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Q.
Two integers $x$ and $y$ are chosen at random from the set $\{x: 0 \leq x \leq 10, x \in I \}$ then probability for $|x-y| \leq 5$ is
Probability
Solution:
There are $11$ integers from $0$ to $10$ , hence total number of ways of choosing $x$ and $y=11 \times 11=121$.
Now $|x-y| \leq 5 \Rightarrow -5 \leq x-y \leq 5$
$\Rightarrow y-5 \leq x \leq y+5$
If $y=0$, then $-5 \leq x \leq 5$
$\Rightarrow x=0,1,2,3,4,5 ;$ total 6 ways.
If $y=1$, then $-4 \leq x \leq 6$
$\Rightarrow x=0,1,2,3,4,5,6 ;$ total 7 ways.
If $y=2$, then $-3 \leq x \leq 7$
$\Rightarrow x=0,1,2, \ldots \ldots .7 ; \,\,\,\,\,$ total 8 ways.
If $y=3$, then $-2 \leq x \leq 8$
$\Rightarrow x=0,1,2 \ldots \ldots .8 ; \,\,\,\,\,$ total 9 ways.
If $y=4$, then $1 \leq x \leq 9$
$\Rightarrow x=0,1,2, \ldots \ldots 9 ; \,\,\,\,\,$ total 10 ways.
If $y=5$, then $0 \leq x \leq 10$
$\Rightarrow x=0,1,2, \ldots \ldots 10\,\,\,\,\,\,\,$ total 11 ways.
If $y=6$, then $1 \leq x \leq 11$
$\Rightarrow x=1,2,3 \ldots \ldots .10 ; \,\,\,\,\,$ total 10 ways.
If $y=7$, then $2 \leq x \leq 12$
$\Rightarrow x=2,3 \ldots \ldots \ldots .10 ; \,\,\,\,\,$ total 9 ways.
If $y=8$, then $3 \leq x \leq 13$
$\Rightarrow x=3,4 \ldots \ldots \ldots .10 ; \,\,\,\,\,$ total 8 ways.
If $y=9$, then $4 \leq x \leq 14$
$\Rightarrow x=4,5 \ldots \ldots \ldots \ldots 10 ; \,\,\,\,\,$ total 7 ways.
If $y=10$, then $5 \leq x \leq 15$
$\Rightarrow x=5,6 \ldots \ldots \ldots \ldots 10 ; \,\,\,\,\,$ total 6 ways.
$\therefore $ Favourable number of ways
$=6+7+8+9+10+11+10+9+8+7+6=91 .$
$\therefore $ Required probability $=\frac{91}{121}$.