Probability that an r-subset contains an element from it's set
Maybe it would make more sense if you write out the probability as
$$\frac{{1 \choose 1}{6 \choose 2}}{7 \choose 3}$$
We choose the $4$ to be selected giving the ${1 \choose 1}$
We choose $2$ of the other $6$ numbers to be selected giving the $6 \choose 2$
View it as $$\frac{\binom62 \binom11}{\binom73}= \frac{\binom62 }{\binom73}=\frac37$$