We could do it more explicitly as a counting problem. An urn has $2$ red balls and $1$ green ball, each with an ID number written on it. We take a ball out of the urn, record its colour and ID number, replace the ball, and do this again and again, a total of $7$ times.
There are $3^7$ sequences of ID numbers, all equally likely.
How many of these sequences have exactly $4$ red? The locations of the reds can be chosen in $\dbinom{7}{4}$ ways.
For each choice of locations, there are $2^4$ possible sequences of ID numbers.
Once the locations of the red ID numbers, and the exact sequence of red ID numbers, are known, the rest of the locations can only be filled in $1$ way.
So there are $\dbinom{7}{4}4^7$ sequences that give us $4$ red and the rest black. The required probability is therefore $\frac{\binom{7}{4}2^4}{3^7}.$ Naturally, this is the same number as the one obtained from the known binomial distribution formula.