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Q.
Two integers x and y are chosen with replacement out of the set $\left\{0, 1, 2, 3, ...........10\right\}$. Then the probability that $|x-y| > 5$ is
Probability
Solution:
Total number of selections $= 11 × 11$
Now $|x - y| > 5$ clearly $x\ne 5$
If $x = 0$, then $y > 5 \Rightarrow 5$ favourable cases
If $x = 1$, then $y > 6 \Rightarrow 4$ favourable cases
If $x = 2$, then $y > 7 \Rightarrow 3$ favourable cases
If $x = 3$, then $y > 8 \Rightarrow 2$ favourable cases
If $x = 2$, then $y > 9 \Rightarrow 1$ favourable cases
Symmetical cases will be obtained for $x = 6, 7, 8, 9,10$
$\therefore $ Favourable cases $=30$
$\therefore $ Desired probability $\frac{30}{121}$