Would it be valid to say that since there were
![](https://latex.codecogs.com/png.latex?\bg_white r )
heads in
![](https://latex.codecogs.com/png.latex?\bg_white n )
tosses, each toss has on average an
![](https://latex.codecogs.com/png.latex?\bg_white r/n)
chance to be heads?
If we consider a situation where for
every n tosses, we get exactly r heads, then we can calculate
![](https://latex.codecogs.com/png.latex?\bg_white P(\text{obtaining a head}) = \frac{\text{total number of heads}}{\text{total number of tosses}} = \frac{r}{n} )
,
which I believe is exactly what you're saying. So yes, you are right.
The question is worded in a way that suggests the interpretation that we are only considering the cases where we obtain exactly r heads, which is why you can think of it as if that's the only possibility each and every time we toss a coin n times.