Question

Two integers $x$ and $y$ are chosen at random from the set $\{x:0\le x\le 10,x\in I\}$ then probability for $|x-y|\le 5$ is

• A. $\frac{91}{121}$
• B. $\frac{91}{100}$
• C. $\frac{87}{100}$
• D. $\frac{87}{121}$

Answer 1

Answer: (A)

There are $11$ integers from $0$ to $10$ , hence total number of ways of choosing $x$ and $y=11\times 11=121$.
Now $|x-y|\le 5\Rightarrow -5\le x-y\le 5$
$\Rightarrow y-5\le x\le y+5$
If $y=0$, then $-5\le x\le 5$
$\Rightarrow x=0,1,2,3,4,5;$ total 6 ways.
If $y=1$, then $-4\le x\le 6$
$\Rightarrow x=0,1,2,3,4,5,6;$ total 7 ways.
If $y=2$, then $-3\le x\le 7$
$\Rightarrow x=0,1,2,\dots \dots .7;$ total 8 ways.
If $y=3$, then $-2\le x\le 8$
$\Rightarrow x=0,1,2\dots \dots .8;$ total 9 ways.
If $y=4$, then $1\le x\le 9$
$\Rightarrow x=0,1,2,\dots \dots 9;$ total 10 ways.
If $y=5$, then $0\le x\le 10$
$\Rightarrow x=0,1,2,\dots \dots 10$ total 11 ways.
If $y=6$, then $1\le x\le 11$
$\Rightarrow x=1,2,3\dots \dots .10;$ total 10 ways.
If $y=7$, then $2\le x\le 12$
$\Rightarrow x=2,3\dots \dots \dots .10;$ total 9 ways.
If $y=8$, then $3\le x\le 13$
$\Rightarrow x=3,4\dots \dots \dots .10;$ total 8 ways.
If $y=9$, then $4\le x\le 14$
$\Rightarrow x=4,5\dots \dots \dots \dots 10;$ total 7 ways.
If $y=10$, then $5\le x\le 15$
$\Rightarrow x=5,6\dots \dots \dots \dots 10;$ total 6 ways.
$\therefore$ Favourable number of ways
$=6+7+8+9+10+11+10+9+8+7+6=91.$
$\therefore$ Required probability $=\frac{91}{121}$.