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

Solution

Maths sir · 4 steps

Fix the gap x - y, count how many pairs give each gap, then double for the other order.

1. Total outcomes. The set has 11 numbers and you choose with replacement, so there are 11 × 11 = 121 ordered pairs (x, y), all equally likely.
2. Count by the gap. Since x and y are integers, |x - y| > 5 means the gap is 6, 7, 8, 9 or 10. First take x > y. For a gap d, the smaller number y can run from 0 up to 10 - d, so there are 11 - d pairs.

   Gap d = x - y	Possible y	Pairs
   6	0 to 4	5
   7	0 to 3	4
   8	0 to 2	3
   9	0 to 1	2
   10	0 only	1
   Total with x > y	all gaps from 6 to 10	15

3. Other order. Swapping x and y gives another 15 pairs with y > x. Pairs with x = y have gap 0, so they do not count. Favourable pairs are 15 + 15 = 30.
4. So. The probability is 30/121, which is option (d).

The idea to keep. For conditions on |x - y|, count pairs gap by gap. With n choices, a gap of d gives n - d pairs in one order, then double for the other order.

Answer from the NTA final answer key. Solution by PrepWiser.