Two integers \(\mathrm{x}\) and \(\mathrm{y}\) are chosen with replacement from the set \(\{0,1,2,3, \ldots ., 10\}\). Then the probability that \(|x-y|>5\) is :

  1. A \(\frac{30}{121}\)
  2. B \(\frac{62}{121}\)
  3. C \(\frac{60}{121}\)
  4. D  \(\frac{31}{121}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{30}{121}\)

Step-by-step Solution

Detailed explanation

\( \text { If } x=0, y=6,7,8,9,10 \) \( \text { If } x=1, y=7,8,9,10 \) \( \text { If } x=2, y=8,9,10 \) \( \text { If } x=3, y=9,10 \) \( \text { If } x=4, y=10\) If \(x=5, y=\) no possible value Total possible ways \(=(5+4+3+2+1) \times 2\) \( =30 \) Required probability…
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