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Q.
Two integers $x$ and $y$ are chosen (with replacement) from the set $\{0,1,2,3, \ldots, 20\}$. Then the probability that $|x-y| >8$ is
Probability - Part 2
Solution:
When $y-x>8$, the following choices can be made:
Suppose $x=0$, then $y$ can be any one of $9,10,11, \ldots, 20 \rightarrow 12$ ways
Similarly $x=1 \rightarrow y$ can be $10,11, \ldots, 20 \rightarrow 11$ ways
$x =11 \Rightarrow y =20 \rightarrow 1 \text { way }$
$\therefore $the total number of ways $=1+2+\ldots . .+12=6 \times 13=78$
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 $=\frac{2 \times 78}{21 \times 21}=\frac{52}{147}$
