Question

Two integers $x$ and $y$ are chosen (with replacement) from the set $\{0,1,2,3,\dots ,20\}$. Then the probability that $|x-y|>8$ is

• A. $\frac{52}{147}$
• B. $\frac{400}{441}$
• C. $\frac{361}{441}$
• D. $\frac{78}{441}$

Answer 1

Answer: (A)

When $y-x>8$, the following choices can be made:
Suppose $x=0$, then $y$ can be any one of $9,10,11,\dots ,20\to 12$ ways
Similarly $x=1\to y$ can be $10,11,\dots ,20\to 11$ ways

$x=11\Rightarrow y=20\to 1\text{ way }$
$\therefore$the total number of ways $=1+2+\dots ..+12=6\times 13=78$
Similarly the number of ways for $x-y>8$ is 78
Total number of ways for $x-y>8$ is 78 .
Total number of all possible ways $=21\times 21$
Hence the required probability $=\frac{2\times 78}{21\times 21}=\frac{52}{147}$