Let $P$, $Q$, $R$ be the sets of first $8$ natural numbers, first $12$ natural numbers,first $20$ natural numbers respectively. A number is selected from each of these sets,say $p$, $q$, $r$ respectively. The probability that $r=|p-q|$ is?
as $r$ can not be zero , then $\Pr (p \ne q)$ is $1-\frac 8 {12 \times 8}$ how to do further? (natural number starts from $1$)
On
Not sure if there is any faster method. One way to move on is to condition on $r$. Starting from $r=1$ (each $r$ with the same probability of $1/20$ to be selected), if $p=1$, then $q$ has to be 2; for other values of $p$, $q$ has two choices. Thus, for this case, the probability is $\frac{1}{20}\cdot \frac{1}{8}[\frac{1}{12}+\frac{1}{6}\cdot 7]$.
Now do the same analysis for $r=2, 3, \cdots, 11$.
On
To have $r = |p-q| \in R$ we must have i) $|p-q|\ne 0$ or in other words that $q\ne p$. and ii) that $r = |p-q|$.
i)As $1\le p \le 8<12$ and $p \in Q$ so for any $p$ the probability that a random $q \in Q$ equals $p$ is $\frac 1{12}$ and the probability that $q \ne p$ is $\frac {11}{20}$
ii) $-11 = 1-12 \le p-q \le 8-1 = 7$ so $0 \le |p-q| \le 11 < 20$ so if $|p-q|\ne 0$ then $|p-q| \in R$ and the probability that a random $r$ in $R$ is equal to a non-zero $|p-q|$ is $\frac 1{20}$.
So the probability of i) and ii) is $\frac {11}{12}\frac {1}{20}=\frac {11}{240}$.
For any choice of $p$ there are exactly $11$ choices of $q$ for which $p\neq q$. Thus the probability that $|p-q|\neq 0$ is $\frac {11}{12}$.
For any choice of $p,q$ such that $|p-q|\neq 0$ there is a unique choice of $r$ such that $r=|p-q|$. Thus the desired answer is $$\boxed {\frac {11}{12}\times \frac 1{20}}$$