A target consists of a disc of unit radius and centre $O$. A certain marksman never misses the target, and the probability of any given shot hitting the target within a distance $t$ from $O$ is $t^{2}$, where $0\leq\ t \leq 1$. The marksman fires $n$ shots independently. The random variable $Y$ is the radius of the smallest circle, with centre $O$, which encloses all the shots. Show that the probability density function of $Y$ is $2ny^{2n-1}$, and show that the expected area of the circle is $\frac{πn}{n + 1}$.
The shot which is furthest from $O$ is rejected. Show that the expected area of the smallest circle, with centre $O$, which encloses the remaining $(n − 1)$ shots is $\frac{n-1}{n+1}\pi$.
In the solutions pertaining to the second part of the question, it says that "we could find the probability that $n − 1$ shots are within a distance $z$ from the centre, which is $nz^{2n-2}(1-z^{2}) + z^{2n}$ (the second term arising because $n−1$ shots will certainly be within within a distance $z$ from the centre if all $n$ are), and differentiate."
I struggle to see why the second term is necessary, given that the question states that we discard the shot furthest from O and then z should be closer than that shot. If all $n$ shots were within $z$, then you haven't rejected the shot that was furthest from $O$.