Randomized agorithm for selecting a $m$-subset out of a $n$-set
There are actually only two random steps in this case:


Start with $\{1,2,3\}$.
With probability $3/4$, replace a random element with $4$.
With probability $3/5$, replace a random element with $5$.


You get $\{1,2,3\}$ if you don't insert $4$ or $5$, which happens with probability $$\left(1-\frac{3}{4}\right)\left(1-\frac{3}{5}\right) = \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10}.$$

You get $\{3,4,5\}$ in two cases:


You replace $1$ with $4$ and then $2$ with $5$.
You replace $2$ with $4$ and then $1$ with $5$.


The probability that you replace a specific element with $4$ is $(3/4)(1/3)=1/4$, and the probability that you replace a specific element with $5$ is $(3/5)(1/3)=1/5$. In total, the probability that we get $\{3,4,5\}$ is

$$ \frac{1}{4}\cdot \frac{1}{5} + \frac{1}{4}\cdot \frac{1}{5} = \frac{1}{10}. $$

So both probabilities are the same.