Probability that an r-subset contains an element from it's set

Question

Let $S = \{1, 2, 3, 4, 5, 6, 7\}$. You choose a uniformly random 3-element subset $X$ of $S$. Thus, each 3-element subset of $S$ has a probability of $1{}/{{7}\choose{3}}$ of being $X$.

Define the event
$A =$ “$4$ is an element of $X$”

What is $Pr(A)$?

The answer is $\frac37$.

My guess was ${{6\choose2}}/{{7\choose3}} = \frac{15}{35} = \frac37$, however I'm not quite sure why ${{6\choose2}}$ actually accounts for an already chosen element of $S$ since all it's doing is counting subsets of size $2$ from $6$. Even though I came up with this answer, I have no idea why it makes sense.

Anyone care to explain?

Answer 1

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 $6 \choose 2$

Answer 2

View it as $$\frac{\binom62 \binom11}{\binom73}= \frac{\binom62 }{\binom73}=\frac37$$

Answer 3

The probability that a uniformly random $k$-element subset of $[n] := \{1,2,\ldots,n\}$ contains a given element of $[n]$ is the same as the probability that a fixed $k$-element subset of $[n]$ contains a uniformly random element $X$ of $[n]$, which is of course $k \times 1/n = k/n$.