UPSC NDA II 2018 Mathematics Solved Paper

Concept:Probability is the ratio of favourable outcomes to total possible outcomes, when two numbers are chosen from 0 to 10 with replacement.Explanation:The set has 11 numbers: $\{0, 1, 2, \dots, 10\}$.Total ordered pairs $(x, y)$ = $11 \times 11 = 121$.We count pairs where $|x - y| > 5$, i.e., the absolute difference is at least 6.List all such pairs systematically:When $x = 0$, $y$ can be 6, 7, 8, 9, 10 → 5 pairs.When $x = 1$, $y$ can be 7, 8, 9, 10 → 4 pairs.When $x = 2$, $y$ can be 8, 9, 10 → 3 pairs.When $x = 3$, $y$ can be 9, 10 → 2 pairs.When $x = 4$, $y$ can be 10 → 1 pair.When $x = 5$, no $y$ gives difference > 5 because maximum $y$ is 10, and $10-5=5$ (not >5).Similarly, for $x = 6,7,8,9,10$, the reverse pairs exist (e.g., $x=6,y=0$ etc.).Thus total favourable ordered pairs = $2 \times (5+4+3+2+1) = 2 \times 15 = 30$.Answer:Probability = $\frac{30}{121}$.Hence the correct option is C.